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https://superuser.com/questions/493389 | StackExchange | Open Web | CC-By-SA | 2,012 | Stack Exchange | Dimitris Katsigiannis, GaTechThomas, Graham Wager, Jeff Silverman, Matthew, Mukesh Choudhary, Venkata Raman, https://superuser.com/users/122843, https://superuser.com/users/1293984, https://superuser.com/users/1293985, https://superuser.com/users/1293986, https://superuser.com/users/1294000, https://superuser.com/users/1294007, https://superuser.com/users/1294016, https://superuser.com/users/1294103, https://superuser.com/users/1294333, https://superuser.com/users/1294334, https://superuser.com/users/140403, https://superuser.com/users/4804, tang duc bao, totolooks1, vedantsap, wodr, 卡片設計 spam | English | Spoken | 300 | 439 | How to get to the Windows 8 Lock Screen on Microsoft Surface?
I just got my Microsoft Surface Tablet. So far it rocks, but I would like to be able to get to the main screen that shows date/time and background image (or, the "Lock Screen"). Thus far I have only been able to get there by hitting the power button twice. Is there a way to get there without hitting the power button?
Currently there are two excellent answers. How does one go about deciding the answer to accept?
Whichever you prefer! Generally, if multiple answers solve my problem, I choose whichever I think will be most useful to future visitors.
This screen is called the "Lock Screen". It is shown when your computer is locked.
To lock your computer, you can:
Go to the start screen, click your name in the top-right corner, and choose "Lock".
Or, press Windows + L.
You can customize your lock screen through the "Personalize" settings page.
First activate the "Settings" Charm and choose "Change PC Settings":
Then select the "Personalize" page:
What if I don't have the keyboard attached and want to use the touch screen?
@GaTechThomas: See my edit.
Hmm, I thought I commented on the other answer in my question about using the touch screen.
@GaTechThomas: Nope you were right, you commented right before I added non-keyboard methods to my answer, when this was the only answer :)
@Matthew Like how both our Start screen screenshots feature Solitaire Collection ;)
@GrahamWager: Gotta keep those Free Cell/Spider skills up-to-date :)
What is called the Lock Screen is available using the profile icon in the top right corner of the Start screen. Simply click it and click "Lock":
Alternatively, if you have a keyboard attached, Windows+L will end in the same result.
| 13,445 |
https://it.wikipedia.org/wiki/HD%2093372 | Wikipedia | Open Web | CC-By-SA | 2,023 | HD 93372 | https://it.wikipedia.org/w/index.php?title=HD 93372&action=history | Italian | Spoken | 208 | 388 | HD 93372 è una stella bianco-gialla nella sequenza principale di magnitudine 6,27 situata nella costellazione della Carena. Dista 104 anni luce dal sistema solare.
Osservazione
Si tratta di una stella situata nell'emisfero celeste australe. La sua posizione è fortemente australe e ciò comporta che la stella sia osservabile prevalentemente dall'emisfero sud, dove si presenta circumpolare anche da gran parte delle regioni temperate; dall'emisfero nord la sua visibilità è invece limitata alle regioni temperate inferiori e alla fascia tropicale. Essendo di magnitudine pari a 6,3, non è osservabile ad occhio nudo; per poterla scorgere è sufficiente comunque anche un binocolo di piccole dimensioni, a patto di avere a disposizione un cielo buio.
Il periodo migliore per la sua osservazione nel cielo serale ricade nei mesi compresi fra febbraio e giugno; nell'emisfero sud è visibile anche per buona parte dell'inverno, grazie alla declinazione australe della stella, mentre nell'emisfero nord può essere osservata limitatamente durante i mesi primaverili boreali.
Caratteristiche fisiche
La stella è una bianco-gialla nella sequenza principale; possiede una magnitudine assoluta di 3,75 e la sua velocità radiale positiva indica che la stella si sta allontanando dal sistema solare.
Voci correlate
Stelle principali della costellazione della Carena
Collegamenti esterni
Stelle di classe spettrale F
Stelle bianco-gialle di sequenza principale | 15,722 |
https://fi.wikipedia.org/wiki/Jalkapallon%20suomenmestaruuskilpailut%201912 | Wikipedia | Open Web | CC-By-SA | 2,023 | Jalkapallon suomenmestaruuskilpailut 1912 | https://fi.wikipedia.org/w/index.php?title=Jalkapallon suomenmestaruuskilpailut 1912&action=history | Finnish | Spoken | 37 | 156 | Jalkapallon suomenmestaruuskilpailut 1912 olivat lajin viidennet SM-kilpailut. Suomen jalkapallomestaruuden voitti Helsingin Jalkapalloklubi. Osanottajia ilmaantui tällä kertaa vain viisi.
Puolivälierä
Välierä
Loppuottelu
Lähteet
Aiheesta muualla
Suomen pääsarjatason sarjataulukot RSSSF:n sivuilla
Palloliiton toimintakertomus
Jalkapallon Suomen mestaruuskilpailut
Vuoden 1912 jalkapallo | 50,584 |
https://ru.wikipedia.org/wiki/%D0%A4%D1%91%D0%B4%D0%BE%D1%80%D0%BE%D0%B2%D0%B0%2C%20%D0%90%D0%BD%D0%B0%D1%81%D1%82%D0%B0%D1%81%D0%B8%D1%8F%20%D0%90%D0%BD%D0%B4%D1%80%D0%B5%D0%B5%D0%B2%D0%BD%D0%B0 | Wikipedia | Open Web | CC-By-SA | 2,023 | Фёдорова, Анастасия Андреевна | https://ru.wikipedia.org/w/index.php?title=Фёдорова, Анастасия Андреевна&action=history | Russian | Spoken | 202 | 550 | Анастасия Андреевна Фёдорова () — российская гандболистка, разыгрывающая «ЦСКА» и «ЦСКА-2», также играет на позиции левой крайней. Ранее играла за дубль команды «Луч».
Биография
Анастасия Фёдорова родилась 20 марта 2001 года.
Начала заниматься гандболом в 2010 году в Бокситогорске на базе МБОУ ДО «БДЮСШ» под руководством Юлии Закатовой. Играет на позиции разыгрывающей и левой крайней.
Выступала в дублях команды «Луч», также играла за сборную Ленинградской области. В сезоне 2018/2019, выступая за клуб первенства России среди женских команд, забила 145 голов при 244 бросках.
В 2019 году перешла в только созданный московский «ЦСКА» и играла в составе дубля до 2021 года. Она по приглашению главного тренера «армейцев» Яна Лесли принимала участие в тренировках с основной сборной в декабре 2019 года. В сезоне 2019/2020 отличилась 53 раза, нанеся 100 бросков.
31 марта 2021 года в последнем матче предварительного раунда женской Суперлиги против «Уфы-Алисы» сыграла в составе основной команды, отличившись шестью голами в ворота соперниц, причём реализовала все свои броски. Она также играла в начале сезона играла в выездных матчах в Уфе и Ижевске. В тех двух играх, прошедших в августе 2020 года, Анастасия забила всего один гол, нанеся пять бросков.
Примечание
Ссылки
Профиль на сайте Федерации гандбола России
Профиль на сайте «ЦСКА»
Гандболистки России | 229 |
https://stackoverflow.com/questions/76128531 | StackExchange | Open Web | CC-By-SA | 2,023 | Stack Exchange | L Tyrone, https://stackoverflow.com/users/2530121 | English | Spoken | 238 | 377 | Why do I get "invalid 'type' (character) of argument" using xtab() in R
The error...
Error in FUN(X[[i]], ...) : invalid 'type' (character) of argument
...is not new and I can find a lot of SO questions about it. But because of low quality MWE's I'm not able to fit the answers to my own problem.
I'm interested not only in a quick code based solution but in a deeper understand why this error happens.
set.seed(0)
df <- data.frame(
gender = sample(c("female", "male"), 100, replace = TRUE),
foo = sample(c(TRUE, FALSE), 100, replace = TRUE),
bar = sample(c(0, 1, 2, 3, 4), 100, replace = TRUE)
)
xtabs(gender ~ foo, data=df)
The right side of the formula is for the classifying variable or variables, the left side is for defining counts. The values in the "gender" column aren't numeric and therefore not valid. Does xtabs(foo ~ gender, data = df) return the result you are expecting?
The left hand side of the formula is usually empty. If you have something there, it should be a vector or matrix of counts. You have gender there, which is a character variable and the error message comes when xtabs() tries to interpret it as counts.
So to fix this, just put the two factors on the right hand side:
xtabs(~ gender + foo, data=df)
BTW, even though neither one of those columns is actually a "factor", xtabs() will automatically convert them.
| 19,189 |
https://de.wikipedia.org/wiki/Baltika%20%28Brauerei%29 | Wikipedia | Open Web | CC-By-SA | 2,023 | Baltika (Brauerei) | https://de.wikipedia.org/w/index.php?title=Baltika (Brauerei)&action=history | German | Spoken | 807 | 1,746 | Baltika ( / ) ist die größte Brauereigruppe in der Russischen Föderation. Sie wurde 1990 in Leningrad (Sankt Petersburg) gegründet und hat sich schnell zu einem wichtigen Wirtschaftsfaktor für die Stadt entwickelt.
Unternehmen
Ursprünglich war Baltika ein Joint-Venture der finnischen Getränkefirma Hartwall und deren norwegischem Partner Orkla. Baltika nahm 1990 die Produktion auf. Mehrheitsaktionär des Unternehmens wurde 1996 die Baltic Beverages Holding (BBH), diese wiederum gehörte seit 2012 der dänischen Carlsberg-Brauerei. Am 28. März 2022 kündigte Carlsberg an, sich aus Russland zurückzuziehen und sich damit auch von Baltika zu trennen.
Braustätten befinden sich in:
Sankt Petersburg
Samara
Chabarowsk
Tula
Rostow am Don
Tscheljabinsk (2006–2015)
Krasnojarsk (1993–2015)
Jaroslawl (seit 2006)
Kiew, Ukraine (seit 2006)
Nowosibirsk (seit 2008)
Diese produzierten 2005 zusammen 22,7 Millionen Hektoliter Bier unter den Markenbezeichnungen Baltika, Arsenalnoje und Leningradskoje. Baltika-Biere werden in 38 Ländern vermarktet. Hauptabsatzmarkt bleiben Russland und die GUS-Staaten. In Tula besitzt Baltika eine Malzfabrik.
Baltika hatte 2006 einen Umsatz von 1,49 Milliarden Euro und erwirtschaftete einen Gewinn vor Steuern von 260 Millionen Euro.
Baltika fusionierte Ende 2006 mit den Firmen „Wena“, „Pikra“ und „Jarpiwo“ und verfügte danach über 10 Brauereien und 4 Mälzereien in 9 russischen Regionen mit mehr als 11.000 Mitarbeitern.
Im Jahr 2007 hatte Baltika in Russland einen Marktanteil von 37,6 %.
Die Baltika Aktiengesellschaft wird von der KPMG beraten.
Präsident der Baltika Aktiengesellschaft war seit Januar 2015 der Pole Jacek Pastuszka. Er war Nachfolger von Isaak Scheps, der die Baltika AG von Dezember 2011 bis zur Eingliederuung in den Tuborg Konzern im Dezember 2014 leitete.
Wegen des rückläufigen russischen Biermarktes schloss Baltika 2009 die ehemalige Wena-Brauerei in Petersburg sowie 2015 die Braustätten in Krasnojarsk und Tscheljabinsk.
Am 16. Juli 2023 wurde Baltika per Dekret vom russischen Staat faktisch enteignet. Die Leitung übernimmt Taimuras Bolloew, der als langjähriger Putin-Freund gilt und bereits früher Manager der Brauerei war.
Produkte
Die meisten Biersorten von Baltika werden durch Nummern unterschieden (Auswahl):
Baltika 0 Besalkogolnoje (Alkoholfreies) ist ein alkoholfreies Bier, gebraut aus leichtem Gerstenmalz
Baltika 1 Ljogkoje (Leichtes) ist ein Light-Bier mit einem physiologischen Brennwert von nur 163 kJ (= 39 kcal) pro 100 g und einem Alkoholgehalt von nicht mehr als 4,4 Volumenprozent (wird in Russland nicht mehr vertrieben, sondern nach Finnland exportiert; in Russland durch Baltika Lite ersetzt)
Baltika 2 Ossoboje (Besonderes) ist ein helles Lagerbier, gebraut aus hellem Gerstenmalz, Reis und „ausgewählten Hopfenarten“ (min. 4,7 % Alkoholgehalt); bis Mitte 2010 hieß Baltika 2 Swetloje (Helles)
Baltika 3 Klassitscheskoje (Klassisches) ist ein helles Lagerbier mit 4,8 % Alkoholgehalt
Baltika 4 Originalnoje (Originales) ist ein 5,6-prozentiges bernsteinfarbenes Lagerbier, gebraut aus dunklem Gerstenmalz und Roggenmalz
Baltika 5 Solotoje (Goldenes) ist ein Lagerbier, gebraut aus hellem Malz
Baltika 6 Porter ist ein untergäriges 7-prozentiges dunkles Bier, das die Brauerei als „nach traditionellen englischen Rezepten gebrautes Porter“ vertreibt
Baltika 7 Eksportnoje (Export) ist ein helles Lagerbier
Baltika 8 Pschenitschnoje (Weizen) ist ein ungefiltertes Weizenbier
Baltika 9 Krepkoje (Starkes) ist ein starkes Lagerbier mit 8,0 %
Baltika Kuler (Cooler) ist ein helles Bier mit leichtem Geschmack, das an Baltika 5 angelehnt ist. Der Alkoholgehalt beträgt nicht weniger als 4,7 Volumenprozent
Baltika Kuler Laim (Cooler Lime) ist Baltika Cooler mit Limettengeschmack
Baltika 20 Jubileinoje (Jubiläumsbier) ist das Bier, das seit Juni 2010 anlässlich des 20-jährigen Jubiläums der Baltika-Brauerei hergestellt wird
Baltika Rasliwnoje ist ein helles Schankbier, das im Jahr 2010 in die Produktpalette aufgenommen wurde
Baltika Rasliwnoje Nefiltrowannoje ist ein helles, ungefiltertes Schankbier, das seit 2011 produziert wird
Baltika München ist ein helles, ungefiltertes Bier nach bayerischer Art, das im Mai 2013 auf den Markt kam
Baltika Praha ist ein nach Prager Vorbild gebrautes Helles, das seit März 2015 erhältlich ist
Folgende Biere von Baltika werden nicht mehr produziert und daher nur noch als Sammlerstücke zu finden:
Baltika 12 Nowogodneje (Neujahrsbier) war ein Bier, das es früher in den Wintermonaten zu kaufen gab. Es handelte sich um ein halbdunkles Bier mit 5,5 % vol.
Baltika Leningradskoje (Leningrader) war ein helles Bier, das im Jahr 2003 anlässlich des 300-jährigen Jubiläums der Stadt Sankt Petersburg hergestellt wurde.
Baltika Medowoje Ljogkoje (Leichtes Honigbier) war ein helles, 4,1-prozentiges Bier mit Honiggeschmack, das es nur in 1,5 Liter-PET-Flaschen zu kaufen gab.
Baltika Medowoje Krepkoje (Starkes Honigbier) war ein starkes, 7,6-prozentiges Bier mit Honiggeschmack, ebenfalls in 1,5 Liter-PET-Flaschen. Die Honigbiere wurden wegen schlechter Verkaufszahlen aus dem Sortiment entfernt.
Im Sortiment befinden sich darüber hinaus weitere Produkte wie beispielsweise ein Kirschbier. Die Brauerei Baltika ist bei europäischen Bierwettbewerben immer wieder erfolgreich, wie zum Beispiel 2015 beim European Beer Star mit Gold für Baltika Nr. 3.
Vertrieb und Herstellung in Deutschland
Baltika ist auch in Deutschland erhältlich und findet sich meist in russischen Supermärkten oder Getränkefachmärkten. Der Sitz der Baltika Deutschland GmbH befindet sich in Hamburg. Seit August 2011 wird die Sorte Baltika 3 auch in Deutschland hergestellt, abgefüllt und vertrieben.
Einzelnachweise
Weblinks
Baltika (russ./engl.)
Lebensmittelhersteller (Russland)
Brauereigruppe
Unternehmen (Sankt Petersburg)
Biermarke
Gegründet 1990
Carlsberg (Brauerei) | 5,976 |
https://stackoverflow.com/questions/52368320 | StackExchange | Open Web | CC-By-SA | 2,018 | Stack Exchange | JNeu, JSmith, https://stackoverflow.com/users/1713208, https://stackoverflow.com/users/8869318, https://stackoverflow.com/users/9337071, tehhowch | English | Spoken | 613 | 1,159 | Find and replace script (Google-apps) for Content grouping
Some of you may be familiar with content grouping in Google Analytics, which basically lets you group any number of URL's in user-specified groups (this is useful for analyzing pages that belong together all at the same time). I'm working on a script to take that to the next level and use it in Google Sheets as well.
Goal: have a working script that rewrites URL's and gives them another name, regardless of whether it uses upper or lower cases in the URL.
So far I have this:
function onOpen() {
var sheet = SpreadsheetApp.getActiveSheet();
var range = sheet.getRange("a1:a10000");
var to_replace = /.*example.*/;
var replace_with = "TEST";
var to_replace2 = /.*another-example.*/;
var replace_with2 = "TEST-Nr2";
replaceInSheet(sheet,range, to_replace, replace_with);
replaceInSheet(sheet,range, to_replace2, replace_with2);
}
This script works in the sense that it rewrites URL's with 'Example' in it to 'Test' and it rewrites 'Another-example' into TEST-Nr2.
However, the final script will probably have thousands of URL's that will need to be rewritten. Furthermore, some URL's have uppercases in them, which I want to ignore and just rewrite.
All of the above leads me to two questions:
How can I write the script in such a way (with regular expressions for example?) that I won't have a Googleplex number of To_replace's and replace_with's?
How can I make my to_replace variables case-incensitive?
If any more information is needed on this matter I will gladly provide so.
Kind regards,
JNeu
for the case insensitive use flag i in your regex
Somehow you know the patterns and the replacement values, yes? You need to impart that knowledge to your script.
The simplest way is to read it from a spreadsheet, e.g. on some sheet in some workbook, you have 1 column with the pattern, and another column with the replacement. Then you just read that data in (Range#getValues()), and then iterate that array to process your data range. Note that the pattern you store in the sheet should not include the literal constructor slashes, i.e. you'd want \d{1,3} and not /\d{1,3}/ in the cell.
Example:
function processAll() {
const source = SpreadsheetApp.openById("id of the spreadsheet with pattern - replacement data"),
info = source.getSheetByName("some sheet name")
.getDataRange().getValues();
const databook = SpreadsheetApp.getActive(),
sheet = databook.getSheetByName("name of the sheet with data to process");
if (!sheet) return; // sheet with that name doesn't exist.
const range = sheet.getRange(1, 1, sheet.getLastRow(), 1);
info.forEach(function (row) {
// Create case-insensitive pattern from the string in Column A, e.g. \d{1,3} and NOT /\d{1,3}/
var pattern = new RegExp(row[0], "i");
var repl = row[1]; // replacement text from Column B
replaceInSheet(sheet, range, pattern, repl);
});
}
Additional Reading:
RegExp
Array#forEach
Thanks a lot for answering! Great idea, using a seperate spreadsheet. However, I cannot get your script to work as of yet. Problem is that the replaceInSheet doesn't know where to get the sheet and range. I tried adding them, but that doesn't seem to work.
I set up 2 spreadsheets, could you maybe point out where I'm going wrong?
Spreadheet with URLS:
https://docs.google.com/spreadsheets/d/1X1b4wpimiFEyEzsY-IMoEx_77Ib_Z8v_2vPkdNeyQGY/edit?usp=sharing
Spreadsheet with replacements:
https://docs.google.com/spreadsheets/d/1YIm_DaJWtIRI1SBUzhSu_ca-w9wYtYmLtNyNbOKXZ90/edit?usp=sharing
@jneu You never showed your replaceInSheets function, so I had to infer the desired arguments. My answer assumes that sheet should be the worksheet that contains urls that need to be replaced, and that range is the range object on sheet that contains all values to be checked for replacement. One issue you may be having is that if you attempt to use a simple trigger to activate this, it will fail, as simple triggers are not allowed to access other documents. Check your Stackdriver logs, via View -> Stackdriver Logs (simple triggers also do not display error messages in the UI)
| 16,888 |
https://ceb.wikipedia.org/wiki/Gunung%20Gelepat | Wikipedia | Open Web | CC-By-SA | 2,023 | Gunung Gelepat | https://ceb.wikipedia.org/w/index.php?title=Gunung Gelepat&action=history | Cebuano | Spoken | 222 | 368 | Bungtod ang Gunung Gelepat sa Indonesya. Nahimutang ni sa lalawigan sa West Nusa Tenggara, sa habagatan-kasadpang bahin sa nasod, km sa sidlakan sa Jakarta ang ulohan sa nasod. metros ibabaw sa dagat kahaboga ang nahimutangan sa Gunung Gelepat, o ka metros sa ibabaw sa naglibot nga tereyn. Mga ka kilometro ang gilapdon sa tiilan niini. Ang Gunung Gelepat nahimutang sa pulo sa Lombok Island.
Ang yuta palibot sa Gunung Gelepat patag sa amihanan, apan sa habagatan nga kini mao ang kabungtoran. Ang usa ka luok sa habagatan, dagat ang pinakaduol sa Gunung Gelepat. Kinahabogang dapit sa palibot ang Gunung Bongak, ka metros ni kahaboga ibabaw sa dagat, km sa amihanan-kasadpan sa Gunung Gelepat. Ang kinadul-ang mas dakong lungsod mao ang Praya, km sa amihanan sa Gunung Gelepat. Sa rehiyon palibot sa Gunung Gelepat, mga bungtod talagsaon komon.
Dunay mga ka tawo kada kilometro kwadrado sa palibot sa Gunung Gelepat nga hilabihan populasyon. Ang klima nga savanna. Ang kasarangang giiniton °C. Ang kinainitan nga bulan Oktubre, sa °C, ug ang kinabugnawan Enero, sa °C. Ang kasarangang pag-ulan milimetro matag tuig. Ang kinabasaan nga bulan Enero, sa milimetro nga ulan, ug ang kinaugahan Agosto, sa milimetro.
Saysay
Ang mga gi basihan niini
Mga bungtod sa West Nusa Tenggara
Mga bungtod sa Indonesya nga mas taas kay sa 200 metros ibabaw sa dagat nga lebel | 42,847 |
https://cs.wikipedia.org/wiki/Lomen%C3%A1%20pyramida | Wikipedia | Open Web | CC-By-SA | 2,023 | Lomená pyramida | https://cs.wikipedia.org/w/index.php?title=Lomená pyramida&action=history | Czech | Spoken | 754 | 1,956 | Lomená pyramida je jednou z egyptských pyramid, nachází se v nekropoli Dahšúr 40 km jižně od Káhiry. Nechal ji postavit faraón Snofru ve 26. století př. n. l. Je vysoká 101 metrů (původně asi 105 m, ale byla snížena působením eroze), což z ní činí čtvrtou největší staroegyptskou pyramidu, každá strana čtvercové základny je dlouhá 188,5 metru. Její interiér byl zpřístupněn veřejnosti roku 2009.
Charakteristka
Pyramida je neobvyklá svým nepravidelným tvarem. Ve spodní části svírají stěny se zemí úhel 55°5', ve výšce 45 metrů se však lámou na mírnější úhel 43°9'17'. Výška spodní části je 47,4 m, horní části 57,67 m, celková výška 105 m. Půdorysná šířka stran čtverce se pohybuje v rozmezí 189,41–189,75 m.
Důvody a diskuze pro tuto isometrii se egyptology vysvětlují jednak historickými souvislostmi a jednak zcela praktickými důvody. Do té doby se stavěly pouze stupňovité pyramidy (Džoser), takže byla vůbec prvním pokusem o pyramidu ve tvaru čtyřbokého jehlanu. Nedostatek zkušeností vedl k tomu, že stavitelé navrhli stěny příliš příkré, což ohrožovalo statiku stavby, také vzhledem k nestabilnímu hlinito-štěrkovému podloží. Proto bylo v průběhu stavby rozhodnuto změnit sklon stěn (kdyby k tomu nedošlo, mohla by pyramida měřit kolem 125 m). Zastánci této domněnky argumentují tím, že Červená pyramida, stavěná krátce po lomené, má také úhel stěn 45°.
Stěny pyramidy jsou pokryty bílým vápencem — jde o nejlépe dochované obložení ze všech egyptských pyramid. Lomená pyramida je ojedinělá také tím, že má uvnitř dvě komory spojené chodbou. Každá komora má vlastní vchod: jeden se nachází na západní straně ve výšce 33 metrů a druhý na severní straně ve výšce 12 metrů. Přes padesát metrů jižním směrem stojí menší satelitní pyramida (pravděpodobně zamýšlená jako sídlo panovníkova Ka), vysoká 26 metrů.
Historický vývoj stavby
Snofru byl jediný panovník, o kterém je známo, že pro sebe nechal vybudovat tři pyramidy (další je pyramida v Médúmu a nedaleká Červená pyramida). Podrobnější výzkum provedený M. Lehnertem naznačuje, že stavitelé hledali výhodnější tvar zajišťující stabilitu stavby v ukládání stavebních bloků spolu s vnitřní konstrukcí prostor k ukládání pohřební výbavy a samotného sarkofágu. Zřejmě první byla Pyramida v Médúmu, jejíž stavbu pravděpodobně zahájil již předchozí faraon Hunej z 3. dynastie, která byla neúspěšná, vnější obložení se zbortilo. Nebyla dokončená. Bývá také označované jako falešná. Druhým stavitelským řešením v Lomené pyramidě se získaly zkušenosti s výstavbou stabilnější formy. Nicméně i zde se v průběhu stavby vyskytly trhliny a dílčí borcení stěn. Úspěch se dostavil až při stavbě třetí Červené pyramidy s upraveným sklonem stěn jehlanu na 45°.
Průzkum
Díky svému neobvyklému tvaru poutala Lomená pyramida evropské cestovatele již od 17. století. Pyramidu ve svých cestopisech zmínil například anglický orientalista Robert Huntington, Edward Melton anglo-holandský cestovatel v 17. století, cestovatel a politik Robert Wood či spisovatel a Richard Pococke. První archeologické průzkumy zde probíhaly v září 1839 pod vedením Johna Shae Perringa. Stavbou se v 19. století zabýval rovněž Karl Richard Lepsius a William Flinders Petrie. Z výzkumu pyramidy prováděného v druhé polovině 40. let 20. století vedeného Abd es-Salámem Husajnem a Alexandrem Varrilem se dokumentace nedochovala. Novodobé komplexní průzkumy pyramidy uskutečnil na počátku 50. let 20. století Ahmed Fachrí. Následovaly výzkumy italské egyptologické expedice Vita Maragioglia a Celeste Rinaldi, rakouské expedice vedené Josefem Dornerem a nakonec německé expedice z německého archeologického ústavu v Káhiře pod vedením Rainera Stadelmanna v roce 1980.
Stavitel a datování
Stavba Lomené pyramidy byla zahájena v 8. roce sčítání skotu, za vlády faraona Snofrua přibližně po dokončení stupňovité pyramidy v Médúmu ~2600 př. n. l. Vystavěním Lomené pyramidy založil Snofru pohřebiště v Dahšúru. Důvod, proč Snofru začal budovat druhou pyramidu, je dosud neznámý. Pravděpodobně se mohlo se také jednat o postupný vývoj konstrukce a stavebních postupů od stupňovité pyramidy k vytvoření symetrické čtyřboké pyramidy (jehlanu) se čtvercovou základnou s dostatečnou statickou stabilitou, včetně vnitřních prostor pro uložení sarkofágu a pohřební výbavy. S tím též souviselo zabezpečí proti následnému vyloupení. Nicméně podrobný průzkum ukázal, že v prostoru pohřební komory nebyl zabudován sarkofág. Předpokládá se, že po zjištění vnitřních prasklin se bezprostředně započalo se stavbou Červené pyramidy, vzdálené od lomené ~1 km, s upraveným sklonem bočních stěn na 45°. Její stavba byla zahájena přibližně v 13.roce vlády Snofrua a dokončena po ~17 letech. Zkušenosti stavitelů při hledání optimálních postupů stavby čtyřbokých pyramidových jehlanů se zužitkovaly, patrně již krátce po dokončení Červené pyramidy, při projektování pyramidy následníka Cheopse Velké pyramidy v Gize.
Odkazy
Poznámky
Reference
Související články
Červená pyramida
Seznam egyptských pyramid
Externí odkazy
A. Varille, A propos des pyramides de Snefrou (Cairo, 1947), 7.
R. Stadelmann, Die agyptischen Pyramiden (Mainz, 1985), 94.
http://guardians.net/egypt/cyberjourney/dahshur/bentpyramid/bent_pyramid_interior.htm
4. dynastie
Egyptské pyramidy
Snofru | 38,036 |
https://eu.wikipedia.org/wiki/Dorans | Wikipedia | Open Web | CC-By-SA | 2,023 | Dorans | https://eu.wikipedia.org/w/index.php?title=Dorans&action=history | Basque | Spoken | 373 | 1,027 | Dorans Frantziako udalerri bat da, Belfort Herrialdea departamenduan dagoena, Borgoina-Franche-Comté eskualdean. 2013an biztanle zituen.
Demografia
Biztanleria
2007an Dorans udalerrian erroldatutako biztanleak 565 ziren. Familiak 235 ziren, horien artean 61 pertsona bakarrekoak ziren (24 bakarrik bizi ziren gizonak eta 37 bakarrik bizi ziren emakumeak), 73 seme-alabarik gabeko familiak ziren, 85 seme-alabak dituzten bikoteak ziren eta 16 seme-alabak dituzten guraso-bakarreko familiak ziren.
Biztanleriak, denboran, ondorengo grafikoan ageri den bilakaera izan du:
Etxebizitza
2007an 243 etxebizitza zeuden, 236 familiaren etxebizitza nagusia ziren, 2 bigarren erresidentzia ziren eta 6 hutsik zeuden. 209 etxeak ziren eta 33 apartamentuak ziren. 236 etxebizitza nagusietatik 194 bere jabearen bizilekua ziren, 38 alokairuan okupaturik zeuden eta 4 doan lagata zeuden; 7 etxek gela bat zuten, 10 etxek bi zituzten, 16 etxek hiru zituzten, 45 etxek lau zituzten eta 157 etxek bost zituzten. 203 etxek euren parking plaza propioa zuten azpian. 95 etxetan ibilgailu bat zegoen eta 124 etxetan bat baino gehiago zituzten.
Biztanleria-piramidea
2009an sexu eta adinaren araberako biztanleria-piramidea hau zen:
Ekonomia
2007an lan egiteko adina zuten pertsonak 367 ziren, horien artean 278 aktiboak ziren eta 89 inaktiboak ziren. 278 pertsona aktiboetatik 264 lanean zeuden (144 gizon eta 120 emakume) eta 14 langabezian zeuden (6 gizon eta 8 emakume). 89 pertsona inaktiboetatik 32 erretiraturik zeuden, 32 ikasten zeuden eta 25 "bestelako inaktibo" gisa sailkaturik zeuden.
Diru sarrerak
2009an Dorans udalerrian 228 unitate fiskal zeuden, 563,5 pertsonek osaturik. Pertsona bakoitzeko diru-sarrera fiskalaren mediana urteko 22.533 euro zen.
Ekonomia jarduerak
2007an zeuden 11 komertzioetatik, 1 erauzte enpresa zen, 1 janari enpresa zen, 2 bestelako produktu industrialen fabrikazioko enpresak ziren, 1 eraikuntza enpresa zen, 1 ibilgailuen saltze eta konpontze enpresa zen, 1 garraio enpresa zen, 1 finantziazio enpresa zen, 2 zerbitzu enpresak ziren eta 1 «beste zerbitzu jarduera batzuk» multzoan sailkatutako enpresa zen.
2009an norbanakoentzat zegoen zerbitzu bakarra igeltseroa zen.
2009an zegoen establezimendu komertzial bakarra okindegi bat zen.
2000. urtean Dorans udalerrian 3 nekazaritza-ustiategi zeuden.
Hezkuntza eta osasun ekipamenduak
2009an haur-eskola 1 eta inguruko herriekin osatutako barreiatutako lehen-hezkuntzako eskola 1 zegoen.
Gertuen dauden herriak
Diagrama honek gertuen dauden herriak erakusten ditu.
Erreferentziak
Kanpo estekak
Résumé statistique INSEEren udalerriko estatistiken laburpena.
Évolution et structure de la population INSEEren udalerriko datuen fitxa.
France par comune Frantziako udalerri guztietako datu zehatzak mapa baten bitartez eskuragarri.
Belfort Herrialdeko udalerriak | 28,377 |
https://fr.wikipedia.org/wiki/Vieux%20Palais | Wikipedia | Open Web | CC-By-SA | 2,023 | Vieux Palais | https://fr.wikipedia.org/w/index.php?title=Vieux Palais&action=history | French | Spoken | 46 | 77 | Vieux Palais peut désigner :
le Vieux Palais de Berlin en Allemagne ;
le Vieux Palais (détruit) à Rouen en France ;
le Vieux Palais d'Espalion en France ;
le Stari dvor (vieux palais) à Belgrade en Serbie.
Voir aussi
Palazzo Vecchio à Florence en Italie. | 17,688 |
https://hu.wikipedia.org/wiki/M%C3%A1rv%C3%A1nyos%20m%C3%A9zev%C5%91 | Wikipedia | Open Web | CC-By-SA | 2,023 | Márványos mézevő | https://hu.wikipedia.org/w/index.php?title=Márványos mézevő&action=history | Hungarian | Spoken | 129 | 457 | A márványos mézevő (Pycnopygius cinereus) a madarak osztályának verébalakúak (Passeriformes) rendjébe és a mézevőfélék (Meliphagidae) családjába tartozó faj.
Rendszerezése
A fajt Philip Lutley Sclater angol ornitológus írta le 1873-ban, a Ptilotis nembe Ptilotis cinerea néven.
Alfajai
Pycnopygius cinereus cinereus (P. L. Sclater, 1874)
Pycnopygius cinereus dorsalis Stresemann & Paludan, 1934
Pycnopygius cinereus marmoratus (Sharpe, 1882)
Előfordulása
Új-Guinea szigetén, Indonézia és Pápua Új-Guinea területén honos. Természetes élőhelyei a szubtrópusi és trópusi síkvidéki és hegyi esőerdők. Állandó, nem vonuló faj.
Megjelenése
Testhossza 20–22 centiméter, testtömege 36–58 gramm.
Életmódja
Gyümölcsökkel és nektárral táplálkozik, de fogyaszt rovarokat is.
Természetvédelmi helyzete
Az elterjedési területe nagy, egyedszáma pedig stabil. A Természetvédelmi Világszövetség Vörös listáján nem fenyegetett fajként szerepel.
Jegyzetek
További információk
Képek az interneten a fajról
Pycnopygius
Madárfajok
Indonézia madarai
Pápua Új-Guinea madarai
Új-Guinea madarai | 48,941 |
https://de.wikipedia.org/wiki/Olympische%20Zwischenspiele%201906/Schie%C3%9Fen | Wikipedia | Open Web | CC-By-SA | 2,023 | Olympische Zwischenspiele 1906/Schießen | https://de.wikipedia.org/w/index.php?title=Olympische Zwischenspiele 1906/Schießen&action=history | German | Spoken | 509 | 1,053 | Bei den Olympischen Zwischenspielen 1906 in Athen wurden 12 Wettbewerbe im Schießen ausgetragen.
Männer
Schnellfeuerpistole (25 m)
Es wurden 30 Schuss in Sechser-Serien auf eine 25 m entfernte Scheibe mit 10 Ringen und einem Durchmesser von 30 cm abgegeben. Gesamtzeit 10 Minuten, drei Probeschüsse.
Duellpistole (20 m)
Es wurden 30 Schuss in Sechser-Serien auf eine 20 m entfernte Scheibe mit 10 Ringen und einem Durchmesser von 19 cm abgegeben. Gesamtzeit 10 Minuten, drei Probeschüsse.
Duellpistole (25 m)
Es wurden 30 Schuss in 10er-Serien nach Kommando auf eine 25 m entfernte Personensilhouette abgegeben, die folgende Trefferpunkte (P) hatte: Rumpf Mitte 5 P, Rumpf vorne und hinten 4 P, Kopf 3 P, Hüften und Oberschenkel 3 P, Knie 2 P, Unterschenkel 1 P
Freier Revolver (50 m)
Es wurden 30 Schuss in Sechser-Serien auf eine 50 m entfernte Scheibe mit 10 Ringen und einem Durchmesser von 50 cm abgegeben. Gesamtzeit 10 Minuten, drei Probeschüsse.
Militärrevolver (20 m)
Es wurden 30 Schuss in Sechser-Serien auf eine 25 m entfernte Scheibe mit 10 Ringen und einem Durchmesser von 30 cm abgegeben. Gesamtzeit 10 Minuten, drei Probeschüsse.
Militärrevolver (20 m, Modell des Jahres 1874)
Es wurden 30 Schuss in Sechser-Serien auf eine 25 m entfernte Scheibe mit 10 Ringen und einem Durchmesser von 30 cm abgegeben. Gesamtzeit 10 Minuten, drei Probeschüsse.
Militärgewehr (300 m)
Es wurden 30 Schuss in Zehner-Serien auf eine 300 m entfernte Scheibe mit 10 Ringen und einem Durchmesser von 100 cm abgegeben. Zeit 30 Minuten je Serie, drei Probeschüsse, Anschlag wahlweise stehend oder kniend.
Militärgewehr (200 m, Modell des Jahres 1874)
Es wurden 30 Schuss in Zehner-Serien auf eine 200 m entfernte Scheibe mit 10 Ringen und einem Durchmesser von 80 cm abgegeben. Zeit 30 Minuten je Serie, drei Probeschüsse, Anschlag wahlweise stehend oder kniend.
Freies Gewehr, Einzel (300 m)
Es wurden 30 Schuss in Zehner-Serien auf eine 300 m entfernte Scheibe mit 10 Ringen und einem Durchmesser von 100 cm abgegeben. Zeit 30 Minuten je Serie, drei Probeschüsse, Anschlag wahlweise stehend oder kniend.
Freies Gewehr, Mannschaft (300 m)
Es wurden von allen fünf Schützen einer Mannschaft 3 × 40 Schuss (stehend, kniend und liegend) auf eine 300 m entfernte Scheibe mit 10 Ringen und einem Durchmesser von 100 cm abgegeben. Gesamtzeit für jede Mannschaft 9 Stunden, 30 Probeschüsse.
Tontaubenschießen (Trap)
30 Schuss auf einzeln hochgeworfene Tonscheiben. Die Entscheidung zwischen Gold und Silber fiel in einem Stechen.
Tontaubenschießen, Doppelschuss (Trap)
20 Schuss (10 Doppelschüsse) auf zwei gleichzeitig hochgeworfene Tonscheiben in 14 m Entfernung.
Inoffizielle Wettbewerbe
Beim Mannschaftswettkampf mit dem freien Gewehr waren für jede Nation 5 Schützen am Start, die jeweils 40 Schuss in stehender, kniender und liegender Stellung abgeben mussten. Die Einzelergebnisse jeder Stellung und das Ergebnis in Summe aller drei Stellungen (Dreistellungskampf) wurden für jeden Schützen gesondert festgehalten. Die besten Schützen wurden anschließend mit einem Preis geehrt (keine Medaillen). Es handelte sich dabei nur um eine inoffizielle Wertung, eine Aufnahme in die offiziellen Siegerlisten der Spiele fand nicht statt.
Freies Gewehr, liegend
Freies Gewehr, kniend
Freies Gewehr, stehend
Freies Gewehr, Dreistellungskampf
Schiessen
Sportschießwettbewerb
Sportschießen (Griechenland)
Sportveranstaltung in Athen | 35,198 |
https://math.stackexchange.com/questions/1237250 | StackExchange | Open Web | CC-By-SA | 2,015 | Stack Exchange | English | Spoken | 98 | 198 | Exercise 2.13 from a computational introduction to number theory and algebra
This is an exercise from V. Shoup. A computational introduction to number theory and algebra.
Let $p=2, e=3, a=b=1, c = 0$, then $p^{2e} = 64, z\in \{0,1,2,\cdots,63\}$, the conclusion is, there are 8 numbers for the following to be true:
$$
\lfloor ((z+1) \mod 64) / 8 \rfloor = 0
$$
Well, there are only 7 such numbers, apparently, $0, 1, 2, 3, 4, 5, 6$.
Is there something wrong with my line of logic?
I think you miss $z=63$. For $z=63$, note that $z+1\equiv 0\pmod{64}$.
| 10,457 | |
https://en.wikipedia.org/wiki/Business%20rules%20approach | Wikipedia | Open Web | CC-By-SA | 2,023 | Business rules approach | https://en.wikipedia.org/w/index.php?title=Business rules approach&action=history | English | Spoken | 927 | 1,150 | Business rules are abstractions of the policies and practices of a business organization. In computer software development, the business rules approach is a development methodology where rules are in a form that is used by, but does not have to be embedded in, business process management systems.
The business rules approach formalizes an enterprise's critical business rules in a language that managers and technologists understand. Business rules create an unambiguous statement of what a business does with information to decide a proposition. The formal specification becomes information for process and rules engines to run.
Advantages
The adoption of business rules adds another tier to systems that automate business processes. Compared to traditional systems, this approach has the following major advantages,
lowers the cost incurred in the modification of business logic
shortens development time
rules are externalized and easily shared among multiple applications
changes can be made faster and with less risk
Business process automation
Business rules represent a natural step in the application of computer technology aimed at enhancing productivity in the workplace. Automated business processes that have business logic embedded inside often take substantial time to change, and such changes can be prone to errors. And in a world where the life cycle of business models has greatly shortened, it has become increasingly critical to be able to adapt to changes in external environments promptly. These needs are addressed by a business rules approach.
Business agility
Business rules enhance business agility. And the manageability of business processes also increases as rules become more accessible.
Technical details
The programs designed specifically to run business rules are called rule engines. More complete systems that support the writing, deployment and management of business rules are called business rules management systems (BRMSs).
Many commercial rule engines provide the Rete algorithm, a proprietary algorithm that embodies many of the principles of Rete. However, there are other execution algorithms such as the sequential algorithm (ILOG and Blaze Advisor terminology), algorithms for evaluating decision tables/trees, and algorithms tuned for hierarchical XML. The Rete algorithm is a stateful pattern matching algorithm designed to minimize the evaluation of repetitive tests across many objects/attributes and many rules. Different fields of usage are best for Rete-based and non-Rete-based execution algorithms. For simple stateless applications with minimal sharing of conditions across rules, a non-Rete-based execution algorithm (such as the sequential algorithm) may be preferable. For evaluating decision tables and trees, an algorithm that exploits the hierarchical relationships between the rule conditions may perform better than a simple Rete or sequential algorithm tuned for discrete rules.
Business rules can be expressed in conventional programming languages or natural languages. In some commercial BRMSs rules can also be expressed in user-friendly rule forms such as decision tables and decision trees. Provided with a suitable interface to design or edit decision tables or trees, it is possible for business users to check or change rules directly, with minimal IT involvement.
When rules are expressed in natural language, it is necessary to first define a vocabulary that contains words and expressions corresponding to business objects and conditions and the operations involving them. To make the rules executable by a rule engine, it is also necessary to implement the operations and conditions in a programming language. With a defined and implemented vocabulary, it is relatively easy to write rules in a BRMS. Changes can also be made quickly as long as they can be expressed in the existing vocabulary. If not, the vocabulary must be expanded accordingly.
Separating the vocabulary from the logic makes it possible for business rules to be modeled, with the business user mapping their business logic and with IT integrating data and the generated code into the target application.
Business rules are also key to the enterprise decision management approach to decision automation.
Increasingly, business rules are also viewed as a critical component of business process management solutions because of the need to ensure flexibility.
Relation to database management
As argued by Christopher J. Date, business rules translate into data integrity constraints when one thinks in terms of the relational model of database management. Thus, a true RDBMS could be thought of in terms of a declarative business rules engine with added data management capability.
However, business rules need not only be regarded as constraints. They can also be used to specify constructive business policies, such as "preferred clients get a discount of 10%". Used in this way, business rules are like SQL queries, rather than data integrity constraints.
Relation to business process management
Some analysts believe the combination of business rules technology with business process management offers an agile approach to workflow and enterprise integration. BPM and BR software support business goals by managing and running business processes and business rules in separate yet complementary ways. A business process is often a complex map of flow controls. It might have many subprocesses, decisions and while loops. Wherever a decision or while loop appears, business rules can evaluate the data provided by the process and control the basis for change in flows.
Often there are separate reasons for updates to processes or rules. New regulations or business strategies may affect the rules without changing core business processes. New applications or procedures might change the business process. In either case, a composite approach to rules and processes can be very flexible.
As more and more BPM vendors either add business rules engines to their BPM engines or OEM business rules management systems, business rules seems to be becoming a subset of BPM.
See also
Business Rules Engine Providers
References
Business software | 38,526 |
https://ar.wikipedia.org/wiki/%D9%85%D8%A7%D9%83%D8%B3%20%D9%84%D9%88%D9%81%D9%8A%D8%B1%D8%A7 | Wikipedia | Open Web | CC-By-SA | 2,023 | ماكس لوفيرا | https://ar.wikipedia.org/w/index.php?title=ماكس لوفيرا&action=history | Arabic | Spoken | 124 | 337 | ماكس لوفيرا (8 يناير 1997 بأندورا لا فيلا في أندورا - ) هو لاعب كرة قدم أندوري في مركز الدفاع. شارك مع منتخب أندورا تحت 17 سنة لكرة القدم ومنتخب أندورا تحت 19 سنة لكرة القدم ومنتخب أندورا تحت 21 سنة لكرة القدم ومنتخب أندورا لكرة القدم. أما مع النوادي، فقد لعب مع .
مراجع
وصلات خارجية
لاعبو كرة قدم رجالية أندوريون
لاعبو كرة قدم رجالية أندوريون مغتربون
مدافعو كرة قدم رجالية
أشخاص من أصل سويسري
أشخاص من أصل نمساوي
أندوريون من أصل ألماني
أندوريون من أصل إسباني
رياضيون أندوريون مغتربون في إسبانيا
لاعبو الدوري الإسباني الدرجة الثانية ب
لاعبو كرة قدم أندوريون
لاعبو كرة قدم أندوريون مغتربون
لاعبو لييدا ايسبورتيو
لاعبو منتخب أندورا لكرة القدم
لاعبو منتخب أندورا لكرة القدم للشباب
مواليد 1997 | 48,720 |
https://dba.stackexchange.com/questions/145887 | StackExchange | Open Web | CC-By-SA | 2,016 | Stack Exchange | Paul, https://dba.stackexchange.com/users/43932, https://dba.stackexchange.com/users/9486, philipxy | English | Spoken | 789 | 1,191 | Is Order part of a result set?
This question got me thinking:
ROW_NUMBER() OVER (PARTITION BY B,A ORDER BY C) doesn't use index on (A,B,C)
In it the questioner states two functions give the same result set, however they don't in that they have identical rows but they are always in different (and consistent) orders.
Is the order of a result set part of the result set? Wikipedia implies no (but doesn't strongly determine an answer, nor does it speak from a position of authority):
https://en.wikipedia.org/wiki/Result_set
Is
Sector Name
A Alpha
B Beta
the same result set as
Sector Name
B Beta
A Alpha
I personally only work with SQL Server, does the answer change by platform? Is the answer a matter of opinion only? (If so please give answers of fact only, or of consensus by professional bodies at any rate- not your personal opinion please)
Possibly the ANSI 92 definition held at
http://www.contrib.andrew.cmu.edu/~shadow/sql/sql1992.txt
gives the answer. Section 3.1 Definitions seems to draw a distinction between the idea of a result set and a result sequence. The distinction was not in my awareness when I wrote the question using the former phrase
v) sequence: An ordered collection of objects that are not necessarily
distinct.
w) set: An unordered collection of distinct objects. The collection
may be empty.
Ansi sql doesn't consider ordering a part of the result set - thus the 2 are identical.
The result set is indeed a set.
Btw, the 'big players' (oracle,Microsoft...) follow the ansi sql quite closely, if they consider them the same so does the standard... (On most things)
The term "result set" [sic] does not appear in the 1992 SQL standard. It's typically used to mean the result of a select statement. Which is a partially ordered sequence of rows from a table:
20.2
Format
< direct select statement: multiple rows >
General Rules
5) If an < order by clause > is specified, then the ordering of rows of the result is effectively determined by the < order by clause > as follows:
From the 2003 SQL standard:
3.1.6.30 result set: A sequence of rows specified by a < cursor specification >, brought into existence by opening a cursor and ranged over by that cursor, and having operational properties of sensitivity, scrollability, holdability, and returnability.
3.1.6.32 returned result set: A result set created during execution of an SQL-invoked procedure and not destroyed when that execution terminates. Such a result set can be accessed by using a cursor other than the one that brought it into existence (a received cursor).
14.3
Format
< cursor specification > ::= < query expression > [ < order by clause > ] [ < updatability clause > ]
22.2
Function
Specify a statement to retrieve multiple rows from a specified table.
Format
< direct select statement: multiple rows > ::= < cursor specification >
Syntax Rules
2) The < cursor specification > shall not contain an < updatability clause >.
General Rules
1) Let Q be the result of the < cursor specification >.
2) Case:
a) If Q is empty, then a completion condition is raised: no data.
b) Otherwise, Q is not empty and Q is returned. The method of returning Q is implementation-defined.
(Clearly the result of a SELECT statement is ordered because that is what its ORDER BY does.)
Tables are unordered. But a result set [sic] is not a table.
Your last sentence undoes you. The standard defines a set as being ordered. Thus you would have to refer to a result collection.
@Paul Read what it says: "result set: A sequence". French toast is not toast. I would agree that the standard has a lot of poor wording. As I say arguments are moot because rows have to be returned per any ORDER BY.
I see my "2003" should be 2006. Quotes are from a draft version downloaded zipped from the Wikipedia SQL page. I see there's a link with text "SQL:2011 draft" there now. Quotes are from file 5CD2-02-Foundation-2006-01.pdf. Title page says ISO/IEC JTC 1/SC 32, Date: 2006-02-01, CD 9075-2:200x(E), ISO/IEC JTC 1/SC 32/WG 3". I had to add spaces after < & before > to defeat markup. PS Again, I don't know how you think a select statement does not return a sequence, although maybe you just don't think "result set" refers to that. The term predates SQL DBMSs & "set" wasn't meant mathematically.
Have you ever worked on a server that had a tempDB data file split into multiple files? And then run simple un-ordered queries on tables on that server? (e.g. select * from tblName) The results will come in different orders each time. Would you say the result is same each time, or different?
| 9,695 |
https://stackoverflow.com/questions/56502483 | StackExchange | Open Web | CC-By-SA | 2,019 | Stack Exchange | Yazeed Sabri, https://stackoverflow.com/users/1256452, https://stackoverflow.com/users/1290731, https://stackoverflow.com/users/6523738, jthill, torek | English | Spoken | 1,079 | 1,459 | Why doesn't Git rebase take exponential time like Darcs?
So according to the this Wikipedia page about merges in version control, Darcs uses patch commutation and so does Git when it rebases.
I'm curious to know why doesn't Git rebase take exponential time in some cases the same way Darcs does?
This page lists a bunch cases of when it could happen with Darcs and how to try and tackle them.
I have not used darcs and could be off base here, but scanning through the Wikipedia article, I observe that its claim here is misleading:
Patch commutation is used in Darcs to merge changes, and is also implemented in git (but called "rebasing"). Patch commutation merge means changing the order of patches (i.e. descriptions of changes) so that they form a linear history. In effect, when two patches are made in the context of a common situation, upon merging, one of them is rewritten so that it appears to be done in the context of the other.
Git neither stores patches, nor reorders them.1 You can reorder patches manually if you like. When you use git rebase to convert commits—which are snapshots—into changesets, the conversion itself is done in a simple manner that is worst-case O(nd) where n is the number of lines and d is the length of the edit script. (Whether this is handled as a patch or as a cherry-pick, which is a merge, depends on which rebase algorithm you choose; the merge case tends to use more CPU time as it also looks for treewide rename operations.)
Hence, if you have C total commits, the worst case for a git rebase -i is roughly O(Cnd).2 Git is not going to try any other order for the commits: it's up to you to do any reordering. If your rebase fails, you could try all possible reorderings, which would be a factorial function on the number of commits, but Git won't do that for you.
1If you do a git rebase on a group of commits that has one or more merges inside itself, Git does "flatten" them and eliminate the merges—but it doesn't try many orders, it just does one topological walk of the whole thing. There is no attempt to be smart about it, and often this will lead to terrible merge conflicts, i.e., it's often a bad idea. If you use the new --rebase-merges flag, Git will try to keep the merges, but does so by re-performing them. It just builds an identical topology, using a mini scripting language, and then lets you edit it manually if you like.
2If rename-detection gets involved, that is O(n2) in the number of files that could be rename-detected. Git tries to limit this in several ways, including having a maximum length for the rename queue, with the current default being 4000 files.
I think I understand what you meant but just to double check, what do you mean by "d is the length of the edit script." what is edit script? Also when say handed as a patch you mean just a normal rebase with no rename resolutions?
(1) The output of git diff (or indeed any solution to the string-to-string edit problem) is an edit script consisting of operations like "insert I at location 17" and "delete D items at location 30". The length of this edit script is d and the length of the input itself is n. (2) Yes, that's exactly right: git rebase without -i or -m or various other options makes Git runs git format-patch and feed the resulting patch-set to git am.
You can see what's going on in an example I recall from somewhere: start with six lines A B C D E F, on one branch insert three lines G G G at the front then insert six more lines A B C D E F at the front; on another branch (from that first patch/commit) change C to C4. Now merge the two branches.
Git will produce A B C4 D E F G G G A B C D E F; darcs will change the second C, based on the premise that its identity tracking produces a better result than Git's context-and-location tracking. If the "4" is initialization code for the block, Git's result is wrong; if it's initialization code for the function or file, darcs's is wrong.
It's that identity tracking that eats the time.
I understand the identity part but I didn't get your last two examples stating from "If the "4" is..."
The question no vcs can hope to answer reliably is whether the change properly belongs with the first abcdef block in the file or with the one that has apparently been progressively been shoved farther down. Darcs spends a lot of rigor and effort and cpu calculating the right result for a bad guess. Git's guess is no better, but it gets to its answer a hell of a lot faster.
Another possibility both vcs's miss is of course that the same change should be made to both blocks.
Question for my own edification: this means that darcs' "merge" operation examines each commit from the merge base to both branch tips? I think that it might be reasonable for Git to do this specifically to look for file renames (perhaps under operator control since it's pretty cpu-expensive).
@torek it does. git imerge (on GitHub) does incremental merging, I think that'd do. But I don't think any automated procedure can be completely trusted here, merging code is always going to require judgment, i.e. awareness of what the automatic merge will produce and a human decision about the right result for this merge. Think about all the possible variations on the "should change both" scenario.
Yeah, I agree that no matter how you do a merge, there are perils. Mostly what I'm thinking of for git merge here is a scan for 100%-identical renames, so that you can make rename-only commits during major refactorings so as to tell Git: when matching up files, these are the right ones. While git-imerge will achieve this (automatically), imerge is painfully slow in a big repository (I tried it on an appropriate situation once).
You can merge a separate rename-only branch wherever it's needed, just as with other conflicts Git's tip-merging can't isolate when they're all mixed together it's pretty easy to isolate then for it--and it makes the history easier for humans to read too @torek
| 33,147 |
https://ja.wikipedia.org/wiki/%E3%83%9F%E3%83%83%E3%82%AF%E3%83%9E%E3%83%83%E3%82%AF%20%28%E6%98%A0%E7%94%BB%29 | Wikipedia | Open Web | CC-By-SA | 2,023 | ミックマック (映画) | https://ja.wikipedia.org/w/index.php?title=ミックマック (映画)&action=history | Japanese | Spoken | 36 | 684 | 『ミックマック』（）は、2009年のフランス映画。
ストーリー
レンタルビデオ店で働くバジルは、偶然居合わせた発砲事件で流れ弾を頭に喰らい、摘出されないまま頭の中に残ってしまった。そのせいで彼は全てを失ってしまったのだが、スクラップ工場の仲間と知り合うことで新たな人生を歩み始めることになった。ある日、頭の中の銃弾を作った会社と30年前に父を殺した地雷を製造した会社を発見する。そこで彼は、工場の仲間と共に2つの軍需企業へのイタズラという名の復讐を企てる。
キャスト
…バジル（頭の中に弾丸の残った男）
ドミニク・ピノン…フラカス（バスター）
ヨランド・モロー…ママ・チャウ（料理番）
ジャン＝ピエール・マリエール…プラカール（刑務所）
…ラ・モーム・カウチュ（軟体女）
…タイニィ･ピート（発明アーチスト）
…カルキュレット（計算機）
オマール・シー…レミントン（民族史学者）
アンドレ・デュソリエ…ド･フヌイエ（オーベルヴィリエ軍事会社社長）
…フランソワ・マルコーニ（ヴィジランテ兵器会社社長）
受賞・ノミネート
第35回セザール賞
ノミネート: 音響賞
ノミネート: 美術賞 - アリーヌ・ボネット
ノミネート: 衣装デザイン賞 - マデリーン・フォンテーヌ
備考
当初バジル役にアメリでリュシアン役を演じたジャメル・ドゥブーズが予定されていたが 、撮影10週間前にドゥブーズが降板したため急きょダニー・ブーンが配役された。
脚注
外部リンク
2009年の映画
フランスのクライム・コメディ映画
ジャン＝ピエール・ジュネの監督映画
パリで製作された映画作品
モロッコで製作された映画作品 | 3,712 |
https://lmo.wikipedia.org/wiki/Gruf | Wikipedia | Open Web | CC-By-SA | 2,023 | Gruf | https://lmo.wikipedia.org/w/index.php?title=Gruf&action=history | Lombard | Spoken | 57 | 103 | El Gruf (italian: Monte Gruf) l'è ona scima di Alp Retegh.
A l'è volt 2.936 meter sora el nivell del mar, e 'l se troeuva in de la Provincia de Sondri, in su la fila che la spartiss la Val Codera de la Val Masen.
Montagne de la Lombardia
Val Codera
Val Masen
Scime de 2900 meter | 37,836 |
https://ko.wikipedia.org/wiki/%ED%8C%94%EB%A0%88%EC%8A%A4%ED%83%80%EC%9D%B8%20%ED%95%B4%EB%B0%A9%EB%8C%80%EC%A4%91%EC%A0%84%EC%84%A0 | Wikipedia | Open Web | CC-By-SA | 2,023 | 팔레스타인 해방대중전선 | https://ko.wikipedia.org/w/index.php?title=팔레스타인 해방대중전선&action=history | Korean | Spoken | 187 | 890 | 팔레스타인 해방대중전선(الجبهة الشعبية لتحرير فلسطين, Popular Front for the Liberation of Palestine) 또는 약칭 PFLP는 팔레스타인의 마르크스-레닌주의 공산당이자 혁명적 사회주의 단체로, 1967년 주르지 하바시가 설립하였다. 1964년 설립된 팔레스타인 해방 기구를 형성하는 2번째로 큰 규모의 단체로, 기구 내 가장 큰 규모의 단체는 1959년 설립된 파타이다. 2015년 PFLP는 팔레스타인 해방 기구 실행위원회와 팔레스타인 국가위원회에 참여하는 것을 거부했다.
오랜 세월 동안 PFLP를 이끈 지도자 아부 알리 무스타파 전 서기장이 이스라엘에 의해 살해된 이후 아흐마드 사다트가 2001년부터 중앙위원회 서기장을 역임하고 있다. PFLP는 파타가 이끄는 요르단강 서안 지구와 하마스가 이끄는 가자 지구가 모두 적법하지 않은 정부로 간주하고 있는데, 이는 팔레스타인 자치 정부가 2006년부터 운영되지 않고 있기 때문이다. 미국, 일본, 캐나다, 오스트레일리아, 유럽 연합은 PFLP를 테러 단체로 규정했다.
PFLP는 마르크스-레닌주의와 게바라주의를 이념으로 삼고 있으며, 파타의 온건적인 자세에 대해 강력히 반대한다. PFLP는 이스라엘을 국가로 인정하지 않으며, 이스라엘 정부와의 협상을 일체 거부하고 이스라엘-팔레스타인 분쟁을 단일 국가안으로 해결하기를 원한다. 아부 알리 무스타파 여단은 PFLP의군사조직이다.
같이 보기
팔레스타인 해방민주전선
각주
1967년 설립된 정당
팔레스타인의 정당
팔레스타인의 무장 단체
아시아의 공산주의
팔레스타인의 공산주의
공산주의 정당
스탈린주의 정당
게바라주의
반수정주의 정당
극좌 정치
극좌 정당
반시온주의 정당
좌익 국민주의 정당
민족 해방 운동
일본이 지정한 테러 단체
팔레스타인의 반시온주의 | 20,028 |
https://ceb.wikipedia.org/wiki/Pialaor%C3%A9 | Wikipedia | Open Web | CC-By-SA | 2,023 | Pialaoré | https://ceb.wikipedia.org/w/index.php?title=Pialaoré&action=history | Cebuano | Spoken | 45 | 79 | Suba ang Pialaoré sa Wanatu. Nahimutang ni sa lalawigan sa Sanma Province, sa amihanan-kasadpang bahin sa nasod, km sa amihanan-kasadpan sa Port-Vila ang ulohan sa nasod. Ang Pialaoré nahimutang sa pulo sa Espiritu Santo.
Saysay
Ang mga gi basihan niini
Mga suba sa Sanma Province | 24,149 |
https://bo.wikipedia.org/wiki/%E0%BD%98%E0%BD%B4%E0%BC%8B%E0%BD%A6%E0%BE%9F%E0%BD%BA%E0%BD%82%E0%BD%A6%E0%BC%8B%E0%BD%94%E0%BC%8B%E0%BD%82%E0%BD%89%E0%BD%B2%E0%BD%A6%E0%BC%8D | Wikipedia | Open Web | CC-By-SA | 2,023 | མུ་སྟེགས་པ་གཉིས། | https://bo.wikipedia.org/w/index.php?title=མུ་སྟེགས་པ་གཉིས།&action=history | Tibetan | Spoken | 3 | 63 |
མུ་སྟེགས་པ་གཉིས།
ཆད་ལྟ་བ་དང་རྟག་ལྟ་བ། ། | 11,813 |
https://it.wikipedia.org/wiki/Cincinnati%20Challenger%201994%20-%20Singolare | Wikipedia | Open Web | CC-By-SA | 2,023 | Cincinnati Challenger 1994 - Singolare | https://it.wikipedia.org/w/index.php?title=Cincinnati Challenger 1994 - Singolare&action=history | Italian | Spoken | 75 | 187 | Il singolare del torneo di tennis Cincinnati Challenger 1994, facente parte dell'ATP Challenger Series, ha avuto come vincitore Vince Spadea che ha battuto in finale Jim Grabb 6-7, 7-6, 7-5.
Teste di serie
Olivier Delaître (primo turno)
Lionel Roux (secondo turno)
Jim Grabb (finale)
Michael Tebbutt (quarti di finale)
Sébastien Lareau (semifinali)
Doug Flach (primo turno)
Todd Woodbridge (quarti di finale)
Shūzō Matsuoka (semifinali)
Tabellone
Finale
Parte alta
Parte bassa
Collegamenti esterni
Cincinnati Challenger 1994 | 28,626 |
https://vi.wikipedia.org/wiki/Lycanades%20antapica | Wikipedia | Open Web | CC-By-SA | 2,023 | Lycanades antapica | https://vi.wikipedia.org/w/index.php?title=Lycanades antapica&action=history | Vietnamese | Spoken | 16 | 41 | Lycanades antapica là một loài bướm đêm trong họ Noctuidae.
Chú thích
Liên kết ngoài
Lycanades | 39,355 |
https://askubuntu.com/questions/1308395 | StackExchange | Open Web | CC-By-SA | 2,021 | Stack Exchange | Khashhogi, guiverc, https://askubuntu.com/users/1171897, https://askubuntu.com/users/469152 | English | Spoken | 437 | 1,316 | Ubuntu 20 wide monitor resolution&scale problem
I'm using ubuntu 20 and have an ultra-widescreen monitor (LG 34 inch). When I connect my monitor with the HDMI cable the display on the monitor is pretty bad.
my screenshot
I can only do 'mirror' mode with the highest resolution of 1920x1080. When I try to do 'join display' or 'single display' I can't get the display on my monitor screen.
The scaling and the resolution is pretty bad for a widescreen monitor, how can I properly set up my monitor.
P.S. The monitor works just fine on Windows, above problems only occur on Linux.
xrandr output:
Screen 0: minimum 8 x 8, current 1920 x 1080, maximum 32767 x 32767
eDP-1-1 connected 1920x1080+0+0 (normal left inverted right x axis y axis) 344mm x 194mm
3840x2160 60.00 + 59.98 59.97
3200x1800 59.96 59.94
2880x1620 59.96 59.97
2560x1600 59.99 59.97
2560x1440 59.99 59.99 59.96 59.95
2048x1536 60.00
1920x1440 60.00
1856x1392 60.01
1792x1344 60.01
2048x1152 59.99 59.98 59.90 59.91
1920x1200 59.88 59.95
1920x1080 60.01* 59.97 59.96 59.93
1600x1200 60.00
1680x1050 59.95 59.88
1600x1024 60.17
1400x1050 59.98
1600x900 59.99 59.94 59.95 59.82
1280x1024 60.02
1440x900 59.89
1400x900 59.96 59.88
1280x960 60.00
1440x810 60.00 59.97
1368x768 59.88 59.85
1360x768 59.80 59.96
1280x800 59.99 59.97 59.81 59.91
1152x864 60.00
1280x720 60.00 59.99 59.86 59.74
1024x768 60.04 60.00
960x720 60.00
928x696 60.05
896x672 60.01
1024x576 59.95 59.96 59.90 59.82
960x600 59.93 60.00
960x540 59.96 59.99 59.63 59.82
800x600 60.00 60.32 56.25
840x525 60.01 59.88
864x486 59.92 59.57
800x512 60.17
700x525 59.98
800x450 59.95 59.82
640x512 60.02
720x450 59.89
700x450 59.96 59.88
640x480 60.00 59.94
720x405 59.51 58.99
684x384 59.88 59.85
680x384 59.80 59.96
640x400 59.88 59.98
576x432 60.06
640x360 59.86 59.83 59.84 59.32
512x384 60.00
512x288 60.00 59.92
480x270 59.63 59.82
400x300 60.32 56.34
432x243 59.92 59.57
320x240 60.05
360x202 59.51 59.13
320x180 59.84 59.32
DP-1-1 disconnected (normal left inverted right x axis y axis)
DP-1-2 disconnected (normal left inverted right x axis y axis)
DP-1-3 connected primary 1920x1080+0+0 (normal left inverted right x axis y axis) 800mm x 335mm
3440x1440 59.97 + 49.99 29.99
2560x1080 60.00 59.94 50.00 60.00
1920x1080 60.00* 60.00 50.00 59.94
1680x1050 59.88
1600x900 60.00
1280x1024 60.02
1280x800 59.91
1280x720 60.00 50.00 59.94
1024x768 60.00
800x600 60.32
720x576 50.00
720x480 60.00 59.94
640x480 60.00 59.94
Please clarify your release. The closest is Ubuntu Core 20 as specialist snap based releases use the yy format, unlike yy.mm used by server & desktops releases, but your question seems to be more a standard yy.mm release not a yy release. (yy releases are different products to the yy.mm releases of Ubuntu)
Ubuntu 20.04.1 LTS @guiverc
| 16,635 |
https://war.wikipedia.org/wiki/Dorylus | Wikipedia | Open Web | CC-By-SA | 2,023 | Dorylus | https://war.wikipedia.org/w/index.php?title=Dorylus&action=history | Waray | Spoken | 149 | 485 | An Dorylus in uska genus han Formicidae. An Dorylus in nahilalakip ha familia nga Formicidae.
Ilarom nga taxa
Dorylus acutus
Dorylus aethiopicus
Dorylus affinis
Dorylus agressor
Dorylus alluaudi
Dorylus atratus
Dorylus atriceps
Dorylus attenuatus
Dorylus bequaerti
Dorylus bishyiganus
Dorylus braunsi
Dorylus brevipennis
Dorylus brevis
Dorylus buyssoni
Dorylus congolensis
Dorylus conradti
Dorylus depilis
Dorylus diadema
Dorylus distinctus
Dorylus ductor
Dorylus emeryi
Dorylus erraticus
Dorylus faurei
Dorylus fimbriatus
Dorylus fulvus
Dorylus funereus
Dorylus furcatus
Dorylus fuscipennis
Dorylus gaudens
Dorylus gerstaeckeri
Dorylus ghanensis
Dorylus gribodoi
Dorylus helvolus
Dorylus katanensis
Dorylus kohli
Dorylus labiatus
Dorylus laevigatus
Dorylus lamottei
Dorylus leo
Dorylus mandibularis
Dorylus mayri
Dorylus moestus
Dorylus montanus
Dorylus niarembensis
Dorylus nigricans
Dorylus ocellatus
Dorylus orientalis
Dorylus politus
Dorylus rufescens
Dorylus savagei
Dorylus schoutedeni
Dorylus spininodis
Dorylus stadelmanni
Dorylus stanleyi
Dorylus staudingeri
Dorylus striatidens
Dorylus termitarius
Dorylus titan
Dorylus westwoodii
Dorylus wilverthi
Dorylus vishnui
Mga kasarigan
Mga sumpay ha gawas
Image gallery
Dorylus | 37,540 |
https://en.wikipedia.org/wiki/Kaaperi%20Kivialho | Wikipedia | Open Web | CC-By-SA | 2,023 | Kaaperi Kivialho | https://en.wikipedia.org/w/index.php?title=Kaaperi Kivialho&action=history | English | Spoken | 85 | 141 | Alli Gabriel (Kaaperi, A. K.) Kivialho (17 July 1884 – 28 April 1957; surname until 1904 Alander) was a Finnish educator, bank director and politician, born in Kisko. He was a member of the Parliament of Finland from 1923 to 1924, representing the National Progressive Party.
References
1884 births
1957 deaths
People from Salo, Finland
People from Turku and Pori Province (Grand Duchy of Finland)
National Progressive Party (Finland) politicians
Members of the Parliament of Finland (1922–1924)
Finnish educators
Finnish bankers
University of Helsinki alumni | 31,834 |
https://zh.wikipedia.org/wiki/%E9%87%91%E4%B8%89%E8%A7%92%20%28%E5%8D%B0%E5%BA%A6%29 | Wikipedia | Open Web | CC-By-SA | 2,023 | 金三角 (印度) | https://zh.wikipedia.org/w/index.php?title=金三角 (印度)&action=history | Chinese | Spoken | 6 | 464 | 印度金三角（India's golden triangle），指印度北部德里、阿格拉、齋浦爾三座城市連成的三角形路線，是印度觀光旅遊的最基本路線。市面許多套裝旅遊行程都以此為基礎，再添加其他目的地。
遊覽順序通常從首都德里出發，前往古都阿格拉，再往西去齋浦爾，最後北返德里。這三座城市擁有印度最著名的地標建築：德里的印度門、印度总统府、德里红堡、贾玛清真寺、胡馬雍陵、顾特卜塔、蓮花寺；阿格拉的泰姬陵、阿格拉堡、阿克巴大帝陵、法泰赫普尔西克里；齋浦爾的風之宮、琥珀堡、简塔·曼塔天文台。德里的建築是德里蘇丹國、莫臥兒帝國及英屬印度三個時代的遺產，阿格拉以莫臥兒帝國為主，齋浦爾則是拉傑普特人土邦的首府，反映了拉傑普特大君的宮廷文化。同時在齋浦爾亦可欣賞拉賈斯坦地區的沙漠自然景觀。
金三角全程約720公里，三座城市間的距離約車程4到6小時。亦有鐵路快車可乘。
印度旅遊 | 41,862 |
https://pt.stackoverflow.com/questions/204111 | StackExchange | Open Web | CC-By-SA | 2,017 | Stack Exchange | Portuguese | Spoken | 284 | 611 | Relacionar 3 tabelas em uma linha - MYSQL
Quero fazer uma tabela unido informações de 3 tabelas. Essa tabela será ordenada pela data (dia/mês/ano Horas:Minutos) e visualmente falando ficaria assim
Imagem da tabela
Para melhor compreensão vou postar a estrutura dos bancos
tabela: realation_notes
id user_id comments comment_date
tabela: historic_attendance
id user_id date contato_p contato_e contato_t contato_w contato_m
tabela: clients_documents
id user_id file_name attachment date
inicialmente pensei em usar um UNION
ficando mais ou menos assim
(SELECT id, user_id, comments, comment_date AS date FROM `realation_notes`
WHERE `user_id` = 582)
UNION
(SELECT * FROM `historic_attendance` WHERE `user_id` = 582)
UNION
(SELECT * FROM `clients_documents` WHERE `user_id` = 582)
ORDER BY date
porém esse SQL me retorna o seguinte erro:
Os comandos SELECT usados têm diferente número de colunas
como eu poderia contornar esse problema?
UNION serve para unir conjunto de dados similares vindo de consultas diferentes.
Para relacionar tabelas entre si a partir de uma chave primária/estrangeira é necessário utilizar alguma forma de JOIN tipo:
SELECT r.id, r.user_id, r.comments, r.comment_date, h.*, d.*
FROM realation_notes AS r
JOIN historic_attendance AS h ON r.user_id = h.user_id
JOIN clients_documents AS d ON r.user_id = d.user_id
WHERE r.user_id = 582
Olá!
O problema de usares o comando "Union" neste caso é que ele requer o mesmo número de colunas na união das tabelas.
Imagina que tens o seguinte esquema:
tabela A (colA, ColB, ColC)
tabela B (colA, ColD, ColE)
Para fazeres Union das duas tabelas precisas que elas tenham os mesmos campos.
Logo,
select colA, ColB, ColC, null, null from table A
union
select colA, null, null, colD, colE from table B
Podes, se quiseres, usar um alias nos valores Null para os dados ficarem mais "limpos".
Boa sorte!
| 41,166 | |
https://fr.wikipedia.org/wiki/Pierre%20de%20L%27Estoile | Wikipedia | Open Web | CC-By-SA | 2,023 | Pierre de L'Estoile | https://fr.wikipedia.org/w/index.php?title=Pierre de L'Estoile&action=history | French | Spoken | 585 | 1,005 | Pierre de L'Estoile, né à Paris en 1546 et mort le , est un mémorialiste et un collectionneur français.
Il a écrit un journal qui sert de chronique aux règnes d' et . Ce document constitue une source unique pour l'étude historique des guerres de religion.
Biographie
Venant de la grande bourgeoisie parisienne, Pierre de l'Estoile (ou de l'Étoile) appartient au cercle de la grande magistrature. Il eut pour précepteur Mathieu Béroalde, chez qui il connut l'écrivain protestant Agrippa d'Aubigné. C'est à Bourges qu'il étudia le droit (1565). Il exerça de 1566 à 1601 la charge d'audiencier à la chancellerie de France. Son emprisonnement en 1589 s'explique par le fait qu'il passait pour un proche des « Politiques ». Il vend sa charge en 1601.
Il fut inhumé dans l'église Saint-André-des-Arts à Paris.
Il est le père de Louis de L'Estoile et de Claude de L'Estoile, un des premiers membres de l'Académie française (1597-1672).
Œuvres
Pierre de L'Estoile a laissé des Mémoires-Journaux (1574-1611) qui sont un précieux témoignage sur les règnes de Henri III et Henri IV. Y sont intégrés d'autres textes (des sonnets, des pamphlets). Ces journaux publiés en 12 tomes ont été numérisés et sont consultables dans Gallica.
On trouve dans le tome 3 une estampe et une description sanglante de l'exécution de Marie Stuart, reine d'Écosse, le .
Ces journaux fournissent également des informations sur la crise de subsistance de 1586-1587 avec des informations météorologiques , des données sur l'évolution du prix du pain à Paris (cherté du blé) et surtout sur les conséquences sociales de cette crise : afflux de mendiants, procession de la châsse de sainte Geneviève pour faire cesser les pluies, augmentation des impôts (aumône spéciale) pour les bourgeois et émeutes, notamment celle du contre les boulangers accusés de vendre trop cher leur pain.
Ce Journal n'était pas destiné à la publication. On en a extrait le Journal de Henri III, publié en 1621 par Louis Servin, et en 1744 par Nicolas Lenglet Du Fresnoy ; et le Journal de Henri IV, paru à La Haye, 1741.
Registre-journal du règne de Henri III, édition établie par Madeleine Lazard et Gilbert Schrenck, Genève, Droz, collection « Textes littéraires français », 1992-2003, 6 volumes.
Journal de Henry III, Roy de France et de Pologne : ou Mémoires pour servir à l'histoire de France.
Registre-journal du règne de Henri IV. Tome I (1589-1591), édition établie par Gilbert Schrenck, Xavier Le Person et Volker Mecking, Genève, Droz, collection « Textes littéraires français », , 2011, 352 p.
Les Belles figures et drolleries de la Ligue, édition établie par Gilbert Schrenck, 2016.
Notes et références
Voir aussi
Bibliographie
Journal de l'Estoile pour le règne de Henri III : 1574-1589, présenté par Louis-Raymond Lefèvre, Gallimard, 1943.
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Margaret M. McGowan, « « La conversation de ma vie » : la voix de L'Estoile dans les Registres / Journaux », dans Travaux de Littérature offerts en hommage à Noémie Hepp, Paris, Publications de l'ADIREL / Les Belles Lettres, 1990, p. 249-259.
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Gilbert Schrenck, « Pierre de L'Estoile et Montaigne, ou la lecture en miettes », dans Esculape et Dionysos. Mélanges en l'honneur de Jean Céard, Genève, Droz, 2008, .
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Liens externes
Les gravures et pièces du recueil des belles Figures et Drolleries de la Ligue sont accessibles sur Gallica Anthologie via le mode recherche (taper L'Estoile dans le champ auteur)
Écrivain français du XVIe siècle
Naissance en 1546
Naissance à Paris
Décès en octobre 1611 | 46,997 |
https://stackoverflow.com/questions/15564084 | StackExchange | Open Web | CC-By-SA | 2,013 | Stack Exchange | Mani, https://stackoverflow.com/users/1654995 | Papiamento | Spoken | 327 | 630 | listbox doesnt appear on device
I am developing an app for windows phone and i had recent problem with the list box. This works fine on my emulator but when i deploy on my device, it doesn't show anything but blank. The list items are created as requested but binding doesn't carry out because each item when clicked doesn't carry information with it.
The code related to it is as follows
XAML
<ListBox Name="NotesListBox" Grid.Row="0" Margin="15,0" >
<ListBox.ItemsPanel>
<ItemsPanelTemplate>
<toolkit:WrapPanel Orientation="Horizontal" />
</ItemsPanelTemplate>
</ListBox.ItemsPanel>
<ListBox.ItemTemplate>
<DataTemplate>
<StackPanel Orientation="Vertical">
<local:NotesTile Name="TileNotes" Tap="NotesTile_Tap_1" Margin="10,10"/>
<TextBlock Text="{Binding FileName}" />
</StackPanel>
</DataTemplate>
</ListBox.ItemTemplate>
</ListBox>
C# code behind
List<Notes> source = new List<Notes>()
{
new Notes(){ Content="This is some text", FileName="one.txt", IsPasswordProtected=true},
new Notes(){ Content="Another text file", FileName="two.txt", IsPasswordProtected=false}
};
//this.DataContext = this;
this.NotesListBox.ItemsSource = source;
In same namespace i got the class as:
class Notes
{
public string Content { get; set; }
public string DateEdited { get; set; }
public string TimeEdited { get; set; }
public string FileName { get; set; }
public bool IsPasswordProtected { get; set; }
}
I got things working fine on emulator and even for devices made for windows phone 8. My app got rejected on marketplace for the same reason.
EDIT
The Binding class should always be public. Making the class Notes as public would resolve this issue.
Hey ya!!! I just found the answer as to modify the class as public and it worked well... Can anyone tell me some reason behind it????
try with setting this.DataContext = this; in the constructor of the xaml.cs before InitilizeComponent() and let us know if its work.
This didn't solve my problem... The app still doesn't show any tile contents. Well this works fine on windows phone 8 as what it is said to me when i submitted to dev center. So my emulator is 8, so i think thats the reason why it is working fine on my emulator. Whats the problem in 7?
| 44,613 |
https://stackoverflow.com/questions/16460151 | StackExchange | Open Web | CC-By-SA | 2,013 | Stack Exchange | Avinash Kumar Pankaj, Pragnani, Varun Nayyar, https://stackoverflow.com/users/1727092, https://stackoverflow.com/users/1885283, https://stackoverflow.com/users/2300237 | Norwegian Nynorsk | Spoken | 480 | 1,635 | make listview clickable setOnItemclicklistener method not working
I made a listView with text and image.Now i want to make its items clickable.I used setOnitemclickListener method but it's not working ..
i think i am doing something wrong
here is my code....
Adapter.java
public class Adapter extends BaseAdapter {
private static final String TAG = "Adapter";
private Activity mActivity;
public ArrayList<Data> mObjects;
static class ViewHolder {
ImageView icon;
TextView title;
TextView name;
TextView review;
DownloadImageTask mTask;
// DownloadImageTask1 mTask1;
// ImageView photo;
}
public Adapter(Activity activity, ArrayList<Data> mObjects) {
this.mActivity = (Activity) activity;
this.mObjects = mObjects;
}
public void setObjects(ArrayList<Data> mObjects) {
this.mObjects = mObjects;
}
@Override
public View getView(int position, View convertView, ViewGroup parent) {
Data item = mObjects.get(position);
View rowView = convertView;
if (rowView == null) {
LayoutInflater inflater = mActivity.getLayoutInflater();
rowView = inflater.inflate(R.layout.item, parent, false);
ViewHolder viewHolder = new ViewHolder();
viewHolder.icon = (ImageView) rowView.findViewById(R.id.image);
// viewHolder.photo = (ImageView) rowView.findViewById(R.id.photo);
viewHolder.title = (TextView) rowView.findViewById(R.id.title);
viewHolder.name = (TextView) rowView.findViewById(R.id.name);
viewHolder.review = (TextView) rowView.findViewById(R.id.status);
rowView.setTag(viewHolder);
}
ViewHolder holder = (ViewHolder) rowView.getTag();
holder.title.setText(item.getmTitle());
holder.name.setText(item.getmConcatinate());
holder.review.setText(item.getmreview());
holder.icon.setBackgroundResource(R.drawable.ic_ab);
// holder.photo.setBackgroundResource(0);
holder.mTask = new DownloadImageTask(item.getmImageUrl(), holder.icon);
if (!holder.mTask.isCancelled()) {
holder.mTask.execute();
}
// holder.mTask1 = new DownloadImageTask1(item.getmImageUrl1(), holder.photo);
// if (!holder.mTask1.isCancelled()) {
// holder.mTask1.execute();
// }
return rowView;
}
@Override
public int getCount() {
return (this.mObjects.size());
}
@Override
public Object getItem(int position) {
return (this.mObjects.get(position));
}
@Override
public long getItemId(int position) {
return (position);
}
public AbsListView.RecyclerListener mRecyclerListener = new RecyclerListener( ){
public void onMovedToScrapHeap(View view) {
ViewHolder viewHolder = (ViewHolder) view.getTag();
DownloadImageTask imagetask = viewHolder.mTask;
// DownloadImageTask1 imagetask1 = viewHolder.mTask1;
if (imagetask != null) {
imagetask.cancel(true);
}
// if (imagetask1 != null) {
// // imagetask1.cancel(true);
// }
}
};
}
using that content in main class
@Override
protected void onCreate(Bundle savedInstanceState) {
StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder()
.detectAll().penaltyLog().build();
StrictMode.setThreadPolicy(policy);
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_profile3);
name = (TextView) findViewById(R.id.textView1);
bmImage2 = (ImageView) findViewById(R.id.imageView1);
address = (TextView) findViewById(R.id.textView2);
gender = (TextView) findViewById(R.id.textView3);
loyalitypoints = (TextView) findViewById(R.id.textView7);
followers = (TextView) findViewById(R.id.textView8);
following = (TextView) findViewById(R.id.textView9);
// list13 = new ArrayList<HashMap<String, Object>>();
mListView = (ListView) findViewById(android.R.id.list);
mListView.setClickable(true);
// mListView=(ListView)findViewById(R.id.list);
mAdapter = new Adapter(this, mSource);
mListView.setAdapter(mAdapter);
Log.w("Parsing JSON Data", "Before Item click");
mListView.setRecyclerListener(mAdapter.mRecyclerListener);
mListView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
public void onItemClick(AdapterView<?> arg0, View arg1, int position,long arg3) {
Log.w("Parsing JSON Data", "After Item click");
}
});
Try setting android:focusable=false for views in your ListRow.
@Pragnani still not working
u will have to make oclicklistener in your adapter only.
viewHolder.name.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View arg0) {
//perform action
}
});
will u please make me explain whole scenario
by using above method i can click on name.Now i want when i click on name it opens a new activity and set the content of name to new activity..I am using the code present in this link
http://stackoverflow.com/questions/14701747/how-to-parse-sub-jsonarray-and-display-image
Try this **
ListView.setOnItemClickListener(new OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View arg1, int position,long arg3)
{
// your action to be performed
}
});
}
| 30,534 |
https://stackoverflow.com/questions/75272708 | StackExchange | Open Web | CC-By-SA | 2,023 | Stack Exchange | Manisha Biswas, https://stackoverflow.com/users/7904647 | English | Spoken | 382 | 977 | Download entire webpage as HTML (including the HTML assets) without save as pop up using Selenium and Python
I am trying to scrape a website and download all the webpages as .html files (including all the HTML assets) so that the locally downloaded page opens just like the same in the server.
Currently using Selenium, Chrome Webdriver, and Python.
Approach:
I tried updating the prefs of the chrome browser. And then login into the website. After logging in I want to download the webpage similarly we do download by clicking ctrl + s from the keyboard.
Below code opens the desired page I want to download but does not disable Windows's save as a pop-up and neither downloads the page to the specified path.
from selenium import webdriver
import pyautogui
chrome_options = webdriver.ChromeOptions()
preferences = {
"download.default_directory":"C:\\Users\\pathtodir",
"download.prompt_for_download": False,
"download.directory_upgrade": True,
"safebrowsing.enabled": True
}
chrome_options.add_experimental_option("prefs", preferences)
driver = webdriver.Chrome(options=chrome_options)
driver.get(***URL to the website***)
driver.find_element("xpath", '//*[@id="id_username"]').send_keys('username')
driver.find_element("xpath", '//*[@id="id_password"]').send_keys('password')
driver.find_element("xpath", '//*[@id="datagrid-0"]/div[2]/div[1]/div[1]/table/tbody/tr[1]/td[2]/a').click()
pyautogui.hotkey('ctrl', 's')
pyautogui.typewrite('hello1' + '.html')
pyautogui.hotkey('enter')
Can somebody please help me to understand what I am doing wrong? Please suggest if there is any other alternative library that can be used in python.
To save a page first obtain the page source behind the webpage with the help of the page_source method.
Then open a file with a particular encoding with the codecs.open method. The file has to be opened in the write mode represented by w and encoding type as utf−8. Then use the write method to write the content obtained from the page_source method.
from selenium import webdriver
import codecs
driver = webdriver.Chrome(executable_path="path to chromedriver.exe")
driver.implicitly_wait(0.5)
driver.get(***URL to the website***)
h = driver.page_source
n=os.path.join("C:\ANYPATH","Page.html")
f = codecs.open(n, "w", "utf−8")
f.write(h)
driver.quit()
Hi Manish, thanks for responding. I tried out your code. However, it is downloading the source code for me but not the HTML assets(CSS, JS file). So, when I open the downloaded file in the browser, it is not the same as we view it on the server.
I was able to fix the issue, the problem was that my code quit before the browser was able to download the file. Adding time.sleep() fixed it.
Updated code:
from selenium import webdriver
import pyautogui
driver.get(***URL to the website***)
driver.find_element("xpath", '//*[@id="id_username"]').send_keys('username')
driver.find_element("xpath", '//*[@id="id_password"]').send_keys('password')
driver.find_element("xpath", '//*[@id="datagrid-0"]/div[2]/div[1]/div[1]/table/tbody/tr[1]/td[2]/a').click()
FILE_NAME = r'C:\ANYPATH\Page.html'
pyautogui.typewrite(FILE_NAME)
pyautogui.press('enter')
time.sleep(10)
driver.quit()
| 28,842 |
https://en.wikipedia.org/wiki/Permutation%20graph | Wikipedia | Open Web | CC-By-SA | 2,023 | Permutation graph | https://en.wikipedia.org/w/index.php?title=Permutation graph&action=history | English | Spoken | 608 | 800 | In the mathematical field of graph theory, a permutation graph is a graph whose vertices represent the elements of a permutation, and whose edges represent pairs of elements that are reversed by the permutation. Permutation graphs may also be defined geometrically, as the intersection graphs of line segments whose endpoints lie on two parallel lines. Different permutations may give rise to the same permutation graph; a given graph has a unique representation (up to permutation symmetry) if it is prime with respect to the modular decomposition.
Definition and characterization
If is any permutation of the numbers from to , then one may define a permutation graph from in which there are vertices , and in which there is an edge for any two indices for which appears before in . That is, two indices and determine an edge in the permutation graph exactly when they determine an inversion in the permutation.
Given a permutation , one may also determine a set of line segments with endpoints and , such that . The endpoints of these segments lie on the two parallel lines and , and two segments have a non-empty intersection if and only if they correspond to an inversion in the permutation. Thus, the permutation graph of coincides with the intersection graph of the segments. For every two parallel lines, and every finite set of line segments with endpoints on both lines, the intersection graph of the segments is a permutation graph; in the case that the segment endpoints are all distinct, a permutation for which it is the permutation graph may be given by numbering the segments on one of the two lines in consecutive order, and reading off these numbers in the order that the segment endpoints appear on the other line.
Permutation graphs have several other equivalent characterizations:
A graph is a permutation graph if and only if is a circle graph that admits an equator, i.e., an additional chord that intersects every other chord.
A graph is a permutation graph if and only if both and its complement are comparability graphs.
A graph is a permutation graph if and only if it is the comparability graph of a partially ordered set that has order dimension at most two.
If a graph is a permutation graph, so is its complement. A permutation that represents the complement of may be obtained by reversing the permutation representing .
Efficient algorithms
It is possible to test whether a given graph is a permutation graph, and if so construct a permutation representing it, in linear time.
As a subclass of the perfect graphs, many problems that are NP-complete for arbitrary graphs may be solved efficiently for permutation graphs. For instance:
the largest clique in a permutation graph corresponds to the longest decreasing subsequence in the permutation defining the graph, so the clique problem may be solved in polynomial time for permutation graphs by using a longest decreasing subsequence algorithm.
likewise, an increasing subsequence in a permutation corresponds to an independent set of the same size in the corresponding permutation graph.
the treewidth and pathwidth of permutation graphs can be computed in polynomial time; these algorithms exploit the fact that the number of inclusion minimal vertex separators in a permutation graph is polynomial in the size of the graph.
Relation to other graph classes
Permutation graphs are a special case of circle graphs, comparability graphs, the complements of comparability graphs, and trapezoid graphs.
The subclasses of the permutation graphs include the bipartite permutation graphs (characterized by ) and the cographs.
Notes
References
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External links
Intersection classes of graphs
Perfect graphs
Geometric graphs | 12,687 |
https://pt.wikipedia.org/wiki/Parkia%20parvifoliola | Wikipedia | Open Web | CC-By-SA | 2,023 | Parkia parvifoliola | https://pt.wikipedia.org/w/index.php?title=Parkia parvifoliola&action=history | Portuguese | Spoken | 38 | 75 | Parkia parvifoliola é uma espécie vegetal da família Fabaceae.
Apenas pode ser encontrada em Palau.
Referências
World Conservation Monitoring Centre 1998. Parkia parvifoliola. 2006 IUCN Red List of Threatened Species. Dados de 10 de Julho de 2007.
Parkia | 23,231 |
https://stackoverflow.com/questions/62242572 | StackExchange | Open Web | CC-By-SA | 2,020 | Stack Exchange | El_Loco, https://stackoverflow.com/users/2225895 | English | Spoken | 331 | 412 | Create anonymous highscorelistor in unity
I have simple game on Google Play and Apple App Store that I have created in Unity. Now I am thinking about creating a global high score, so if you are in top 10 for example you can enter a made up user name.
The thing is I do not want the users to sign in since I want the game to be easy to access. One user can also enter a new username each time a record is beaten. So you can for example have the first three positions.
In principle, the app should read the high score from database and one should be able to write to it with {userID, new_score}, but no data is connected to any particular user. I am used to work with Firebase, what would be the easiest way to achieve this?
Thanks,
Erik
You could try having a read and create only set of records, where write requires a valid anonymous auth key. I haven't tried that setup with Anonymous only auth before though.
Otherwise you could also try setting up a cloud function to validate and write the high score records.
Well, the fact that you don't have data connected to any particular user do not means that you can't store the highscores. Cause one user can beat himself.
So if your database (I don't know if it's realtime or firestore) knows which are the top scores, when a user (indpendently of his ID) beats the score threshold, you can give him the option to put a name on that highscore, and update your database with this new score.
I will also recommend to bind users with data, his highscore or the name it's using to register the highscore, but you can achieve the same result without it.
Yes, but I was also thinking about the security rules for firebase. They usually complain if you let anyone write to the database. Maybe I was unclear with that thought.
| 5,387 |
https://ca.wikipedia.org/wiki/Novi%20Ligure | Wikipedia | Open Web | CC-By-SA | 2,023 | Novi Ligure | https://ca.wikipedia.org/w/index.php?title=Novi Ligure&action=history | Catalan | Spoken | 24 | 58 | Novi Ligure és un municipi italià, situat a la regió del Piemont i a la província d'Alessandria. L'any 2006 tenia 28.370 habitants.
Municipis d'Alessandria | 12,482 |
https://war.wikipedia.org/wiki/Aphytis%20nigripes | Wikipedia | Open Web | CC-By-SA | 2,023 | Aphytis nigripes | https://war.wikipedia.org/w/index.php?title=Aphytis nigripes&action=history | Waray | Spoken | 36 | 65 | An Aphytis nigripes in uska species han Hymenoptera nga syahan ginhulagway ni Compere hadton 1936. An Aphytis nigripes in nahilalakip ha genus nga Aphytis, ngan familia nga Aphelinidae. Waray hini subspecies nga nakalista.
Mga kasarigan
Aphytis | 39,022 |
https://de.wikipedia.org/wiki/Arcfox%20Alpha-S | Wikipedia | Open Web | CC-By-SA | 2,023 | Arcfox Alpha-S | https://de.wikipedia.org/w/index.php?title=Arcfox Alpha-S&action=history | German | Spoken | 299 | 596 | Der Arcfox Alpha-S ist eine batterieelektrisch angetriebene Kombilimousine der zum BAIC-Konzern gehörenden Automarke Arcfox.
Geschichte
Das Fahrzeug wurde nach dem SUV Alpha-T als zweites Modell der Marke im Januar 2021 offiziell vorgestellt. Auf dem chinesischen Heimatmarkt kam der Alpha-S im April 2021 in den Handel. Als Konkurrenzmodelle werden unter anderem der BYD Han oder der Xpeng P7 genannt.
Im Juni 2022 wurde das Fahrzeug genauso wie der Alpha-T als limitierte Version Le Petit Prince vorgestellt. Beide zeichnen sich durch diverse gestalterische Elemente des kleinen Prinzen aus.
Technik
Das Fahrzeug wurde gemeinsam von BAIC und Magna International entwickelt. Hinsichtlich der Technik wurde die Entwicklung von Huawei unterstützt. Die Fließhecklimousine soll von Anfang an in bestimmten Situationen nach Autonomiestufe 3 fahren können, wenn das Ausstattungspaket HI geordert wurde. Eine neue Generation von Lidar könnte später auch die Autonomiestufe 4 ermöglichen.
Technische Daten
Wie der Alpha-T ist der Alpha-S ist mit Hinterrad- oder Allradantrieb erhältlich. Die heckangetriebene Variante hat einen Elektromotor an der Hinterachse, die maximale Leistung wird mit 160 kW (218 PS) angegeben. Die Version mit Allradantrieb hat einen zweiten Elektromotor an der Vorderachse. Die Leistung verdoppelt sich dadurch auf 320 kW (435 PS). Mit dem Ausstattungspaket HI, das nur für die Allradversion verfügbar ist, wird die maximale Leistung mit 473 kW (643 PS) angegeben.
Für den Wagen steht ein Lithium-Ionen-Akkumulator mit drei Größen zur Wahl. Der kleinste hat einen Energieinhalt von 67,3 kWh, der größte 93,6 kWh. Während für die hinterradgetriebene Version der kleinste und der größte zur Auswahl steht, ist die Allradversion nur mit den beiden größeren Akkus erhältlich. Die Reichweite nach NEFZ wird bei den hinterradgetriebenen Modellen mit 525 km bzw. 708 km angegeben. Die allradgetriebene Version erreicht 603 km bzw. 500 km. Der Strömungswiderstandskoeffizient cw beträgt 0,26.
Einzelnachweise
Weblinks
Offizielle Website (chinesisch)
Beijing Automotive Group
Kombilimousine
Elektroautomodell | 31,075 |
https://ceb.wikipedia.org/wiki/W%C4%81d%C4%AB%20Ma%E1%B8%A9lah | Wikipedia | Open Web | CC-By-SA | 2,023 | Wādī Maḩlah | https://ceb.wikipedia.org/w/index.php?title=Wādī Maḩlah&action=history | Cebuano | Spoken | 86 | 150 | Wadi ang Wādī Maḩlah sa Yemen. Nahimutang ni sa distrito sa Bani Sa'd ug lalawigan sa Al Maḩwīt, sa kasadpang bahin sa nasod, km sa kasadpan sa Sanaa ang ulohan sa nasod.
Ang klima init nga kamadan. Ang kasarangang giiniton °C. Ang kinainitan nga bulan Oktubre, sa °C, ug ang kinabugnawan Enero, sa °C. Ang kasarangang pag-ulan milimetro matag tuig. Ang kinabasaan nga bulan Agosto, sa milimetro nga ulan, ug ang kinaugahan Pebrero, sa milimetro.
Ang mga gi basihan niini
Mga suba sa Al Maḩwīt (lalawigan) | 39,248 |
https://fr.wikipedia.org/wiki/Roman%C3%A8che%20%28homonymie%29 | Wikipedia | Open Web | CC-By-SA | 2,023 | Romanèche (homonymie) | https://fr.wikipedia.org/w/index.php?title=Romanèche (homonymie)&action=history | French | Spoken | 18 | 57 | Romanèche-Thorins, commune de Saône-et-Loire.
Romanèche, ancienne commune de l'Ain, intégrée à Hautecourt-Romanèche.
Romanèche, hameau de Montluel, dans l'Ain. | 47,833 |
https://es.wikipedia.org/wiki/Guerra%20de%20las%20Arenas | Wikipedia | Open Web | CC-By-SA | 2,023 | Guerra de las Arenas | https://es.wikipedia.org/w/index.php?title=Guerra de las Arenas&action=history | Spanish | Spoken | 2,729 | 4,621 | La Guerra de las Arenas (en árabe: حرب الرمال, en francés: Guerre des Sables) fue un conflicto militar entre Marruecos y Argelia que se inició en octubre de 1963. Tras varios meses de incidentes fronterizos, la guerra abierta estalló en la región argelina de Tinduf y Hassi Beida y posteriormente se extendió a Figuig en Marruecos. La lucha entre ambos países finalizó el 5 de noviembre, logrando así la Organización para la Unidad Africana un alto el fuego el 20 de febrero de 1964, dejando las fronteras sin cambios.
Antecedentes
Varios son los factores que contribuyeron al estallido del conflicto entre Marruecos y Argelia, incluida la falta de un acuerdo de demarcación precisa de las fronteras entre ambos países. El Tratado de Lalla Mughnia el 18 de marzo de 1845, que establece la frontera entre Marruecos y la Argelia francesa, estipula que es un «territorio seco, inhabitable y su delineación es superflúa». Esta frontera dibujada representaba solo 150 km, comenzando desde el Mar Mediterráneo en el norte hasta la ciudad marroquí de Figuig al sur. El resto son frontera delimitadas por las diferentes tribus pertenecientes a cada uno de los países.
Tras la ocupación de Marruecos en 1912 , la administración francesa decidió estabilizar las fronteras entre los dos países, sin embargo estas fronteras variaban de un mapa a otro debido a su mala definición (la Línea Varnier 1912 y la Línea Trincke 1938). El descubrimiento de minas de hierro y manganeso en la región hizo que Francia decidiera en 1950 examinar la demarcación de las fronteras e incluir Tinduf y Colomb Béchar en las provincias francesas de Argelia.
Desde la independencia en 1956, Marruecos reclama la soberanía sobre estos territorios, así como sobre otros, siempre bajo el prisma no histórico ni mucho menos político, si no de la ambición del nacionalismo latente y su grial "Gran Marruecos". Para poner fin al apoyo de Marruecos al Frente de Liberación Nacional Argelino, Francia propone el principio de "restitución" de estos territorios contra el establecimiento de una "organización común de las regiones del Sáhara" (OCRS), encargada de explotar los depósitos mineros del Sáhara descubiertos recientemente, pero el rey Mohammed V rechazó la oferta francesa, subrayando que el problema fronterizo se resolvería con las autoridades argelinas tras la independencia de Argelia de Francia (años más tarde se vería que no sería así). El objetivo de Francia con esta propuesta era detener el apoyo de Marruecos a la revolución argelina, ya que proporcionaba armas a los revolucionarios y albergaba a sus líderes en la ciudad de Uchda, y de paso, mantener un cierto control del Protectorado Francés de Marruecos «anexionando» mediante un redibujo de las fronteras los nuevos yacimientos minerales encontrados en lado argelino.
El 13 de marzo de 1963, el rey Hassan II, sucesor de Mohamed V muerto en 1961, llevó a cabo su primera visita a Argelia tras su independencia, donde le recordó a su homólogo argelino, Ben Bella, el acuerdo firmado con el gobierno interino argelino sobre el estado de las fronteras entre los dos países creados por el colonialismo francés. Tras la visita del Rey de Marruecos a Argelia, estalló una guerra mediática entre ambos países, ya que Argelia declaró que Marruecos tiene ambiciones expansionistas en la región, mientras que Marruecos vio las acusaciones argelinas respaldadas por Egipto liderado por Gamal Abdel Nasser, lo que suponía una amenaza de la unidad del país. Gamal Abdel Nasser, el presidente egipcio de la época, clasificó a las monarquías árabes como regímenes reaccionarios y apoyó los movimientos revolucionarios contra ellos, saltando así la chispa que faltaba.
Estallido de la guerra
Las tensiones entre Marruecos y Argelia están aumentando gradualmente, y ninguna de las partes quiere correr el riesgo de retroceder. Desde 1962, Tinduf es la sede de los incidentes: en el referéndum de independencia, los habitantes indican en su boletín: "SÍ a la independencia, pero somos marroquíes". Los
dos países comienzan a reforzar su presencia militar a lo largo de la frontera y la prensa comienza a difundir abusos: los guardias fronterizos argelinos expulsan a 50 marroquíes asentados en la zona de Colomb Bechar. Como represalia, los marroquíes hicieron lo mismo con unos 60 argelinos en la zona de
Uchda. La situación se fue complicando, llegando el Gobierno argelino a declarar el cierre de la frontera en las zonas de Tinduf y Colomb Bechar, así como a anunciar la nacionalización de unos marroquíes.
La primera fase de la guerra se caracteriza por la lucha por el control de los puestos fronterizos de Hassi Beida y Tinyub. Durante el mes de septiembre suceden hechos entre ambos países que provocan que las relaciones se enrarezcan gravemente. Ben Bella, ve su posición amenazada por una insurrección originada en la zona de Cabilia y liderada por el Frente de las Fuerzas Socialistas (FFS) de Husein Aït Ahmed. De inmediato, Ben Bella en un discurso pronunciado a raíz de la insurrección cabileña el 30 de septiembre acusa a Hassan II de aportar tropas en la frontera como apoyo a estos rebeldes, sin embargo, estas acusaciones serán negadas por el gobierno marroquí. Por su parte, la prensa marroquí había acusado al presidente argelino de apoyar el supuesto complot de julio de 1963 por el que se detuvieron a numerosos miembros de la UNFP y la prensa argelina, a su vez, atacó al régimen monárquico.
A principios del mes de octubre ambos países refuerzan su presencia militar en la frontera. Ben Bella alegando un conflicto inminente invoca el 3 de octubre una cláusula constitucional que le dota de poderes absolutos ante una emergencia nacional. Ante la creciente tensión, Argelia y Marruecos deciden reunirse y sus respectivos ministros de Asuntos Exteriores, Abdelaziz Buteflika y Ahmed Guedira, mantienen un encuentro en Uchda (Marruecos) el 5 de octubre, acompañados de los embajadores Mohamed Aouad y Saad Dahlab, así como del representante especial de Ben Bella para asuntos intermagrebíes, Muhammad Yazid. En dicho encuentro ambas partes acordaron dar punto final a las campañas de prensa hostiles, permitir a las comunidades fronterizas llevar una vida normal de nuevo y no hacer nada que agravase la situación. Por último, se concertó una nueva cita en Tremecén el 10 de octubre para continuar las conversaciones.
El 7 de octubre el ministro de defensa argelino anuncia que Marruecos ha aceptado retirarse de la frontera, de manera que Ben Bella puede concentrar sus fuerzas en reprimir la rebelión de Cabilia que ha estallado una semana antes, sin embargo, al día siguiente, Argelia denuncia un incidente fronterizo en la zona de Colomb Bechar provocado por los marroquíes y que deja diez bajas. Estos hechos provocan una serie de idas y venidas entre Rabat y Argel de emisarios oficiales.
El 14 de octubre Argelia denuncia claramente que las fuerzas marroquíes invaden territorio argelino. Según los argelinos, se trata de una invasión por parte de miles de soldados apoyados por tanques y aviones, mientras que fuentes militares marroquíes insisten en negar que se trate de la invasión de
territorio argelino, sino que se trata de la recuperación de dos puestos en territorio marroquí. Esta divergencia tan patente entre las versiones marroquí y argelina de lo que está ocurriendo se debe a que los mapas de la zona publicados en los años 50 lo sitúan dentro de Marruecos, pero que en los últimos
dibujados por Francia el límite está en el río Draa, más al norte. Es decir, los mapas fueron modificados después de la independencia de Marruecos y antes de la de Argelia.
El 18 de octubre se abre un nuevo frente en el conflicto fronterizo al atacar Argelia la ciudad de Ich, una localidad al norte de Figuig, lejos de la zona de Hassi Beida y Tinyub. El punto de inflexión en el conflicto tuvo lugar el 20 de octubre, el mismo día en que la Liga Árabe y Gamal Abdel Nasser toman la iniciativa de intentar mediar en el conflicto, salta la noticia de que Marruecos ha capturado un helicóptero argelino con nueve oficiales, de los cuales cuatro son egipcios y entre ellos se encontraba el futuro presidente Hosni Mubarak. La intervención de la RAU a partir de entonces es explotada por las autoridades marroquíes como una prueba determinante de que la actitud de Argelia, con ayuda militar exterior, es la de derrocar al monarca alauí.
El día 23, soldados marroquíes atacaron el puesto de Hassi-Taghucht, a 90 km al sur de Tauz, mientras que unidades argelinas intentaron cercar el puesto de Usada, a 10 km de Zedgu. Los dos ataques fueron rechazados. Por su parte, el ejército argelino lanza su último contraataque contra Hassi Beida con material pesado y al ser rechazado, se retira dejando casi un centenar de muertos sobre el terreno. El 27 de octubre se anunció en Marrakech el cese del fuego mientras, en la depresión de Hassi Beida, las tropas marroquíes ocupaban todas las crestas circundantes. Se anunció entonces, una conferencia de paz en Bamako entre Ben Bella y Hassan II, con la mediación de Haile Selassie y el presidente Modibo Keita de Mali. La conferencia tuvo lugar el día 30 de octubre y puso fin oficialmente a las hostilidades.
Internacionalización del conflicto
Al poco tiempo de estallar el conflicto surgen varias iniciativas de mediación, consistentes básicamente en la celebración de una reunión de los jefes de Estado implicados, acompañados de otros dirigentes que realizarían la labor de propiciar la negociación de un alto el fuego. Entre otras, destacan las propuestas de la Liga Árabe, Túnez y Libia serán consideradas, pero finalmente descartadas. La principal razón por la que estas iniciativas no se consolidan es que alguna de las dos partes, ya sea Marruecos o Argelia, no la considera imparcial. Así, los marroquíes no estiman como neutral a una Liga Árabe demasiado “nasserista”, ni los argelinos ven con buenos ojos que el presidente Burguiba medie en un asunto fronterizo muy similar a otro que Túnez tiene pendiente con Argelia.
Una vez empezado el conflicto, el 19 de octubre la Liga Árabe se reúne de emergencia para adoptar una resolución llamando al alto el fuego entre ambas partes, lo que se hace llegar a Ben Bella y Hassan II por telégrafo. Una segunda reunión tiene lugar al día siguiente en el que se pide la retirada a posiciones anteriores al estallido del conflicto y se crea una comisión de mediación compuesta por Túnez, Líbano, Libia, la RAU, el entonces presidente de la organización y el secretario general de la Liga Árabe. Además, se solicita a las dos partes que faciliten la labor de la comisión en todo lo que sea posible, incluyendo el cese de las hostilidades a nivel mediático. Gamal Abdel Nasser interviene de manera directa liderando la iniciativa de la Liga Árabe y enviando una misiva a Hassan II el 20 de octubre en la que le urge a detener el conflicto. En dicha carta, después de reprochar el ataque marroquí y justificar el apoyo de la RAU a Argelia, el rais repite en lo esencial lo decidido por la Liga, es decir, propone el cese de hostilidades, el retorno a posiciones anteriores al 8 de octubre y la celebración de una reunión de los jefes de Estado norteafricanos.
Sin embargo, la labor más persistente en favor de un arreglo pacífico es la de Haile Selassie, emperador de Etiopía y padre fundador de la Organización para la Unidad Africana (OUA). El emperador enseguida trató de promover una negociación y emprende una intensa actividad diplomática, desafortunadamente, su labor de mediación se verá obstaculizada por las posiciones enfrentadas de marroquíes y argelinos, que se muestran irreconciliables. Como condición previa a un alto el fuego, los marroquíes solicitaban que Argelia reconociese que el problema fronterizo existía y accediese a negociar la frontera. Por su parte, los argelinos rechazaban tal extremo, escusándose en la Carta fundacional de la OUA que defiende Internacionalización del conflicto el statu quo en materia de fronteras, pues aunque éstas eran un legado colonial se consideraban un mal menor en prevención de conflictos regionales, y exigían que Marruecos se retirase sin condiciones de lo que consideraban territorio argelino, sí no lo hacía, Argelia no aceptaría un alto el fuego.
Por otra parte, Habib Burguiba fue especialmente insistente con su propuesta de mediación, que fundamentalmente consistía en la celebración de una conferencia de ministros de Asuntos Exteriores de los Estados magrebíes el 28 de octubre y, posteriormente, una cumbre entre los tres Jefes de Estado en Bizerta. El gobierno argelino, además, consideraba que Túnez era parte interesada en este conflicto, pues como Marruecos reivindicaba la soberanía de ciertos territorios del Sáhara que en aquel momento estaban bajo el control de Argelia. Finalmente, los tunecinos propusiesen abandonar su reclamación si se conseguía alcanzar un acuerdo entre los países magrebíes para realizar una explotación conjunta de los recursos del Sáhara.
El secretario general de la ONU, U-Thant, era partidario de evitar la intervención de este organismo internacional a toda costa, pues éste argumentaba que mientras existiesen iniciativas regionales, ya fuesen magrebíes o africanas, prefería no actuar. Por su parte, Estados Unidos también compartía la
valoración que hacía U-Thant de los riesgos de involucrar a la organización internacional, pues temían que llevar la cuestión fronteriza a la ONU, solo conseguiría internacionalizar el conflicto, sometiéndolo a la dinámica bipolar de tensiones entre Bloques. Al lado opuesto de estas consideraciones se encontraban Marruecos y España, que sí veían necesaria la intervención de la ONU, aunque no se atreverían a solicitarlo oficialmente sin el apoyo de los países occidentales con peso en el Consejo de Seguridad.
Francia optó por intentar mantenerse neutral, pues al margen de otras razones, tenía importantes intereses en Argelia. La política del gobierno francés, por tanto, fue la de esforzarse por aparentar una total neutralidad a la vez que intentaba ejercer su influencia para que la situación no se agravase y terminase minando sus intereses en el Norte de África. Por todo ello, Francia prefería la discreción a la crítica abierta al régimen revolucionario argelino.
Uno de los factores que hicieron saltar las alarmas en las cancillerías occidentales fue la intervención de Cuba en este conflicto. Semejante intervención suponía el riesgo de internacionalización del conflicto y su agravamiento en el marco bipolar propio de la Guerra Fría. El apoyo cubano al FLN argelino
comienza en octubre de 1961, cuando un representante del gobierno cubano se entrevistó con el presidente del Gobierno Provisional de la República Argelina en Túnez y acordaron lo que se convertiría en la primera ayuda militar cubana a un país africano, y que consistió en armamento de manufactura estadounidense. Las armas se desembarcaron en Casablanca y fueron transportadas hasta Uchda. Enseguida comenzaron también relaciones de cooperación argelo-cubana en otros ámbitos como el envío de médicos cubanos, la atención médica a heridos en Cuba, la acogida de niños huérfanos argelinos, etc. Así, cuando estalló la Guerra de las Arenas, días después de la toma de Hassi Beida y Tinyub por los marroquíes, Ben Bella pidió ayuda a Cuba, que fue concedida en apenas unas horas por Fidel Castro. La reacción de Marruecos ante esto es la de romper sus relaciones diplomáticas con Cuba el 31 de octubre y llamar a consultas a sus embajadores en Siria y en la RAU.
En cuanto a España, el gobierno empezó a mostrar preocupación por la expansión de regímenes e ideologías revolucionarias y el consiguiente potencial de riesgo de desestabilización en la región norteafricana. El gobierno español mantuvo su postura de que solo actuaría en concierto con las naciones occidentales, sobre todo Francia y Estados Unidos y que, de momento, pedía formalmente a ambas partes en conflicto que negociasen un arreglo pacífico.
Véase también
Frontera entre Argelia y Marruecos
Relaciones Argelia-Marruecos
Conflicto del Sahara Occidental - Primera batalla de Amgala
Referencias
Bibliografía
GIL BENUMEYA, Rodolfo (1963). “El conflicto fronterizo argelino-marroquí, antes y después de la Conferencia de Addis-Abeba”. Revista Política Internacional, n.º 70, p. 133.
LAÍNEZ, Fernando María (1976). “Marruecos-Argelia, primer asalto”. Historia 16, n.º1, pp.31-36.
TORRES GARCÍA, Ana (2013). “La frontera terrestre argelino-marroquí: de herencia colonial a instrumento de presión”. Historia Actual Online, n.º31, pp.7-19.
TORRES GARCÍA, Ana (2012). La Guerra de las Arenas. Conflicto entre Marruecos y Argelia durante la Guerra Fría (1963). Barcelona: Bellaterra.
Enlaces externos
Emisión de la Tv francesa sobre el conflicto
Arenas
Guerras de Argelia
Guerras de Marruecos
Intervenciones militares de Cuba
Relaciones Argelia-Marruecos
Conflictos en 1963
Conflictos en 1964
Argelia en 1963
Argelia en 1964
Marruecos en 1963
Marruecos en 1964 | 6,821 |
https://pl.wikipedia.org/wiki/Fania%20Lewando | Wikipedia | Open Web | CC-By-SA | 2,023 | Fania Lewando | https://pl.wikipedia.org/w/index.php?title=Fania Lewando&action=history | Polish | Spoken | 380 | 995 | Fania Lewando z domu Fiszelewicz, jid. פאַנני לעװאַנדאָ (ur. 1887/1888 we Włocławku, zm. 1941) – polsko-żydowska restauratorka, autorka książki kucharskiej dla wegetarian pod tytułem Wegetariańsko-dietetyczna książka kucharska. 400 przepisów przygotowanych wyłącznie z warzyw.
Urodziła się we Włocławku jako jedno z sześciorga dzieci sprzedawcy ryb Haima Efraima Fiszelewicza oraz Estery Malki z domu Stulzaft; źródła jako datę urodzenia Fani Lewando podają 1887 lub 1888 rok. Wyszła za mąż za pochodzącego z terenu Białorusi Lazara Lewando, który zajmował się handlem jajami. Małżeństwo planowało emigrację do USA, do której nie doszło, z powodu stanu zdrowia Lazara. Niegojące się rany jego nogi uniemożliwiły uzyskanie wizy.
Lewandowie w latach 20. XX w. osiedlili się w Wilnie, gdzie prowadzili (Lazar jako właściciel, Fania jako szefowa kuchni) restaurację „Dietojarska Jadłodajnia” przy ul. Niemieckiej 14 (obecnie lit. Vokiečių gatvė). W lokalu gościli artyści – literaci i plastycy (w tym Marc Chagall), którym oferowano wegetariańskie potrawy kuchni Żydów aszkenazyjskich. Fania Lewando założyła i prowadziła w Wilnie szkołę gotowania dla gospodyń domowych.Żydowskie jadłodajnie wegetariańskie nie należały wówczas do rzadkości, co wynikało z przyczyn ekonomicznych – wymóg zachowania koszerności znacząco podnosił koszty podawania dań mięsnych. Lokal Fani Lewando wyróżniał się jednak na ich tle dostosowaniem do wymagań zamożnych klientów, ona sama uznawała zaś dietę bezmięsną za lepszą oraz apelowała, aby nie postrzegać takiego sposobu odżywiania jako oznaki biedy.
Lewandowa była szefową kuchni koszernej również na transatlantyku MS Batory na trasie Gdynia–Nowy Jork.
Jest autorką książki kucharskiej w języku jidysz Wegetariańsko-dietetyczna książka kucharska. 400 przepisów przygotowanych wyłącznie z warzyw (jid. Vegetarisch-Dietischer Kochbuch. 400 Speiser Gemach Oisshlishlech fun Grinsen; wyd. 1937/38); w przedmowie do publikacji Lewando przekonywała, że wegetarianizm jest zgodny z duchem judaizmu. Książka uchodziła za utraconą; ostatecznie udało się ją odnaleźć, a w 2015 roku ukazał się jej przekład na język angielski. Polskojęzyczna Dietojarska kuchnia żydowska, wydana w 2020 roku, zawiera 80 wybranych przepisów z książki Fani Lewando tłumaczonych bezpośrednio z jidysz.
Dokładne okoliczności śmierci Fani Lewando nie są znane; najprawdopodobniej zginęła podczas próby wydostania się z zajętego przez Niemców Wilna w 1941 roku. W Wilnie znajduje się Stolperstein upamiętniający jej osobę.
Przypisy
Polscy kucharze
Polscy przedsiębiorcy XX wieku
Żydowscy przedsiębiorcy
Restauratorzy
Autorzy książek kucharskich XX wieku
Pisarze jidysz
Ludzie związani z Wilnem (II Rzeczpospolita)
Polscy Żydzi zamordowani podczas Zagłady
Ludzie urodzeni we Włocławku
Urodzeni w XIX wieku
Zmarli w 1941 | 17,190 |
https://stackoverflow.com/questions/3314739 | StackExchange | Open Web | CC-By-SA | 2,010 | Stack Exchange | Andrey Cls, Haumich, Jackson Gonzalez, https://stackoverflow.com/users/6887278, https://stackoverflow.com/users/6887279, https://stackoverflow.com/users/6887280, https://stackoverflow.com/users/6887611, pat | English | Spoken | 160 | 236 | How to incorporate a batch file with vb.net
I wanted to make mysql dump files using a batch file and an app made through visual studio 2008.
How can I incorporate this batch file or call it from vb?
There is a code like this in vb.net but its using an absolute address:
Process.Start("C:exe\execute.exe")
How do I modify this so that I could just execute the file without providing the exact address. Or is there any place in the file system(windows 7) where I could just copy the exe file and just call it this way?
Process.Start("execute.exe")
You could copy execute.exe somewhere
that's within your PATH environment
variable.
Your .net code could use System.IO.Directory.SetCurrentDirectory to change the working directory at run time
You could right click on a shortcut to your .net script and set the "Run in" box to somewhere where your execute.exe is.
You could ship the two together so that they'll always be in the same directory.
| 35,129 |
https://uk.wikipedia.org/wiki/%D0%9B%D0%B0-%D0%95%D1%81%D0%BA%D0%BE%D0%BD%D0%B4%D1%96%D0%B4%D0%B0%20%28%D0%A2%D0%B5%D1%85%D0%B0%D1%81%29 | Wikipedia | Open Web | CC-By-SA | 2,023 | Ла-Ескондіда (Техас) | https://uk.wikipedia.org/w/index.php?title=Ла-Ескондіда (Техас)&action=history | Ukrainian | Spoken | 184 | 542 | Ла-Ескондіда () — переписна місцевість (CDP) в США, в окрузі Старр штату Техас. Населення — 153 особи (2010).
Географія
Ла-Ескондіда розташована за координатами (26.380650, -98.873728). За даними Бюро перепису населення США в 2010 році переписна місцевість мала площу 0,15 км², уся площа — суходіл.
Демографія
Згідно з переписом 2010 року, у переписній місцевості мешкали 153 особи в 42 домогосподарствах у складі 36 родин. Густота населення становила 1017 осіб/км². Було 48 помешкань (319/км²).
Расовий склад населення:
До двох чи більше рас належало 0,0 %. Частка іспаномовних становила 100,0 % від усіх жителів.
За віковим діапазоном населення розподілялося таким чином: 37,3 % — особи молодші 18 років, 53,5 % — особи у віці 18—64 років, 9,2 % — особи у віці 65 років та старші. Медіана віку мешканця становила 28,1 року. На 100 осіб жіночої статі у переписній місцевості припадало 91,2 чоловіків; на 100 жінок у віці від 18 років та старших — 77,8 чоловіків також старших 18 років.
Цивільне працевлаштоване населення становило 11 осіб. Основні галузі зайнятості: освіта, охорона здоров'я та соціальна допомога — 100,0 %.
Джерела
Переписні місцевості Техасу
Населені пункти округу Старр (Техас) | 32,888 |
https://bg.wikipedia.org/wiki/%D0%A7%D0%B5%D1%82%D0%B2%D0%B5%D1%80%D0%BE%D0%B5%D0%B2%D0%B0%D0%BD%D0%B3%D0%B5%D0%BB%D0%B8%D0%B5 | Wikipedia | Open Web | CC-By-SA | 2,023 | Четвероевангелие | https://bg.wikipedia.org/w/index.php?title=Четвероевангелие&action=history | Bulgarian | Spoken | 412 | 1,160 | Четвероевангелие (четириевангелие, тетраевангелие) (, ) е християнска богослужебна книга, оформена като кодекс, съдържаща и четирите евангелия на Новия завет – Евангелието според Матей, Марк, Лука и Йоан.
За да се използват в богослужението, към четвероеангелията са били добавяни литургически указания за времето на четене. По този начин текстът се приближава до т. нар. изборни или служебни еванглия, където отделните четива са подреден според времето на тяхното произнасяне. Освен текстовете на евангелистите обикновено има синаксар () – указател на четивата за подвижните празници и месецослов (менологий) – указател на четивата за неподвижните празници.
Четвероевангелието може да играе ролята на напрестолно евангелие, най-често украсено, което се изнася при Малкия вход на службата.
Най-известните български средновековни ръкописи с текста на четвероевангелията са:
Зографско четвероевангелие
Мариинско евангелие
Добромирово евангелие
Четвероевангелие на цар Иван Александър – прочуто заради миниатюрите, с които е илюстрирано.
Библиография
Алексеев, Анатолий А. Евангелие от Иоанна в славянской традиции. (Novum Testamentum Palaeoslovenice). Санкт-Петербург: Росийское библейское общество, 1998.
Алексеев, Анатолий А. Текстология славянской библии. Санкт-Петербург: Дмитрий Буланин, 1999.
Дограмаджиева, Екатерина. Показалецът „Евангелия различни на всяка потреба“ в славянските ръкописни евангелия. София: Академично издателство „Проф. М. Дринов“, 1998.
Мирчев, Кирил и Иван Добрев. Евангелие. В: Петър Динеков (гл. ред.), Кирило-Методиевска енциклопедия, Т. 1. София: БАН, 1985, 631 – 645.
Славова, Татяна. Преславска редакция на Кирило-Методиевия старобългарски евангелски превод. Кирило-Методиевски студии. Т. 6, 1989, 15 – 129
Трифуновиħ, Ђорђe. Jеванђеље. В: Азбучник српских средњовековних књижевних поjмова. Београд: NOLIT, 1990, 119 – 122.
Aland, K., M. Black, B. Fischer, H. J Frede, Christian Hannick, J. Hofmann, K. Junack (Hrsg.). Die alten Übersetzungen des Neuen Testaments, die Kirchenväterzitate und Lektionare. Der gegenwärtige Stand ihrer Erforschung und ihre Bedeutung für die griechische Textgeschichte (Arbeiten zur neutestamentlichen Textforschung 5). Berlin – New York: Walter de Gruyter, 1972.
Alekseev, Anatolij. The Last but Probably not the Least: The Slavonic Version as a Witness of the Greek NT Text. In: Evangelos Konstantinou (Hrsg.), Methodios und Kyrillos in ihrer europäischen Dimension. Frankfurt am Main, et al.: Peter Lang, 2005, 247 – 260.
Ehrman, Bart D and Michael W Holmes (eds.). The Text of the New Testament in Contemporary Research. Essays in Honour of B.M. Metzger. Grand Rapids, Michigan: Willam B. Eerdmans Publishing Company., 1995.
Garzaniti, Marcello. Die altslavische Version der Evangelien. Forschungsgeschichte und zeitgenössische Forshung. Köln, Weimar, Wien: Bohlau Verlag, 2001.
Gregory, Caspar René. Canon and Text of the New Testament. Edinburgh: T & T Clark, 1907.
Вижте също
Диатесерион
Външни препратки
За езика на Видинското четвероевангелие
Четириевангелие
Издания и преводи на Библията | 40,329 |
https://it.wikipedia.org/wiki/Why%20Make%20Sense%3F | Wikipedia | Open Web | CC-By-SA | 2,023 | Why Make Sense? | https://it.wikipedia.org/w/index.php?title=Why Make Sense?&action=history | Italian | Spoken | 23 | 40 | Why Make Sense? è il sesto album in studio del gruppo di musica elettronica inglese Hot Chip, pubblicato nel 2015.
Tracce
Collegamenti esterni | 25,360 |
https://ceb.wikipedia.org/wiki/Quebrada%20Arepa | Wikipedia | Open Web | CC-By-SA | 2,023 | Quebrada Arepa | https://ceb.wikipedia.org/w/index.php?title=Quebrada Arepa&action=history | Cebuano | Spoken | 61 | 114 | Suba ang Quebrada Arepa sa Kolombiya. Nahimutang ni sa departamento sa Departamento del Cauca, sa sentro nga bahin sa nasod, km sa habagatan-kasadpan sa Bogotá ang ulohan sa nasod. Ang Quebrada Arepa mao ang bahin sa tubig-saluran sa Río Magdalena.
Ang mga gi basihan niini
Río Magdalena (suba sa Kolombiya, lat 11,10, long -74,85) tubig-saluran
Mga suba sa Departamento del Cauca | 44,983 |
https://vi.wikipedia.org/wiki/Ospriocerus%20painterorum | Wikipedia | Open Web | CC-By-SA | 2,023 | Ospriocerus painterorum | https://vi.wikipedia.org/w/index.php?title=Ospriocerus painterorum&action=history | Vietnamese | Spoken | 31 | 74 | Ospriocerus painterorum là một loài ruồi trong họ Asilidae. Ospriocerus painterorum được Martin miêu tả năm 1968. Loài này phân bố ở vùng Tân nhiệt đới.
Chú thích
Tham khảo
Ospriocerus | 45,079 |
https://uk.wikipedia.org/wiki/%D0%9A%D0%B0%D0%BB%D0%BE%D0%BC%D0%B5%D0%BB%D1%8C | Wikipedia | Open Web | CC-By-SA | 2,023 | Каломель | https://uk.wikipedia.org/w/index.php?title=Каломель&action=history | Ukrainian | Spoken | 192 | 648 | Каломель — пластичний мінерал, дихлорид димеркурію ланцюжкової будови, що містить 84.98 % ртуті.
Етимологія та історія
Перша згадка про Каломель була вже в записах 1608 р. Бегіна і 1609 р. Освальда Кролла. Мінерал був відомий древнім тибетцям. Проте науково описаний лише в 1612 році Теодором Туреке де Майерне († 1655), який назвав мінерал «красивий чорний». Назва складається зі стародавніх грецьких слів καλός — «красивий» і μέλας — «чорний», тому що він стає чорним у реакції з аміаком.
Загальний опис
Утворюється при зміні кіноварі, ртутного срібла, амальгами, ртутної бляклої руди, селенового метацинабариту й інших мінералів, які містять меркурій, а також нагріванням суміші хлорного меркурію з металічним меркурієм. Під впливом світла темнішає.
Використовують у медицині, у більшості, як проносний і дезінфікувальний засіб. У техніці йде на виготовлення фарб із золотом для порцеляни та для бенгальських вогнів.
Зустрічається разом із самородною ртуттю, еглестонітом, терлінгуаїтом і монтроїдитом.
Дія на організм
Хлориста ртуть(І) є проносним, а також сечо- й жовчогінним засобом.
Див. також
Хлорид ртуті(I)
Список мінералів
Примітки
Література
Посилання
КАЛОМЕЛЬ //Фармацевтична енциклопедія
International Chemical Safety Card 0984
National Pollutant Inventory — Mercury and compounds Fact Sheet
NIOSH Pocket Guide to Chemical Hazards
Хлориди (мінерали)
Мінерали ртуті | 12,381 |
https://math.stackexchange.com/questions/4736951 | StackExchange | Open Web | CC-By-SA | 2,023 | Stack Exchange | Jyrki Lahtonen, https://math.stackexchange.com/users/11619 | English | Spoken | 179 | 410 | Transitive Group and its Normal Subgroup
I had some trouble handling the question below:
Let $G$ be a group acting on a set $S$ containing at least two elements. Assume $G$ is transitive and $\{g\in G:gx=x\ \text{for all}\ x\in S\}=\left<e\right>$. If $N\lhd G$ and $N<G_x$ for some $x\in S$, then $N=\left<e\right>$.
Some terminologies are explained here:
(i) We say $G$ is transitive, if given any $x,y\in S$, there exists $g\in G$ such that $gx=y$.
(ii) $G_x$, the stablizer of $x$, which also denoted as $\mathrm{Stab}(x)$, is the set $\{g\in G:gx=x\}$.
Here is my attempt. For any $n\in N$, since $n$ is a subgroup of $G_x$ we have $nx=x$. Then $N$ is normal in $G$ implies $gng^{-1}\in N$ for all $g\in G$. But I can't go any further. Can someone offer me any suggestions? Thanks in advance.
If $n\in G_x$ and $gx=y$ then $gng^{-1}\in G_y$.
$Nx = x$ for some $x \in S$.
Hence $gNx = Ngx = gx \implies N$ fixes $Gx$.
But $G$ is transitive. Hence $Gx = S$.
Hence $N$ fixes $S$.
Hence $N = \langle e\rangle$.
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Foy5gcur8_o_1 | Youtube-Commons | Open Web | CC-By | null | Mod-11 Lec-05 Turing Decidable Languages | None | English | Spoken | 5,951 | 6,564 | Yes, so far I have discussed in Turing machines the Turing acceptable languages, Turing computable functions and to construct Turing machines as for using the Turing machines that you have already constructed, some composite Turing machines so far we have discussed. So, you know the standard Turing machine, definition of Turing machine and what is Turing acceptable language, Turing computable functions, in Turing machines we have discussed these two concepts and we have constructed Turing machines, some little more complicated using composition of Turing machines that we have introduced the concept of composition of Turing machines combining Turing machines we have discussed. Now, in case of Turing computable functions for multi input Turing computable functions also I have discussed, in particular computing with natural numbers also I have introduced where natural numbers will be considered in the unary representation that means, natural numbers for one element alphabet that singleton I star, here the correspondence is 0 is empty string and each number n is instead to be i power n, this is the string we consider and we give the input to the Turing machine. So, these are the things that I have already introduced, now let me talk about another concept that is Turing decidable languages, Turing decidable languages, here Turing acceptable languages we have already named them as recursively enumerable languages, this is recursively enumerable and in this case we called this is recursive languages, the recursive languages Turing decidable languages, now first let me define what is that the definition a language L or some alphabet let me write or an alphabet sigma naught is said to be Turing decidable, is said to be Turing decidable if its characteristic function, its characteristic function let me write that as chi L, this is from sigma naught star to 0 1 or I can essentially take two symbols here talking about S or no, this chi L that is defined by this characteristic function you know the definition, any string you take you will give 1 if that string is in L 0 else. So, the string if it is in L we are giving 1 I may you may put say S or no sort of for 0 you can take the symbol no n and for 1 you can take y and you can assign accordingly. So, this characteristic function is Turing computable, so a language is Turing decidable if its characteristic function is Turing computable that means, you give a string from sigma naught star as an input to a Turing machine which computes this chi L, if x is in L that returns 1 if x is not in L then it returns 0 that means, essentially it decides it says whether a particular string from the alphabet whether over an alphabet whether it is in the language or not in the language this is a decision making Turing machine. Let me give you an example you know the Turing computable means we should have a Turing machine computing this function a very simple example that you know if you consider the language set of all strings say over a b such that the length of x is even you know this is a regular language for some natural number m. So, length is even this is a Turing decidable language how do we say this we have to construct a Turing machine if you give x as an input any x from over a b it has to say 1 or 0 it has to say s or no. Let me construct a Turing machine for that purpose, so to start with because as earlier we give the input in this form any string you give if x is in L this returns 1 if x is in L this returns 0 if else if x is not in L this is what is the expectation. So, we will start with the reading and writing at here and this termites with this, so if this is the input given you take a left move and you have to understand whether it is of even length that is a cross checker. So, you will be switching between because you know the fire rate automaton accepting an even length string you will be switching between two states you will be having two states if it is in the initial state when in the first state that you are accepting if it is in the second state you are rejecting. So, that is how you are you are constructing finite automaton. So, let us construct the Turing machine following that logic, so first you take a left move and a cross check whether it is blank or not blank. So, there are two situations if this is blank I am coming to this place if it is not blank let me now see what is the next symbol that means I will go left. So, now if this is blank that means I have odd length. So, this is one situation if it is not blank further I will go to left and a cross check what is happening with this. So, now in the first place when you get the blank you know clearly that is empty string and therefore, you can accept it. So, you take a right move and print one and take a right move and halt if this is the place when you are getting blank that means after one symbol you are reaching to this and you are going to left move that means if one length string is there then after that we are reaching this blank. So, at this place you will print 0 in the next and take a right move and halt. Now, for even length we will always be getting empty symbol here for odd length strings we are always be getting empty symbol at this location after taking this left move and ultimately you are getting this printing 0 if it is odd length printing 1 if it is even length. Let me just demonstrate this to an example for example you have empty string as input if this is the input then that means you are giving this is the input format for the empty string and you take a left move here and since it is blank here it takes one right move. So, that means you are cross checking and this will be transformed. So, you are taking one left move and then take a right move at this place you print one and take a right move and halt here. So, this is accepting if you take for example the string a this is one length string for this how the computation. So, the input is given in this format input is given in this format now take a left move cross check this is not blank. So, I will take one more left move at this place I have got blank. So, then take a right move here print 0 and take a right move and halt. So, this will be transformed to 0 this. So, that means the string is rejected now let me consider the input a b let me consider the input a b in this machine what is happening a b is considered with the input format like this how this will be transformed that is a question now. So, from here first left move will it will take left move and it will come here it is not blank it will take another left move this is not blank and now take another left move you are getting blank here. So, this is since it is even length string when you are getting blank you are at this position at this position you are coming to this branch. So, it takes now a right move print one and take a right move what is happening now you have to cross check this situation. So, the thing is this from here you are taking a right move take one take right move. So, at this place you have b and it halts here of course you have several blanks here. So, we are not going to represent this. So, we are the our target is to prepare you know the output in this format. So, when I am giving empty string when I am giving one length string I am getting appropriately 1 and 0, but if I have more than one length string if I have more than one length string then what is happening it is printing 1 and it is halting in the next symbol where this is not blank. So, simultaneously when we are reading you arise this you arise the symbols because that is not happening here you arise this and ultimately that machine will give you the appropriate output. So, the modifications in this machine now if you have more than one length string the symbols are lying over there. So, let us modify this way take a left move take a left move and check whether this is non-empty if it is non-empty then print blank then take a left move. Now, if it is blank then as earlier if it is not blank sigma not equal to blank then print blank there take a left move. Now, this continues as earlier. So, if it is blank take a right move print 1 take right move here print 0 take a right move. Now, what is happening with this what are the input that we are giving as long as we are getting non-empty symbols here non-empty symbols here then it is that particular cell it is we are making it blank then taking a left move again if you are having non-empty here non-blank then you are making the particular cell blank then you are taking a left move. So, keep doing this now while coming from this place to this place we are visiting each and every symbol and you are cross checking whether it is empty or not if it is blank if it is blank that means you are coming to this place. So, what you have to do here at this place we are making them blank. So, while coming you are making this blank you are coming to till this end take a right move you print appropriately 1 and take a right move and halt here. So, this is how we can construct a decider this is a Turing machine that halts that halts on any input by printing yes or no. Now, let me take one more example if I consider the language such that palindromes. Now, this is a regular language for which we have finite automaton this is a context free language for which you have a push down automaton you have constructed those things and for this I hope you have constructed Turing machine to show this is a Turing acceptable language. Now, a decider let us look at a decider for this purpose. So, the mechanism you know as earlier this is say a 1 a 2 a n what did we do we are reading this symbol and cross checking with this symbol and going to this point reading this symbol cross checking with this symbol and so on. At that point of time what did we do we have arrest this and we have arrest this after cross checking when the match is there we are continuing further n minus 1 a n minus 1 whether it is equal to a 2 we are cross checking we are arising and continuing like this. In the middle if till this middle if you continue and if the matching is happening we are declaring that it is a palindrome. Otherwise we are going to you know left keep going to left move and hanging the machine or we are going to the infinite loop that is how we have constructed to show this language is Turing acceptable language the palindromes. But, here when I am finding that proper match at every stage that means if it is a palindrome I am somewhere on the tape in between. Now, how do we recognize that you know to give the output of this format whether it is 1 or 0 because when the input is given this way we have to leave either 0 or 1 s or no. So, when I am finding the situation when I am finding the mismatch by the time if I have made certain blank symbols here if I have made certain blank symbols here somewhere mismatch if it is happening I am not clear that how many cells I have to go left and where exactly you know I have to print in the second cell 0 or 1. Because once you made certain cells blank we will not be clear where exactly you are because there is no counting business here to understand that yes this many blanks are there I will go back and do that we are not cross checking that. So, what we have to do in the beginning before arising this tape whatever in the beginning what you do you first print some special symbol and then once you do this cross checking wherever the mismatch happens of course, you arrange everything and then you have to and then you come to this place and print 0 if there is match till the end of the story then what you do you come to this and take a right move and print 1. So, this is how the decider can be given here this is one important point one has to understand if you arrange certain symbols here if certain number of blanks are created and after cross checking the input if there is a match or mismatch we will not be clear that to what extent you have to go to left and print 1 or 0. So, for that purpose first we have to print at this place some special symbol. So, that what are the rest of the business as earlier we can continue. So, what we do from here I will come to this blank that means a lash. So, you come to this place you print a special symbol there say for example, dollar I am printing here now the situation is at this place I have dollar and dollar is not part of this input and thus you can distinguish the symbols between these two. Now, maybe you can start from here this symbol you can cross check with this and keep cross checking. So, take a right move now sigma that is not blank if that is not blank then you can print blank here print blank. Now, I have arranged this place I will go to this end and cross check whether with this symbol they are equal or not. So, what I have to do after printing blank here I go to this blank that is r hash then take a left move I am coming to this position and now here you cross check what is that symbol whether it is sigma. So, if the same sigma is occurring here then I am happy you print blank there you print blank there now come to this blank and then take a right move. So, here you print blank once you print blank then you can take a Lash come to this place and then take a right move. So, at this position we have to connect look here this is a composite mission this is not pointing to this entire mission if it is pointing to the entire mission that means it is actually starting from here, but the situation here is it is not pointing to the entire mission I am just pointing to this component that means after this I have to just pursue right move. So, take a right move and see if it is not equal to blank then you print blank here again then go till the end because we have already made blank here now I have say a n minus 1 is available then take a left move here if it is a 2 is matching with a n minus 1 then print blank then come back to this and keep doing this. This is the logic we have followed to cross check this language is doing acceptable the rest of the things are same. Now, here in the beginning suppose you have empty that means that means this is a symbol that you have here this is the string you know even length string that is a palindrome. So, in which situation you get blank here when you take a right move you get blank here we are just to distinguish may be after right move you get blank after right move if you get blank then of course this is an accepting situation. Now, you have consume one symbol here and then when you take a left move when you are getting the same symbol that is a non empty symbol then you are arising it and continuing. Now, here there are two situations one you get a blank symbol or some non empty non empty symbol some non blank symbol that is different from sigma. So, there are two situations in one situation you have to accept it in another situation you have to reject it. So, what do I do when you have this is here blank when you have here blank then you will accept that is an odd length a palindrome and in the situation that is different from sigma and it is not blank also if this is the situation. Now, here we have to reject. So, these are the situations that arise now when I am getting blank in the beginning we know left of this if you have arise certain symbols they are all arise. So, you simply go to this dollar this dollar you can convert it to blank and take a right move print 1 and accept this is a situation even in this case also the similar machine can be connected, but in this case we have to arise whatever is the left over input when the mismatch is happening if it is not sigma if it is not blank that means some other non blank symbol is available in which situation whatever left to this we have to arise it and then you have to you have to print you have to print 0. So, here this is easy take L dollar that means you go till end of the tape left end of the tape there you print blank there you print blank take a right move here print 1 here print 1 and then take a right move and halt we are happy here. Similarly, if you receive this I can connect to the same machine go till L dollar print blank here then take a right move print blank here take a right move print 1 and take a right move and halt. So, this is the halting situation with the acceptance now when you want to reject there may be some more symbols on the tape first you have to erase them then go till dollar then take a right move because dollar you will make it blank take a right move there you print 0 which is not accepting. So, that means at this situation to reject the string you have to erase and go till dollar then you have to give the output in this format. So, this portion now I hope you understand what one has to do here at this portion you have to do this business the rest of the symbols you have to erase then go till dollar then print 0 finally as output. So, this particular machine like here whatever we have mentioned take an exercise and construct that particular component to give a decider for this language palindromes. So, a decider is one once again I emphasize you give any input from the desired alphabet you give any input the situation is earlier in case of Turing acceptable languages or recursive renewable languages what we are doing you give an input if it is in the language that simply halts otherwise you are creating whether it is hanging or you are putting it in infinite loop. So, that the string is not accepted that is how we have managed, but in case of this Turing decidability that means to say a language Turing decidable you give any input from the alphabet over the alphabet there are two situations you will have precisely either 1 you will print finally or 0 you will print finally that means you may print y for example, you print s or you print no to indicate. So, that means on any input this Turing machine whatever that we are constructing that has to halts. So, you have one has to observe this remark that this type of Turing machine the Turing machine that is existing for a Turing decidable language that halts on any input a decider let me call a decider is a Turing machine that halts on any input by printing by accepting or rejecting the input by accepting or rejecting the input. So, let me present this way here printing 1 we are calling it is accepting by printing 0 we are calling it is rejecting. So, this kind of Turing machine which is halting on any input this is the notion of algorithm chess Turing thesis what chess Turing thesis claims is an algorithm is a decider. That means an algorithm is a Turing machine that halts on any input that is how it is considered and this is the formal notion of giving algorithm. Now, as I have promised earlier let me now consider some examples of Turing computable languages or natural numbers we have already discussed a successor function addition their Turing computable functions and this using Turing computable compatibility we have introduced the notion of decider. Now, let me cross check this property for example, you consider a function this is as I have mentioned monos this is from n cross n to n in short let me use may be m n is m minus n if m greater than equal to n 0 n. So, this is a non negative number we have to give the natural numbers. So, instead of considering minus we consider monos in this computation. So, a Turing machine which can perform this job let me give you certain hints to construct a Turing machine to show this is a Turing computable function monos defined like this is a Turing computable function the Turing computable function the Turing machine for this let me give you some idea in this direction you may be giving input i m i n like this this is the input format. After finally, many steps that Turing machine has to leave this we use the shorthand for this m minus with a dot here called monos let me use m monos n with a minus or which I am putting a dot i power m monos n this should be the left over thing this may be 0 if this is smaller if this is bigger or equal then m minus n that is what we are performing in this machine what you have to do you are s 1 here for example, you are s 1 here and you go and are s 1 here are s 1 here you go and are s 1 more i here and keep doing this. If you receive blank after erasing this symbols then you cross check what is the situation here whatever is left over that is if this is bigger then something will be left over those many eyes will be there if this is smaller then when you are erasing and you are cross checking with this you will encounter this blank quickly. So, you just understand that when I am doing this business erasing 1 i here erasing 1 corresponding to that 1 i here if I am reaching to this blank first then m is smaller than n if I am reaching to this blank first then n is smaller than m. So, appropriately you have to leave the output. So, can we start with this logic. So, take a left move here you have i or blank that is how we will be considering. So, take a left move and see whether you have i or blank that is the situation. If you are already receiving blank if I have by that time we have erased certain things whatever is left over till that point I have to go and see, but now the bottleneck is like this if they are equal if they are equal I have erased till this point and then I have erased these things. Then how do I recognize that to this point I will come and whatever is left over I have to understand. So, for which again you can bring some notion of special symbol put it here put it here and for which you will be cross checking. So, you can now consider some special symbol printing here then you do this business you go to this another special symbol here cross check whether that particular special symbol is a arrived or not. So, that is how you can manage because when I am making them blanks I have to be clear like to what extent I am going to what extent that you are going. So, to understand these things may be at this place you can have a special symbol and at this place also you can have a special symbol and continue this business by cross checking to construct the Turing machine which computes this function. One more example these are little easy let me put say capital M to talk about multiplication M n this is also Turing computable function multiplication of two numbers is Turing computable is Turing computable. What you have to do in this case again the input will be in this form the input will be in this form. Now what I have to do here I have to multiply these two and I have to leave that as output. So, what what is the procedure that we can follow here this i power n for example, you copy this M number of times. So, that you will be creating i power m n and then of course, then you can say for example, in this case i power n is here you erase 1 i you erase 1 i and this is the i power n you can copy it here and once again erase one more i copy i power n here and so on. So, this is the logic if you follow then we can consider this business let me. So, what we do you just again certain hints I am giving from here I come to this I come to this place say L hash take a left move if I have either if I have either then I may just convert into a special symbol for example, say print say some scent just to indicate this then I will use R hash I am here I will come to this first blank then once again coming here R hash that means R hash square I can use now you apply the copying mission which can copy this i power n and print it here that is what happens here. So, now I will apply say copying mission. So, I am at this place I am applying copying mission now what you do after copying this what will be left over this i power m minus 1 1 cent is here blank i power n this is the situation we see now in the next place what I will do I have to come to this point and a skip sense if you are getting here you have to skip sense if you are getting blank that means this block is over if you are getting i then again you print send there then take right move take right move now in this situation I have to go to this point and then I have to copy. So, here I have to go till this end because here I have to copy this. So, what do you do because this copying now we have to give because copying not this string because I will be generating afterwards another i power n and so on. So, copying one string that is that is leaving. So, that means you have to now have a Turing machine which may for example, where I have this kind of situation a Turing machine which prints this because the second block of x need to be printed here say x let it be there and y x if you create a Turing machine which can perform this business then you can appropriately modify this and construct multiplication Turing machine I hope this is this hint is sufficient for you to you know construct this Turing machine once again to show multiplication Turing machine Turing computable function you take i power n i power m as inputs and I have to leave this as output I am just starting the business now you see the logic is from here I will count the number of i's here that means essentially cross taking each i I am just cutting here may be by printing some center whatever I will go to the end and every time I mark one i I copy this block to the end and keep doing this at the end of course, we have to erase these two and move. So, that means essentially what I am doing if I have some situation like this when I am here x y is input if I can create something like this that means i power n will always be copied that many number of times and create i power m n here. So, end of the story once you create that machine and connect it appropriately end of the story what will be the situation when I am giving this as input i power m i power n this is the input with this kind of process you will have i power m i power n i power m n I will get i power m n here now you have you create one machine that can erase two blocks of strings before one particular string that means given. So, another type of machine that you can look at this I am calling m one for example, m two is a machine which you can try when I am giving x y as input two strings this machine leaves by erasing x this leaves y as output for example, if I create such a machine then what I have to do at this point I will apply this machine once then I am left with i power m i power m n on the tape once again I will apply this machine and then I will be left with i power m n. So, using this you can using this phenomena using this machines m one and m two connecting appropriately you can construct a multiplication machine not only multiplication you can consider you know little more complicated arithmetic and construct appropriately this machines a Turing machine which computes this arithmetic operations and certain manipulations for natural number that are computable what is the time 50. Now, let me look at some important properties related to Turing decidable languages that is as we have understood there is a Turing machine that is a decider suppose this is a Turing machine you give any input from sigma star it is resulting s r now it is printing s r now. So, this is how it is working now using this phenomena we can quickly understand this result if l is Turing decidable that means recursive l compliment is also recursive the Turing decidable language how it is what are the Turing machine that you are constructing whenever it is printing one because you have the Turing machine m now you construct the Turing machine m dash the transition map is exactly same except that whenever it is printing one finally, you take the input whenever it is printing one. So, you give it to m so whenever it is saying s then you ask it to say no whenever it is saying no you ask it to say s. So, this is how you can manipulate that means the Turing machine the Turing machine m is used only thing is before finally, it is halting you just take a left move and see whether it has printed one if it has printed one you print 0 there and take a right move and halt and vice versa. Thus you can understand quickly that if l is recursive its compliment is also a recursive language. So, this is the construction that one can make and now another result if l and l compliment are recursively enumerable that means Turing acceptable languages if they are Turing acceptable languages then both of them both are recursive what we have to do if l is recursive enumerable you have a Turing machine for those strings which are in l it says s and since l compliment is also recursive enumerable you have a Turing machine for those strings which are in l compliment that says s. So, given input what you do you manipulate it like this. So, let me say m 1 is a machine Turing machine that takes input and says s for l. So, this is for the machine l and now what you do simultaneously you give the input to m as well as m 1 as well as m 2 that is for l compliment that is for l compliment take this and for the input x either it this will say s or this will say s both cannot say s. So, what you do whenever it says s. So, you simply report that if this is saying s you say s if it is saying s your report no now given an input one of them only will precisely say s both cannot respond. For example, in this case it may hang or it may go to infinite loop like this. So, what we have to do whenever it is saying s that means after a finitely many steps you are getting this information if it is not in l that means if it is in l compliment this machine will say s this will halt. So, in which case you ask it to say no thus you can understand that if l and its compliment if l and its compliment both are recursive and innuble languages that means Turing acceptable languages both of them are recursive because l is recursive that I have shown here again using this result if l is recursive l compliment is also recursive you can understand that both of them are recursive languages Turing decidable languages. | 13,378 |
https://cs.wikipedia.org/wiki/Divize%20B%201970/1971 | Wikipedia | Open Web | CC-By-SA | 2,023 | Divize B 1970/1971 | https://cs.wikipedia.org/w/index.php?title=Divize B 1970/1971&action=history | Czech | Spoken | 142 | 352 | V soubojích 5. ročníku České divize B 1968/69 se utkalo 14 týmů dvoukolovým systémem podzim - jaro. Tento ročník začal v srpnu 1970 a skončil v červnu 1971.
Nové týmy v sezoně 1970/71
Z 3. ligy – sk. A 1969/70 sestoupilo do Divize B mužstvo VTJ Dukla Slaný. Z krajských přeborů ročníku 1969/70postoupila vítězná mužstva TJ Lokomotiva Bílina ze Severočeského krajského přeboru. Také sem byla přeřazena mužstva TJ Spartak Radotín z Divize A a TJ Tatra Smíchov, TJ Motorlet Praha a TJ Bohemians ČKD Praha "B" z Divize C.
Výsledná tabulka
Zdroj:
Poznámky:
Z = Odehrané zápasy; V = Vítězství; R = Remízy; P = Prohry; VG = Vstřelené góly; OG = Obdržené góly; B = Body
Reference
Externí odkazy
Česká Divize B (ČD-B) 1970/71, archiv RP 1970-1971
Divize B
Fotbal v Česku v roce 1970
Fotbal v Česku v roce 1971 | 23,679 |
https://stackoverflow.com/questions/29489066 | StackExchange | Open Web | CC-By-SA | 2,015 | Stack Exchange | Ryan Vincent, emihir0, https://stackoverflow.com/users/3184785, https://stackoverflow.com/users/4712455 | English | Spoken | 713 | 888 | Database design with multiple types of products
I am currently making a new database for a manufacturing company and the problem is as follows:
1) Each design (unique ID) is produced on a specific mould (unique ID) and with specific set of cutting knives.
2) Each design has several operations specified for that design only which need to be done in order to produce the design (product).
Basically I have (I believe) completed 1) step, that is, each design having a specific mould and cutting knives assigned in DesignMouldCuttingKnives table.
The entry there would be something like this:
And so I can query for Design 9017 and make a list of all cutting knives used in the specific design.
I think this would work quite efficiently (I would appreciate suggestions for improvement/rework to this part too if you have any).
My current problem, however, is that each design has specific set of operations required to be done in order to produce it.
For instance, 9017 has 10 unique operations (e.g. 2 of them are cutting some material, 3 of them are stitching the cuts, 2 of them are gluing some stuff together, 2 of them are stitching extra accessories, 1 is packing the stuff up).
Each operation is done by 1 person and so every operation has specific monetary value attached to it (the wage will be calculated using this).
How do I make this as simple as possible, without having to create a unique table for each design itself? The problem is that a specific stitching operation has different monetary value on one design compared to other design and so I cannot simply make a table of all operations and attach the operation ID to specific design as I did with the cutting knives (I think, if it is somehow possible, please let me know).
Any suggestions to make this as simple as possible, assuming it would need to be easily queried for?
I plan to make this in MS Access as the database is aimed to be accessed by 1-5 users locally (not online).
You say: 'each design has specific set of operations required to be done in order to produce it'. I cannot see that 'set of operations' table in your database design. If you do not create it explicitly then you will have to 'logically generate' it from your current structure each time you need it.
I have not made a table with 'set of operations' simply because I am not sure how to approach it in a sensible way. The problem is that the operations are unique for a specific design and so I would think the best way is to expand the DesignList table into multiple tables for each design, that is, table e.g. "9017" and then there will be a list of specific operations. But the problem is that there would be about 2000 of those tables (designs) and that is not really a sensible way.
Would it help to think of a separate table that holds a list of 'specific operations', each of which has a unique id. Also, a table that holds a list of 'specific designs', each with a unique id. Then a table that relates the 'specific design' to the ''specific operations' with a 'operation order number'?
I thought of that and I guess it is the only way to do it without having to create a specific table for each Design. Thanks.
May I also ask: is using MS Access for this a good idea or would you suggest different approach? The database will be accessed locally by 1-5 users with, hopefully, different user access priviledges.
Could the number of users expand in the future? would it cope with, say, 100 users? I would be tempted to work with a database that handles 'users' naturally. I think there are versions of ms 'sql server' that are cheap and also have the tools to convert from 'ms access'. I would look at those as well as other rdbms' systems. It depends on what you are familiar with more than which of the rdbms's as they all will be fine for what you want to do
The database is unlikely to be accessed by more than 5 users within any foreseen time period.
| 37,623 |
https://ja.wikipedia.org/wiki/%E5%B7%9D%E5%8F%88%E6%99%BA%E8%8F%9C%E7%BE%8E | Wikipedia | Open Web | CC-By-SA | 2,023 | 川又智菜美 | https://ja.wikipedia.org/w/index.php?title=川又智菜美&action=history | Japanese | Spoken | 94 | 1,073 | 川又 智菜美（かわまた ちなみ、1994年2月4日 - ）は、フリーアナウンサー。元女優、ファッションモデル、タレント。学生時代は弓川留奈、川上リサの名義で芸能活動をしていた。
来歴・人物
神奈川県横浜市出身。
東京宝塚劇場前で宝塚スターの出待ちをしている時にスターダストプロモーションにスカウトされ芸能界入りする。
アイドルグループ・ももいろクローバー（現・ももいろクローバーZ）のメンバーにも構想段階では選ばれていた。中学3年生時にファンタジー小説『月の啓示』を刊行。趣味はテニス。特技は英会話で、実用英語技能検定準1級の資格を所持している。
慶應義塾大学法学部法律学科卒業後、2016年山陽放送にアナウンサーとして入社。2017年3月、山陽放送を退社。セント・フォース所属となる。
2021年4月からWOWOW専属アナウンサーとして2年間活動した。
2023年6月18日、結婚を報告。
アナウンサーとしての担当番組
フリー転向後
はやドキ!（TBSテレビ、2018年1月4日 - 2019年9月26日） - 火〜木曜→水〜金曜→水〜木曜担当
日テレNEWS24（2020年10月 - 2021年3月） - キャスター
週刊テニスワールド（WOWOW、2021年4月 - 2023年3月） - ナレーション
WOWOWテニス グランドスラム中継（WOWOW、2021年5月 - 2022年1月） - 進行
チャンピオンズリーグダイジェスト!（WOWOW、2022年2月19日 - 2022年11月4日） - 進行
山陽放送時代
RSKイブニング5時（テレビ） - 火曜サブ司会
あもーれ!マッタリーノ（ラジオ／2016年8月29日 - 2017年3月27日）- 月曜パートナー
芸能人としての活動歴
出演
テレビ
土曜ワイド劇場「法律事務所2」（テレビ朝日）
テストの花道（NHK教育／2011年4月 - 2012年3月）
ラジオ
ASIAN美少女
CM
NIKKEスクールユニフォーム2007
明治製菓チョコレートンの冒険
須磨学園パンフレット
ANIMAX 第3回全日本アニソングランプリ 募集告知
栄光ゼミナール
雑誌
ピュアピュア
なかよし
ディズニーガールズ
レモンティーン
PV
lela「樹海の月」
著書
『月の啓示』（SDP、2009年、、弓川留奈名義）
脚注
注釈
出典
外部リンク
セント・フォース｜川又智菜美
あもーれ!マッタリーノ - archive.is
フリーアナウンサー
セント・フォース
WOWOWのアナウンサー
RSK山陽放送のアナウンサー
日本の女優
日本の女性ファッションモデル
過去のスターダストプロモーション所属者
ももいろクローバーZのメンバー
慶應義塾大学出身の人物
横浜市出身の人物
1994年生
存命人物 | 37,389 |
https://ceb.wikipedia.org/wiki/Buttu%20Pana | Wikipedia | Open Web | CC-By-SA | 2,023 | Buttu Pana | https://ceb.wikipedia.org/w/index.php?title=Buttu Pana&action=history | Cebuano | Spoken | 174 | 279 | Bukid ang Buttu Pana sa Indonesya. Nahimutang ni sa lalawigan sa Provinsi Sulawesi Barat, sa sentro nga bahin sa nasod, km sa sidlakan sa Jakarta ang ulohan sa nasod. metros ibabaw sa dagat kahaboga ang nahimutangan sa Buttu Pana.
Ang yuta palibot sa Buttu Pana kasagaran kabungtoran, apan sa amihanan nga kini mao ang kabukiran. Ang kinahabogang dapit sa palibot dunay gihabogon nga ka metro ug km sa amihanan sa Buttu Pana. Dunay mga ka tawo kada kilometro kwadrado sa palibot sa Buttu Pana medyo hilabihan populasyon.. Sa rehiyon palibot sa Buttu Pana, kabukiran talagsaon komon.
Hapit nalukop sa lasang ang palibot sa Buttu Pana. Ang klima tropikal nga kasalupan. Ang kasarangang giiniton °C. Ang kinainitan nga bulan Abril, sa °C, ug ang kinabugnawan Pebrero, sa °C. Ang kasarangang pag-ulan milimetro matag tuig. Ang kinabasaan nga bulan Abril, sa milimetro nga ulan, ug ang kinaugahan Septiyembre, sa milimetro.
Saysay
Ang mga gi basihan niini
Kabukiran sa Provinsi Sulawesi Barat
Kabukiran sa Indonesya nga mas taas kay sa 200 metros ibabaw sa dagat nga lebel | 30,989 |
https://ru.wikipedia.org/wiki/%D0%91%D0%BE%D1%80%D0%B8%D0%B4%20%D1%82%D1%80%D0%B8%D0%BF%D0%B0%D0%BB%D0%BB%D0%B0%D0%B4%D0%B8%D1%8F | Wikipedia | Open Web | CC-By-SA | 2,023 | Борид трипалладия | https://ru.wikipedia.org/w/index.php?title=Борид трипалладия&action=history | Russian | Spoken | 54 | 184 | Борид трипалладия — бинарное неорганическое соединение
палладия и бора
с формулой PdB,
кристаллы.
Получение
Сплавление стехиометрических количеств чистых веществ:
Физические свойства
Борид трипалладия образует кристаллы
ромбической сингонии,
пространственная группа P nma,
параметры ячейки a = 0,5463 нм, b = 0,7567 нм, c = 0,4872 нм, Z = 4
.
Примечания
Литература
Соединения палладия
палладия | 35,640 |
https://stackoverflow.com/questions/76135485 | StackExchange | Open Web | CC-By-SA | 2,023 | Stack Exchange | English | Spoken | 231 | 366 | multer removes the body from the incomming request
I send an http post request from a simple Angular app in a form of FormData which is set the content type to multipart/form-data, the back-end receives the incoming request and parses it using multer
after multer parses the request I always find that req.body is null
here is a minimal reproduction
steps to reproduce
install the dependencies npm install
run the backend part npm start from the backend dir
run the frontend part npm start from the frontend dir
open your browser on localhost:4200, it will send an HTTP request directly to the backend without filling any form, and the received response is displayed on the same page.
the problem here is that it displays {body:null,..} where it should be {body: {content: "test"}}
The problem is on the frontend with the toFormData method, which constructs incorrect formData, which is why it's not parsed/empty on the server.
Two ways to fix it:
change payload to an object, instead of an array:
payload = { content: 'test' };
To use payload as an array, you need to loop an array and then loop object keys.
Try replacing it with this:
export function toFormData(data: any): FormData {
let formData = new FormData();
for (let obj of data) {
for (let key in obj) {
let element = obj[key];
formData.append(key, element);
}
}
return formData;
}
| 39,888 | |
https://sv.wikipedia.org/wiki/Momordica%20rumphii | Wikipedia | Open Web | CC-By-SA | 2,023 | Momordica rumphii | https://sv.wikipedia.org/w/index.php?title=Momordica rumphii&action=history | Swedish | Spoken | 30 | 70 | Momordica rumphii är en gurkväxtart som beskrevs av W.J.de Wilde. Momordica rumphii ingår i släktet Momordica och familjen gurkväxter. Inga underarter finns listade i Catalogue of Life.
Källor
Gurkväxter
rumphii | 47,753 |
https://vi.wikipedia.org/wiki/Prodasineura%20incerta | Wikipedia | Open Web | CC-By-SA | 2,023 | Prodasineura incerta | https://vi.wikipedia.org/w/index.php?title=Prodasineura incerta&action=history | Vietnamese | Spoken | 33 | 70 | Prodasineura incerta là loài chuồn chuồn trong họ Platycnemididae. Loài này được Pinhey mô tả khoa học đầu tiên năm 1962.
Chú thích
Tham khảo
Prodasineura
Động vật được mô tả năm 1962 | 35,546 |
https://stackoverflow.com/questions/3270589 | StackExchange | Open Web | CC-By-SA | 2,010 | Stack Exchange | Adam S, Chris Miller, DJC, Furkan Şen, Joel Perez, Jon Skeet, Michael Borgwardt, Pi Yue, Rodrigo Rangel, Sachin Gupta, Sivaprasad M, bryndenrivers, https://stackoverflow.com/users/16883, https://stackoverflow.com/users/16945347, https://stackoverflow.com/users/19719584, https://stackoverflow.com/users/22656, https://stackoverflow.com/users/294973, https://stackoverflow.com/users/6788732, https://stackoverflow.com/users/6788733, https://stackoverflow.com/users/6788734, https://stackoverflow.com/users/6788755, https://stackoverflow.com/users/6788777, https://stackoverflow.com/users/6788788, https://stackoverflow.com/users/6788805, https://stackoverflow.com/users/6788961, rishi | English | Spoken | 694 | 1,081 | SQLite query where clause with floating point numbers fails?
I'm putting a float in an Android based SQLite database, like so:
private static final String DATABASE_CREATE =
"create table " + DATABASE_TABLE + " ("
+ KEY_ID + " integer primary key autoincrement, "
+ KEY_FLOAT + " REAL, " + ...
...
content.put(KEY_FLOAT, 37.3f);
db.insert(DATABASE_TABLE, null, content);
When I query the table:
Cursor cursor = db.query(false, DATABASE_TABLE,
new String[] { KEY_ID, KEY_FLOAT, ... },
KEY_LATITUDE + "=37.3",
null, null, null, null, null);
the cursor comes back empty. If I change the value of the float to 37.0 it works properly, returning the record in the cursor.
I've tested this at some length, changing the database column spec from "REAL" to "float", etc. AS long as I have any fractional portion after the decimal point, the cursor returns empty. What is going on?
Thanks in advance!
suggested reading: http://floating-point-gui.de/
37.0 is exactly representable in binary floating point, so you don't have any of the normal issues. 37.3 is not exactly representable - your query isn't matching because the value in the database isn't exactly 37.3.
Options:
Use a decimal-based numeric type, if such a thing exists in SQLite (I'll check in a minute)
Make your query use a tolerance, e.g. "LATITUDE > 37.29 AND LATITUDE < 37.31". Comparing binary floating point values for exact equality is just asking for trouble, I'm afraid.
EDIT: I've just checked the docs, and it looks like SQLite doesn't support decimal non-integer types. I'll leave that suggestion above as it would be relevant for the same issue in other databases though.
EDIT: Pascal's solution of effectively rolling your own numeric type is a good one in many cases. Work out what level of precision you want, and multiply/divide accordingly... so if you want 2 decimal places of precision, multiply by 100 when you store a value and then divide by 100 after fetching it. Your queries will need to be in the "multiplied" form, of course.
Thank you Jon. I've been aware of the floating-point demons but refused to accept that they could enter in at such shallow precision: I'd tried applying val = (Math.round(val*1000)) / 1000; before going into the DB, which is useless... more @ pascal.
@DJC: Just try working out how 0.1 should be represented in binary... you can get very close, but you'll never get exactly to 0.1.
Granted, it's a bad idea to compare float values for equality.
Howver, I see that SQLite uses 8-byte floating point values (which is like a DOUBLE), so it's odd that 37.0 is considered equal to 37.3. Unless you modified for your example the values used in the actual code?
You could store your LATITUDE as integers, in tenth of degree, applying the precision yourself, and converting the value on read/write...
I assumed that the "it works with 37.0" bit was having changed both parts.
Thank you pascal. I had applied your solution, multiplying values by 1000 and placing the integral portion into the DB, which works fine. Unfortunately the android.widget.SimpleCursorAdapter that I am using to populate a list for the UI is of course unaware of a need to convert the number back, so I have either to tweak that or else to shift to storing strings. I am tempted to do the latter, thinking that at my required precision they are probably comparable in storage and precision to the doubles...
Oh yes.. FYI the problems were occurring at the hundredths.. not tenths.. I had simplified the example :)
I tried with <column> BETWEEN <min> AND <max> and it worked, tolerance LATITUDE > 37.29 AND LATITUDE < 37.31 didn't work for me. But I tried BETWEEN with this hint.
My observations:
When I have a single record, greater than (>) didn't work.
If there is more than one record, greater than (>) worked perfectly.
The following thing worked for me:
Set the Column type to Double and insert rows using Double data type for Java. Somehow Double worked...
I faced the same issue. My Sqlite column was of type FLOAT. I had to cast my Java variable of type Float as double to make the comparison work.
| 11,434 |
https://azb.wikipedia.org/wiki/%D9%BE%D8%B1%D9%88%D9%82%D8%B1%D8%B3%E2%80%8C%D9%88%DB%8C%D9%84%D8%AC%D8%8C%20%D9%81%D9%84%D9%88%D8%B1%DB%8C%D8%AF%D8%A7 | Wikipedia | Open Web | CC-By-SA | 2,023 | پروقرسویلج، فلوریدا | https://azb.wikipedia.org/w/index.php?title=پروقرسویلج، فلوریدا&action=history | South Azerbaijani | Spoken | 53 | 273 | پروقرسویلج، فلوریدا (اینگیلیسجه:Progress Village, Florida ) آمریکا اؤلکهسینده بیر یاشاییش منطقهسیدیر. سوْن نۆفوس ساییمی اساسيندا ۵۳۹۲ نفر ایمیش و ۸٫۴ کیلومتر موربّع ساحهسی وار دنیز سوُلاری سَویهسیندن ۵ متر یۇکسکلیکده یئر آلیب و فلوریدا ایالتینده یئرلشیب.
قایناقلار
اینگیلیسجه ویکیپدیاسینین ایشلدنلری طرفیندن یارانمیش«Progress Village, Florida»، مقالهسیندن گؤتورولوبدور.( ۸ آقوست ۲۰۱۸ تاریخینده یوْخلانیلیبدیر).
آمریکا شهرلری | 42,655 |
https://pnb.wikipedia.org/wiki/%DA%A9%D8%B1%D9%88%D9%B9%D8%A7%D9%84%D8%B3%20%D9%BE%D9%88%D8%B3%DB%8C%D9%84%D8%B3 | Wikipedia | Open Web | CC-By-SA | 2,023 | کروٹالس پوسیلس | https://pnb.wikipedia.org/w/index.php?title=کروٹالس پوسیلس&action=history | Western Punjabi | Spoken | 14 | 47 | کروٹالس پوسیلس وائپر سپ دے ٹبر دا اک سپ اے۔
بارلے جوڑ
وائپر سپ | 36,406 |
https://pt.wikipedia.org/wiki/Arno%20Klare | Wikipedia | Open Web | CC-By-SA | 2,023 | Arno Klare | https://pt.wikipedia.org/w/index.php?title=Arno Klare&action=history | Portuguese | Spoken | 66 | 134 | Arno Klare (nascido em 1 de fevereiro de 1952) é um político alemão. Nasceu em Oberhausen, na Renânia do Norte-Vestfália, e representa o SPD. Arno Klare é membro do Bundestag pelo estado da Renânia do Norte-Vestfália desde 2013.
Vida
Ele tornou-se membro do bundestag após as eleições federais alemãs de 2013. É membro do Comité de Transporte e Infraestrutura Digital.
Políticos do Partido Social-Democrata da Alemanha | 15,096 |
https://es.wikipedia.org/wiki/Municipio%20de%20Fossum%20%28condado%20de%20Norman%2C%20Minnesota%29 | Wikipedia | Open Web | CC-By-SA | 2,023 | Municipio de Fossum (condado de Norman, Minnesota) | https://es.wikipedia.org/w/index.php?title=Municipio de Fossum (condado de Norman, Minnesota)&action=history | Spanish | Spoken | 167 | 274 | El municipio de Fossum (en inglés: Fossum Township) es un municipio ubicado en el condado de Norman en el estado estadounidense de Minnesota. En el año 2010 tenía una población de 156 habitantes y una densidad poblacional de 1,67 personas por km².
Geografía
El municipio de Fossum se encuentra ubicado en las coordenadas . Según la Oficina del Censo de los Estados Unidos, el municipio tiene una superficie total de 93.32 km², de la cual 92,43 km² corresponden a tierra firme y (0,95 %) 0,89 km² es agua.
Demografía
Según el censo de 2010, había 156 personas residiendo en el municipio de Fossum. La densidad de población era de 1,67 hab./km². De los 156 habitantes, el municipio de Fossum estaba compuesto por el 93,59 % blancos, el 2,56 % eran amerindios y el 3,85 % eran de una mezcla de razas. Del total de la población el 0 % eran hispanos o latinos de cualquier raza.
Referencias
Enlaces externos
Municipios de Minnesota
Localidades del condado de Norman | 39,555 |
https://en.wikipedia.org/wiki/Texas%20Medal%20of%20Honor%20Memorial | Wikipedia | Open Web | CC-By-SA | 2,023 | Texas Medal of Honor Memorial | https://en.wikipedia.org/w/index.php?title=Texas Medal of Honor Memorial&action=history | English | Spoken | 1,247 | 2,027 | The Texas Medal of Honor Memorial is a statue commemorating recipients of the Medal of Honor from the state of Texas. Sculpted by Doyle Glass and Scott Boyer, it was dedicated on Memorial Day of 2008 in Midland, Texas at the Commemorative Air Force International Headquarters. In July 2018 the Memorial was assigned to the Ground Forces Detachment of the Commemorative Air Force and moved to Gainesville, Texas. Since May of 2021 it is currently on display at the International Artillery Museum in Saint Jo, Texas and open to the public.
The Memorial depicts Medal of Honor recipient George H. O'Brien, Jr. as he would appear on the day he earned the Medal of Honor for his service during the Korean War. Perched on a rock, the heroic-sized bronze figure rises above a granite base, which displays the names of most recipients of the medal from Texas. The Model for this statue was Buck Hartlage a Louisville Kentucky native.
Other Recipients of the Medal of Honor from Texas
Indian Wars
John Connor – Army, Galway, Ireland Jefferson, TX
James B. Dozier – Army, Warren County, TN Fort Richardson, TX
Pompey Factor – Army, Arkansas Fort Duncan, TX
Robert Lee Howze – Army, Overton, TX Overton, TX
George Loyd (or Lloyd) – Army, County Tyrone, Ireland Canton, TX
William McCabe – Army, Belfast, Ireland Fort Duncan, TX
Franklin M. McDonald – Army, Bowling Green, KY Fort Griffin, TX
Adam Paine – aka: Adan Payne – Army, Florida Fort Duncan, TX
Isaac Payne – Army, Mexico Fort Duncan, TX
John Ward (or John Warrior) – Army, Arkansas Fort Duncan, TX
Spanish–American War
George Frederick Phillips – Navy, Saint John, NB, Canada Galveston, TX
Philippine–American War
George Mathews Shelton – Army, Brownwood, TX Bellington, TX
Mexican Campaign
William Kelly Harrison – Navy, Waco, TX Texas
World War I
David B. Barkeley, aka: David Bennes Barkley – Army, Laredo, TX San Antonio, TX
Daniel Richmond Edwards – Army, Mooreville, TX Bruceville, TX
David Ephraim Hayden – Navy, Florence, TX Texas
World War II
Lucian Adams – Army, Port Arthur, TX Port Arthur, TX
William James Bordelon – USMC, San Antonio, TX Texas
John Duncan Bulkeley – Navy, San Antonio, TX Texas
Horace Seaver Carswell, Jr. – Army AC, Fort Worth, TX San Angelo, TX
Robert George Cole – Army, Fort Sam Houston, TX San Antonio, TX
Samuel David Dealey – Navy, Dallas, TX Texas
Forrest Eugene Everhart, Sr. – Army, Bainbridge, OH Texas City, TX
James H. Fields – Army, Caddo, TX Houston, TX
Thomas Weldon Fowler – Army, Wichita Falls, TX Wichita Falls, TX
Marcario Garcia – Army, Villa De Castano, Mexico Sugar Land, TX
William George Harrell – USMC, Rio Grande City, TX Mercedes, TX
James Lindell Harris – Army, Hillsboro, TX Hillsboro, TX
William Deane Hawkins – USMC, Fort Scott, KS El Paso, TX
Lloyd Herbert Hughes – Army AC, Alexandria, LA San Antonio, TX
Johnnie David Hutchins – Navy, Weimer, TX Texas
Neel Ernest Kearby – Army AC, Wichita Falls, TX Dallas, TX
George D. Keathley – Army, Olney, TX Lamesa, TX
Truman Kimbro – Army, Madisonville, TX Houston, TX
Jack Llewellyn Knight – Army, Garner, TX Weatherford, TX
Raymond Larry Knight – Army AC, Houston, TX Houston, TX
Turney White Leonard – Army, Dallas, TX Dallas, TX
James Marion Logan – Army NcNeil, TX Luling, TX
Jose Mendoze Lopez – Army, Mission, TX Brownsville, TX
Jack Lummus – USMC, Ennis, TX Texas
Jack Warren Mathis – Army AC, San Angelo, TX San Angelo, TX
Audie Leon Murphy – Army, Kingston, TX Dallas, TX
Charles Howard Roan – USMC, Claude, TX Texas
James E. Robinson, Jr. – Army, Toledo, OH Waco, TX
Cleto L. Rodriguez – Army, San Marcos, TX San Antonio, TX
Herman C. Wallace – Army, Marlow, OK Lubbock, TX
Eli Lamar Whiteley – Army, Florence, TX Georgetown, TX
Korean War
George Andrew Davis, Jr. – USAF, Dublin, TX Lubbock, TX
Ambrosio Guillen – USMC, La Junta, CO El Paso, TX
Jack G. Hanson – Army, Escatawpa, MS Galveston, TX
John Edward Kilmer – Navy, Highland Park, IL Houston, TX
Benito Martinez – Army, Fort Hancock, TX Fort Hancock, TX
Frank Nicias Mitchell – USMC, Indian Gap, TX Roaring Springs, TX
Whitt Lloyd Moreland – USMC, Waco, TX Austin, TX
George Herman O'Brien, Jr. – USMC, Fort Worth, TX Big Spring, TX
Charles F. Pendleton – Army, Camden, TN Fort Worth, TX
James Lamar Stone – Army, Pine Bluff, AR Houston, TX
Travis E. Watkins – Army, Waldo, AR Texas
Vietnam War
Richard Allen Anderson – USMC, Washington, DC Houston, TX
Roy Perez Benavidez – Army, Cuero, TX Houston, TX
Thomas Elbert Creek – USMC, Joplin, MO Amarillo, TX
Alfredo "Freddy" Gonzalez – USMC, Edinburg, TX San Antonio, TX
Robert David Law – Army, Fort Worth, TX Dallas, TX
Milton Arthur Lee – Army, Shreveport, LA San Antonio, TX
Finnis Dawson McCleery – Army, Stephenville, TX San Angelo, TX
David Herbert McNerney – Army, Lowell, MA Fort Bliss, TX
Clarence Eugene Sasser – Army, Chenango, TX Houston, TX
Russell Albert Steindam – Army, Austin, TX Austin, TX
Alfred Mac Wilson – USMC, Olney, IL Abilene, TX
Marvin Rex Young – Army, Alpine, TX Odessa, TX
Texas-born honorees based elsewhere
The following Medal of Honor Recipients were born in the State of Texas, but either moved to or enlisted in military service in another state to which their award is accredited.
William Grafton Austin – Army, Galveston, TX New York, NY – Indian Wars
John McLennon – Army, Fort Belknap, TX Fort Ellis, MT – Indian Wars
Samuel M. Sampler – Army, Decatur, TX Altus, OK – World War I
Silvestre Santana Herrera – Army, El Paso, TX Phoenix, AZ – World War II
John Riley Kane – Army AC, McGregor, TX Shreveport, LA – World War II
John Cary "Red" Morgan – Army AC, Vernon, TX London, England – World War II
George Benton Turner – Army, Longview, TX Los Angeles, CA – World War II
Oscar Palmer Austin – USMC, Nacogdoches, TX Phoenix, AZ – Vietnam War
Steven Logan Bennett – USAF, Palestine, TX Lafayette, LA – Vietnam War
Frederick Edgar Ferguson – Army, Pilot Point, TX Phoenix, AZ – Vietnam War
Terrence Collinson Graves – USMC, Corpus Christi, TX New York, NY – Vietnam War
Miguel Keith – USMC, San Antonio, TX Omaha, NE – Vietnam War
Exception
One notable exception to the list may be the earliest Texan recipient, Milton M. Holland. He was born a Texas slave in 1844 and served with the Fifth U.S. Colored Troops during the Civil War. On September 29, 1864, during an attack in which all his unit's officers were killed or disabled, Sergeant Major Holland took over despite his own wounds, and led his comrades to take the Confederate position near Richmond, Virginia. He was awarded the Medal of Honor on April 6, 1865, and at his death in 1910 was interred at Arlington National Cemetery.
See also
Medal of Honor Memorial (Indianapolis)
Kentucky Medal of Honor Memorial
Oregon Veterans Medal of Honor Memorial
References
External links
Texas Medal of Honor Memorial
Commemorative Air Force
Doyle Glass Sculpture website
Lions of Medin website
War Hero Memorials
Kentucky Medal of Honor Memorial
Kentucky Medal of Honor Memorial (Kentucky Educational Television)
Medal of Honor
Military monuments and memorials in the United States
Monuments and memorials in Texas
Midland, Texas
2008 sculptures
Bronze sculptures in Texas
Statues in Texas
Sculptures of men in Texas
2008 establishments in Texas | 6,663 |
https://ja.wikipedia.org/wiki/%E3%82%B7%E3%82%A8%E3%83%BC%E3%83%8A%E7%9C%8C | Wikipedia | Open Web | CC-By-SA | 2,023 | シエーナ県 | https://ja.wikipedia.org/w/index.php?title=シエーナ県&action=history | Japanese | Spoken | 68 | 828 | シエーナ県（シエーナけん、）は、イタリア共和国トスカーナ州の県の一つ。県都はシエーナ（シエナ）。カナ転記としては「スィエーナ」が現地音に近い。
地理
位置・広がり
トスカーナ州東南部に位置し、南東にラツィオ州およびウンブリア州と境界を接する。県都シエーナは県域の北東部に位置し、州都フィレンツェから南へ約50km、グロッセートから北北東へ約64km、ペルージャから西北西へ約89km、首都ローマから北北西へ約183kmの距離にある。
隣接する県は以下の通り。
北 - フィレンツェ県
北東 - アレッツォ県
南東 - ヴィテルボ県 （ラツィオ州）、テルニ県、ペルージャ県（以上ウンブリア州）
南西 - グロッセート県
北西 - ピサ県
県内の地域と主要な都市
2001年の国勢調査に基づく居住地区（）別人口統計によれば、人口5000人以上の都市は以下の通り。
シエーナ - 43,515人
ポッジボンシ - 21,369人
コッレ・ディ・ヴァル・デルザ - 15,243人
アッバディーア・サン・サルヴァトーレ - 6,652人
シナルンガ - 6,123人
キアンチャーノ・テルメ - 6,065人
トッリータ・ディ・シエーナ - 5,065人
行政区画
シエーナ県には36のコムーネが属する。主要なコムーネ（人口8000人以上）は下表の通り。左端の数字はISTATコードを示す。人口は2011年1月1日現在。
右の地図中の番号は、コムーネのISTATコード下3桁を示す。下表に掲げた主要なコムーネのうち、地図中に名称を記さなかったものについては、番号を太字で示した。
21世紀に入って以降、以下のコムーネ統廃合が行われている。
2017年
サン・ジョヴァンニ・ダッソがモンタルチーノに編入
文化・観光
県には以下のユネスコ世界遺産がある。
ピエンツァ市街の歴史地区 （ピエンツァ）
サン・ジミニャーノ歴史地区 （サン・ジミニャーノ）
シエーナ歴史地区 （シエーナ）
ヴァル・ドルチャ
脚注
関連項目
ピチ - 当地が発祥とされる太麺パスタ
外部リンク
県公式サイト
イタリアの県
トスカーナ州 | 3,641 |
https://arz.wikipedia.org/wiki/%D8%A7%D8%AF%D9%8A%D8%AB%20%D8%B1%D9%88%D8%B2%D9%81%D9%84%D8%AA | Wikipedia | Open Web | CC-By-SA | 2,023 | اديث روزفلت | https://arz.wikipedia.org/w/index.php?title=اديث روزفلت&action=history | Egyptian Arabic | Spoken | 42 | 110 | اديث روزفلت كانت سياسيه من امريكا.
حياتها
اديث روزفلت من مواليد يوم 6 اغسطس سنة 1861 فى نيو يورك.
وفاتها
اديث روزفلت ماتت فى 30 سبتمبر سنة 1948.
لينكات برانيه
مصادر
سياسيه من امريكا
وفيات 1948
مواليد 1861
مواليد فى نيو يورك | 42,157 |
https://stackoverflow.com/questions/52389392 | StackExchange | Open Web | CC-By-SA | 2,018 | Stack Exchange | Clayton Lewis, https://stackoverflow.com/users/8390735 | English | Spoken | 130 | 254 | File Rename With Date (-1 Day)
I'm writing a PowerShell script to rename a file, with the ending being yesterdays date in yyyymmdd.
I'm currently adjusting the date manually each day to be yesterday's date, but is someone able to advise how I could go about scripting this in? I've tried using Get-Date and AddDays(-1) but I'm receiving errors.
gci D:\Temp\ |
? {$_.Name -match 'Example'} |
Rename-Item -NewName {$_.Name -replace 'Example','Example_Example_20180917'}
What error are you getting? I think you will also need to make sure the date you have it using the [DateTime] format to be able to use Get-Date
If I understood your question correctly, you need to format your [datetime] object as a string.
Here is a basic example:
$date = (Get-Date).AddDays(-1)
Rename-Item -Path C:\Example.txt -NewName "C:\Example_$($date.ToString("yyyyMMdd")).txt"
| 32,893 |
https://stackoverflow.com/questions/54022741 | StackExchange | Open Web | CC-By-SA | 2,019 | Stack Exchange | Pork Chop, https://stackoverflow.com/users/2753526, https://stackoverflow.com/users/6059239, saurabh shah | English | Spoken | 116 | 345 | Shiny on Ubuntu machine
[2019-01-03T18:20:08.983] [INFO] shiny-server - Shiny Server v1.5.9.923 (Node.js v8.11.3)
[2019-01-03T18:20:08.986] [INFO] shiny-server - Using config file "/etc/shiny-server/shiny-server.conf"
[2019-01-03T18:20:09.047] [WARN] shiny-server - Running as root unnecessarily is a security risk! You could be running more securely as non-root.
[2019-01-03T18:20:09.055] [INFO] shiny-server - Starting listener on http://[::]:3838
[2019-01-03T18:20:09.060] [ERROR] shiny-server - HTTP server error (http://[::]:3838): listen EAFNOSUPPORT :::3838
[2019-01-03T18:20:09.060] [INFO] shiny-server - Shutting down worker processes
This is the shiny server status:
shiny-server stop/waiting
How can I resolve this problem?
what version of ubuntu are you using?, also have a look at this https://github.com/rstudio/shiny-server/issues/153
using Ubuntu 14.04 version.
I tried the steps given in the link you have mentioned but it did not work
| 16,460 |
oUlOzSDUj1I_1 | Youtube-Commons | Open Web | CC-By | null | Yours Truly, Johnny Dollar - 560919 479 The Imperfect Alibi Matter - Ep 3 | None | English | Spoken | 1,885 | 2,296 | From Hollywood it's time now for Johnny dollar Joe Rastelli at homicide Johnny I was out when you called anything new on the Harvey stone killing Joe not a thing But maybe we've already got all we need meaning Helen Barrett. We're still holding her Joe. I don't think she did it No, oh, I know it all adds up to her, but we'll just call it a hunch I'm sure they're fine Johnny, but facts are better. You want to hear some facts? I'll be right over Tonight and every weekday night Bob Bailey in the transcribed adventures of the man with the action-packed expense account America's fabulous freelance insurance investigators truly Johnny dollar Expense account submitted by special investigator Johnny dollar to the home office northeast and dem to the associates Hartford, Connecticut Assignment the imperfect alibi matter location New York City Expense account continued Item 8 a dollar 40 cab fare to police headquarters from my hotel to talk to lieutenant Joe Rastelli Sit on Johnny. Thanks Facts you said facts number one Harvey stone was shot in the left side of the forehead at closed range with the 38 caliber Smith and Wesson the gun was near the body Any prints on it? No, it was clean, but Helen Barrett had gloves with it Helen says she left Harvey's apartment and went to her own to pack up They were going to a lope when she got back to his apartment. He was dead, so she told me Johnny I'd like to believe her too. She seems like a pretty nice kid, but but what not enough facts in her favor Who saw her leave stones apartment? We can't find anyone who did what time she leaves. She can't remember Did anyone see her return? What time? That's a lot of questions not to be able to answer Johnny Yeah, yeah, I know What was the time of death medical examiner figures at somewhere between 1130 and midnight? Well Helen told me she thought it was about 11 when she left Harvey's apartment and about midnight when she returned Yeah about Even if she did leave she only lives a few blocks away. It's a lot of time unaccounted for it Johnny. Yeah Better fill me in on what you know Well as I get it Harvey stone took over the management of his father's corporation when old E.J Took to a wheelchair about a year ago. Yeah. Yeah, I know Two years ago old E.J married an ex chorus girl named deaf. Yeah. Yeah Well, she's about Harvey's age The two of them were apparently pretty friendly and the old man was bothered by it occasionally Incidentally he and Daphne are joint beneficiaries on Harvey's insurance policy 150,000 worth sounds like you're trying to tie the old man into the killing Be quite a stretch Johnny from his wheelchair up in westchester county to harvey's apartment on east 57 I know but right now i'm more interested in Daphne. Oh I told you last night. I thought dutch kreger was mixed up in this I went to his office afterward and spotted a picture of Daphne stone on his wall Inscription all my love Signed Daphne You think Daphne got dutch to do her and himself a favor, huh? That's a possibility, isn't it? Sure sure it's a possibility Trouble is there are all kinds of possibilities Right now. I got a stick with a probability. Helen Barrett. Uh-huh How are you doing on motive for her? Not good. Not bad We know there was some question as to whether they were going to be married or not Helen says the hesitation was on her part But suppose it was the other way around harvey decided not to go through with a marriage. Yeah Getting cut out of the stone money would hurt some girls plenty Maybe this was her way of getting even with him for breach of promise You know joe for a guy who loves facts seems to me you're edging over into hunches, too I admit it isn't a closed case by a long shot So let's get back to facts Harvey was shot in the forehead with a 38 smith and wesson sometime between 11 30 Excuse me rostelli speaking who What about Oh, we'll send him in Somebody wants to see me about the killing. Oh Your lieutenant rostelli that's right Your hand on the stone killing trying to I want to talk to you about it. Sit down. Thank you. What's your name? Gentry alvin gentry. So what about the stone killing? I killed harvey stone. What? Let's have that again I said I killed stone. I want to make a statement. Why did you kill him? He's making a play for my girl. I didn't like it your girl. You mean helen barrett Who helen barrett harvey stones fiance. No, I don't know her. I mean my girl daris a hatchet girl at barnes. Go on go on The stone was on to make for her every time he came in barnes. He'd make a play for her I told him to lay off and he wouldn't He asked her to go away with him I went to his apartment. I killed him. How'd you kill him? I shot him where I told you in his apartment. I mean where did the bullet hit him? Oh In the chest. What kind of gun do you use? 45 cold. What'd you do with the gun? I threw it in the river Okay, gentry get out What I said get out But I told you yeah, you told me all right now. I'm telling you get out Look, I don't understand. I'll tell you what you do You just go on out of here and think it over when you come back with a few facts straight facts Yeah, like the caliber of the gun and where stone was shot in the location of the gun You get the fact straight and I'll be glad to listen to you now get out Okay Confessing sam number one. Yeah, there's always a string of them That's one reason we don't usually release the caliber of the gun to the papers to help weed out these confessing sands Wonder why they do it Ah, psychiatrist was explaining it to me once something to do with repressed feelings of guilt. I think he said Next one will probably say he stabbed harvey stone with a letter opener. Yeah Well, I'm gonna run on to have a talk with Daphne Stully speaking All right Now look mike you take the statement, huh? Thanks Well, I was wrong about the letter opener johnny. Oh We got a guy now who claims he used a razor on harvey slittest throat from ear to ear As I left I spent about three minutes feeling sorry for rostelli and his crank confessions But then I dropped that routine and started feeling sorry for my own problems The case against hellen barrett looked pretty bad, but I still kept thinking of daphne stone's picture in dutch kreger's office Expense account item 9 320 transportation to the stoner state in westchester coddy I was shown into the king size drawing room again to wait for daphne But then I saw a very interesting site that wiped daphne out of my mind for a moment It was old ej's wheelchair at the door to the solarium And what was unusual about it was that it was empty I edged toward the door then I got a glimpse of ej puttering around his orchids He spotted me though and hobbled quickly to his wheelchair With an abrupt wave he wheeled into the hall and out of sight A couple of minutes later in came daphne Hello johnny daphne You said it was important that you talk to me, but I really don't feel much like talking after What's happened? I'm sure you understand. I think so How's ej taking it my husband is reacting as I suppose any father would who'd just lost his son Be well and hurt You didn't tell me ej could navigate without his wheelchair I saw him a minute ago inspecting his orchids The wheelchair was parked near the door I I didn't think it was important johnny It's true. He can be out of his chair for short periods, but it's rather uncomfortable for him out of his chair for how long Not long enough to get into new york and back if that's what you're wondering Thanks You told me was ej who was opposed to harvey's plans to mary ellen barrett But I found out that you were the one who was fighting it I suppose it was foolish of me to pretend otherwise I guess I just didn't want you to get any wrong ideas about what about the reason I opposed it. What's the right idea? The name of stone means something johnny Dignity tradition breeding I doubt if someone like ellen barrett an entertainer nice as she is Could keep that tradition alive Are you kidding? I'm completely serious Something like this happened once before with harvey secretary martha winters And you stopped it just like you were trying to stop him from marrying ellen. I don't like the way you put that I merely persuaded him to think of the family name You know you kill me deafney What do you mean by that this dose of blue blood you've picked up aren't you a deafney come later yourself? How dare you save it Ej told me he lifted you out of a chorus line when he married you Now how about it? Yes, it's true. So where do you get off with this? I don't suppose you'd ever understand this johnny But there are chorus girls and chorus girls This I know I had to support my mother somehow But all the while I knew that life wasn't for me And when I got a chance at this life, I took it And since I married edward I've lived the way anyone with the name of stone should live I've put my past behind me Even dutch crigger dutch. Yeah Yeah, I saw your picture on his office wall He was part of the past it doesn't exist anymore Isn't it kind of a coincidence? He tried to worm his way into one of harvey's business deals I had nothing to do with it And harvey acted correctly in refusing to have anything to do with dutch. I see Then you opposed harvey's marriage to ellen to protect the family name, huh? Just as I opposed the previous attachment to his secretary Sure, it wasn't because you didn't like the idea of Competition It's a pretty low thing to say under the circumstances. Well, just what are the circumstances It's It's very simple I've lost someone who was Very dear to me Even though I was harvey's stepmother. We were practically the same age Sure I know people talked about it made crude jokes about it, but I didn't care Because I found in harvey something I'd never had in my life before. | 28,298 |
https://stackoverflow.com/questions/6804125 | StackExchange | Open Web | CC-By-SA | 2,011 | Stack Exchange | Darc, Dor, El Yobo, Ibu, Jared Farrish, MD NAZMUL ISLAM, Marcel Brozek, Nargess Akbari, Nathan, OMG Ponies, Seçil , TehShrike, Trinovalino Risamasu, direct00, dulke, genesis, https://stackoverflow.com/users/135152, https://stackoverflow.com/users/14884336, https://stackoverflow.com/users/14884337, https://stackoverflow.com/users/14884338, https://stackoverflow.com/users/14884362, https://stackoverflow.com/users/14884430, https://stackoverflow.com/users/14884461, https://stackoverflow.com/users/14884470, https://stackoverflow.com/users/14884479, https://stackoverflow.com/users/14884491, https://stackoverflow.com/users/201789, https://stackoverflow.com/users/207812, https://stackoverflow.com/users/217588, https://stackoverflow.com/users/286757, https://stackoverflow.com/users/451969, https://stackoverflow.com/users/560299, https://stackoverflow.com/users/635129, https://stackoverflow.com/users/764846, razirazrazraz, sean-7777 | English | Spoken | 496 | 864 | Limiting a SQL Select query
I am connecting to an SQL database in my PHP script and am having trouble with the LIMIT command:
$result = mysql_query("
SELECT *
FROM product
WHERE `category` like \"" . $_GET['category'] . "\"
LIMIT 0, 16
");
This all works, except that if I only have 10 rows then $result contains rows 0~10 and then 0~6 as well.
I am using a a while loop while($row = mysql_fetch_assoc($result)) to check if there is a result and then run an action. Is there any way of having it limit the select statement to only show rows 0~10?
please type sql injection in the search field
You should NEVER inject unwashed text from "unknown" sources (read: anything from the GET/POST/COOKIE globals) into a SQL query. Either use mysql_real_escape_string() on the variable, or use PDO prepared statements.
Also, you should avoid using MySQL's non-standard double quote for strings; it does support the proper usage of single quotes, you might as well use standards compliant code if you can.
I can't understand what is the problem?
@Dor: The OP is claiming that MySQL is padding the results to make 16 rows when there's only 10 -- the first six are repeated. I've never seen that behavior, personally...
@OMG Ponies - it's possible the LIKE statement is returning unexpected results
@direct00: That's more plausible than what I understand the OP to believe is happening.
$result = mysql_query("SELECT * FROM product
WHERE `category` like '" . mysql_real_escape_string($_GET['category']) . "' LIMIT 0, 10");
is it what are you looking for? It will give you ten rows maximally..
Additionally, please read this article about SQLi
including the appropriate reference to mysql_real_escape_string(), which in my opinion should be part of the answer... :)
A better idea would be to use prepared statements (but I digress).
Not exactly... I know about the edit timeout. ;) @rfw - Please see my comment below the question. :)
@rfw: I do not use any prepared statements and I never saw reason why would I do this
@JaredFarrish: not exactly? what do you think?
http://stackoverflow.com/questions/1742066/why-is-pdo-better-for-escaping-mysql-queries-querystrings-than-mysql-real-escape
@genesis: It looks prettier (so you don't have to stare at code with concatenations everywhere), and you can more easily see what you're doing. :)
@rfw: I do not like it because my logs has double size as normal :p. Btw how does it know that it should be int/str?
@rfw: Not necessarily, Prepared Statements requires an extra round-
trip to the server.
@Dor: It's a lot more useful if you're doing bulk operations with the same query, I guess.
@rfw: so in case you're inserting 20000 lines into table, right?
@genesis: Yes, but I prefer using prepared statements over sanitization of data since it's less human error prone.
This is what i was already doing, didnt notice i had inserted the data twice. Thanks to everyone for pointing out the security issues in this, i hadn't even considered that.
"Why you should not be using mysqli::prepare" - http://joshduff.com/270/why-you-should-not-be-using-mysqli-prepare
| 1,608 |
https://my.wikipedia.org/wiki/%E1%80%9B%E1%80%BD%E1%80%BE%E1%80%94%E1%80%BA%E1%80%B8%E1%80%90%E1%80%AD%E1%80%AF%E1%80%81%E1%80%AF%E1%80%99%E1%80%AD%E1%80%81%E1%80%BB%E1%80%AD%E1%80%98%E1%80%B0%E1%80%90%E1%80%AC | Wikipedia | Open Web | CC-By-SA | 2,023 | ရွှန်းတိုခုမိချိဘူတာ | https://my.wikipedia.org/w/index.php?title=ရွှန်းတိုခုမိချိဘူတာ&action=history | Burmese | Spoken | 15 | 186 | ရွှန်းတိုခုမိချိဘူတာ (俊徳道駅 ရွှန်းတိုခုမိချိ- အဲခိ ) သည် ဂျပန်နိုင်ငံ၊ အိုဆာကာခရိုင်၊ ဟိဂရှိအိုဆာကာမြို့တွင် တည်ရှိသည့် ဘူတာတစ်ခု ဖြစ်သည်။
ကိုးကား
ဂျပန်နိုင်ငံရှိ ဘူတာရုံများ | 18,901 |
https://travel.stackexchange.com/questions/86006 | StackExchange | Open Web | CC-By-SA | 2,017 | Stack Exchange | Gayot Fow, https://travel.stackexchange.com/users/1737 | English | Spoken | 89 | 115 | Does an Indian need a visa to visit the Schengen area as a tourist?
I am an Indian passport holder but Permanent Resident (PR) of Australia.
Do I need a visa for Switzerland and France to visit as a tourist?
Yes you need a visa. Did you attempt to research this?
Yes, you will need a Schengen visa. No special exemptions exist for holders of any other visas or permanent residencies, however it should be fairly straightforward to receive a Schengen visa as a PR of a stable country.
| 3,951 |
https://zh.wikipedia.org/wiki/%E6%B2%99%E7%BA%B3%E8%8E%B1%E8%80%B6 | Wikipedia | Open Web | CC-By-SA | 2,023 | 沙纳莱耶 | https://zh.wikipedia.org/w/index.php?title=沙纳莱耶&action=history | Chinese | Spoken | 15 | 286 | 沙纳莱耶（，；）是法国上卢瓦尔省的一个市镇，属于布里尤德区。
地理
（）面积，位于法国奥弗涅-罗讷-阿尔卑斯大区上盧瓦爾省，该省份为法国中南部省份，北起多姆山省和卢瓦尔省，西接康塔爾省，南至洛泽尔省，东临阿尔代什省。
与接壤的市镇（或旧市镇、城区）包括：。
的时区为UTC+01:00、UTC+02:00（夏令时）。
行政
的邮政编码为，INSEE市镇编码为。
政治
所属的省级选区为。
人口
于时的人口数量为人。
参见
上卢瓦尔省市镇列表
参考文献
上卢瓦尔省市镇 | 5,611 |
https://stackoverflow.com/questions/10235952 | StackExchange | Open Web | CC-By-SA | 2,012 | Stack Exchange | Cheo Cheang Ming, D. Vedha Leo Adithya, Hajo, Katrina Wong, dan-lee, https://stackoverflow.com/users/1342021, https://stackoverflow.com/users/23482130, https://stackoverflow.com/users/23482131, https://stackoverflow.com/users/23482132, https://stackoverflow.com/users/23482840, https://stackoverflow.com/users/612202, xu ban | English | Spoken | 139 | 185 | How do I display all categories on a page with a corresponding image?
I'm trying to display a list of all the categories from a product custom post on a "product" page with thumbnail images. What would the code be to do that and could I use the category description as the image source for the thumbnail?
I'm not sure what code I should show but I have a wordpress theme I'm trying to modify. I need a page that says "Product Categories" That displays all the different categories of products I've tagged. So I tagged say "bikes" and "cars" as "transportation" I need to display transportation with a thumbnail that represents transportation that links to the category page.
you should provide additional details, e.g. the wordpress plugin used or sourcecode samples.
This question is so out of context.
| 9,182 |
https://en.wikipedia.org/wiki/Pseudopleospora | Wikipedia | Open Web | CC-By-SA | 2,023 | Pseudopleospora | https://en.wikipedia.org/w/index.php?title=Pseudopleospora&action=history | English | Spoken | 54 | 113 | Pseudopleospora is a genus of fungi in the class Dothideomycetes. The relationship of this taxon to other taxa within the class is unknown (incertae sedis). A monotypic genus, it contains the single species Pseudopleospora ruthenica.
See also
List of Dothideomycetes genera incertae sedis
References
External links
Index Fungorum
Enigmatic Dothideomycetes taxa
Monotypic Dothideomycetes genera | 5,127 |
https://codegolf.stackexchange.com/questions/93509 | StackExchange | Open Web | CC-By-SA | 2,016 | Stack Exchange | AdmBorkBork, Bassdrop Cumberwubwubwub, CAD97, DLosc, Jordan, KarlKastor, Luis Mendo, Neil, Suever, Wheat Wizard, Yytsi, https://codegolf.stackexchange.com/users/11261, https://codegolf.stackexchange.com/users/16766, https://codegolf.stackexchange.com/users/17602, https://codegolf.stackexchange.com/users/29325, https://codegolf.stackexchange.com/users/36398, https://codegolf.stackexchange.com/users/41257, https://codegolf.stackexchange.com/users/4162, https://codegolf.stackexchange.com/users/41754, https://codegolf.stackexchange.com/users/4198, https://codegolf.stackexchange.com/users/42963, https://codegolf.stackexchange.com/users/51825, https://codegolf.stackexchange.com/users/51939, https://codegolf.stackexchange.com/users/56557, https://codegolf.stackexchange.com/users/56656, manatwork, orlp, sergiol | English | Spoken | 2,581 | 6,217 | Recreate the Windows ME screensaver as ASCII
This challenge is inspired from this answer at the Ask Ubuntu Stack Exchange.
Intro
Remember the Windows ME screensaver with the pipes? Time to bring the nostalgia back!
Challenge
You should write a program or function which will output an ASCII representation of the screensaver. In the screensaver there should be a single pipe which will grow in semi-random directions.
The start of the pipe will be randomly placed at any of the borders of the screen and the pipe piece should be perpendicular to the border (corner first-pipes can either be horizontal or vertical). Each tick the pipe will grow in the direction it is facing (horizontal/vertical) at an 80% chance or take a corner at a 20% chance.
Pipe representation
To create the pipe, 6 unicode characters will be used
─ \u2500 horizontal pipe
│ \u2502 vertical pipe
┌ \u250C upper left corner pipe
┐ \u2510 upper right corner pipe
└ \u2514 lower left corner pipe
┘ \u2518 lower right corner pipe
Input
The program / function will take 3 values of input, which can be gathered through function parameters or prompted to the user.
Amount of ticks
Screen width
Screen height
Amount of ticks
For every tick, a piece of pipe will be added to the screen. Pipes will overwrite old pipe pieces if they spawn at the same position.
For example, take a screen of size 3x3
ticks == 3
─┐
┘
ticks == 4
─┐
└┘
ticks == 5
│┐
└┘
Whenever a pipe exits the screen, as in the last example at 5 ticks, then a new pipe will spawn at a random border. For example:
ticks == 6
│┐
└┘
─
The new pipe should have a 50% chance of being horizontal or vertical.
Screen width/height
The screen width and height can be combined into a single value if that's preferrable in your language of choice. The screen width and height will always have a minimum value of 1 and a maximum value of 255. If your language of choice supports a console or output screen which is smaller than a 255x255 grid of characters then you may assume that the width and height will never exceed the boundaries of your console. (Example: Windows 80x25 cmd window)
Output
The output of your program/function should be printed to the screen, or returned from a function. For every run of your program, a different set of pipes should be generated.
Test cases
The following test cases are all random examples of valid outputs
f(4, 3, 3)
│
─┘
│
f(5, 3, 3)
│
─┘┌
│
f(6, 3, 3)
─│
─┘┌
│
f(7, 3, 3)
──
─┘┌
│
Obviously, the more ticks that have occured, the harder it becomes to prove the validity of your program. Hence, posting a gif of your output running will be preferred. If this is not possible, please post a version of your code which includes printing the output. Obviously, this will not count towards your score.
Rules
This is code-golf, shortest amount of bytes wins
Standard loopholes apply
If you use the unicode pipe characters in your source code, you may count them as a single byte
This is quite a hard challenge that can possibly be solved in many creative ways, you are encouraged to write an answer in a more verbose language even though there are already answers in short esolangs. This will create a catalog of shortest answers per language. Bonus upvotes for fancy coloured gifs ;)
Happy golfing!
Disclaimer: I am aware that Unicode characters aren't ASCII, but in lack of a better name I just call it ASCII art. Suggestions are welcome :)
The unicode characters that you want in the output are not ASCII.
@Wheat Wizard I know, I use different terms throughout this post. However, the collection of text characters to create art can be called ASCII art (frankly, I didn't know what else to call it).. Edited in a disclaimer xD
I think this should be tagged ascii-art instead of graphical-output -- reference
Nostalgia and Windows ME don't fit well on the same line
@TimmyD you are right, updated the tags
I sort of wish this were [tag:popularity-contest] so I could go a little overboard on solving this, but I'm definitely taking a look at this this weekend. (Golf is cool too, of course!)
@CAD97 That was my initial choice but those are generally frowned upon, couldn't really think of solid winning criterium for such a challenge either, other than "make it look fancy"
The 3D Pipes screensaver predated Windows ME.
"The start of the pipe will be randomly placed at any of the borders of the screen." Must the first pipe be perpendicular to the edge of the screen?
@Jordan yes, the first pipe should always "come out of the border", the direction of corner first-pipes doesn't matter
@BassdropCumberwubwubwub Do we have any options if our language doesn't support unicode? i.e. substituting ASCII chars
@Suever You could substitute the unicodes with (any) ascii characters to demonstrate your language but it would make the answer non-competing
“Whenever a pipe exits the screen (…), then a new pipe will spawn at a random border.” In the animation you posted seems to wrap around and only the color change is random. Why not keep it like that in the rule?
"The screen width and height can be combined into a single value if that's preferrable in your language of choice." Do you mean we can use the same value for width and height, i.e. always have a square screen?
May we output infinitely, instead of a fixed amount of ticks, like a screensaver?
@Jordan I thought he meant tuples.
Obligatory reference: Ballmer Peak
JavaScript (ES6), 264 266 274 281
(t,w,h,r=n=>Math.random()*n|0,g=[...Array(h)].map(x=>Array(w).fill` `))=>((y=>{for(x=y;t--;d&1?y+=d-2:x+=d-1)x<w&y<h&&~x*~y?0:(d=r(4))&1?x=r(w,y=d&2?0:h-1):y=r(h,x=d?0:w-1),e=d,d=r(5)?d:2*r(2)-~d&3,g[y][x]="─└ ┌┐│┌ ┘─┐┘ └│"[e*4|d]})(w),g.map(x=>x.join``).join`
`)
Counting unicode drawing characters as 1 byte each. (As specified by OP)
Less golfed
(t,w,h)=>{
r=n=>Math.random()*n|0; // integer range random function
g=[...Array(h)].map(x=>Array(w).fill(' ')); // display grid
for (x=y=w;t--;)
x<w & y<h && ~x*~y||( // if passed boundary
d = r(4), // select random direction
d & 1? (x=r(w), y=d&2?0:h-1) : (y=r(h), x=d?0:w-1) // choose start position
),
e=d, d=r(5)?d:2*r(2)-~d&3, // change direction 20% of times
g[y][x]="─└ ┌┐│┌ ┘─┐┘ └│"[e*4|d], // use char based on current+prev direction
d&1 ? y+=d-2 : x+=d-1 // change x,y position based on direction
return g.map(x=>x.join``).join`\n`
}
Animated test
Note: trying to keep the animation time under 30 sec,more thicks make animation pace faster
f=(t,w,h,r=n=>Math.random()*n|0,g=[...Array(h)].map(x=>Array(w).fill` `))=>
{
z=[]
for(x=y=w;t--;d&1?y+=d-2:x+=d-1)
x<w&y<h&&~x*~y?0:(d=r(4))&1?x=r(w,y=d&2?0:h-1):y=r(h,x=d?0:w-1),
e=d,d=r(5)?d:2*r(2)-~d&3,g[y][x]="─└ ┌┐│┌ ┘─┐┘ └│"[e*4|d],
z.push(g.map(x=>x.join``).join`\n`)
return z
}
function go() {
B.disabled=true
var [t,w,h]=I.value.match(/\d+/g)
var r=f(+t,+w,+h)
O.style.width = w+'ch';
var step=0
var animate =_=>{
S.textContent = step
var frame= r[step++]
if (frame) O.textContent = frame,setTimeout(animate, 30000/t);
else B.disabled=false
}
animate()
}
go()
#O { border: 1px solid #000 }
Input - ticks,width,height
<input value='600,70,10' id=I><button id=B onclick='go()'>GO</button>
<span id=S></span>
<pre id=O></pre>
Just when I thought QBasic might actually win a golf challenge. ;) Have an upvote.
Nothing says nostalgia quite like...
QBasic, 332 bytes
INPUT t,w,h
RANDOMIZE
CLS
1b=INT(RND*4)
d=b
IF b MOD 2THEN c=(b-1)/2*(w-1)+1:r=1+INT(RND*h)ELSE c=1+INT(RND*w):r=b/2*(h-1)+1
WHILE t
LOCATE r,c
m=(b+d)MOD 4
IF b=d THEN x=8.5*m ELSE x=13*m+(1<((b MOD m*3)+m)MOD 5)
?CHR$(179+x);
r=r-(d-1)MOD 2
c=c-(d-2)MOD 2
b=d
d=(4+d+INT(RND*1.25-.125))MOD 4
t=t-1
IF(r<=h)*(c<=w)*r*c=0GOTO 1
WEND
QBasic is the right language for the task because:
Its encoding includes box drawing characters--no need for Unicode
LOCATE allows you to print to any location on the screen, overwriting what was there previously
Microsoft®
Specifics
This is golfed QBasic, written and tested on QB64 with autoformatting turned off. If you type/paste it into the actual QBasic IDE, it will add a bunch of spaces and expand ? into PRINT, but it should run exactly the same.
The program inputs three comma-separated values: ticks, width, and height. It then asks for a random-number seed. (If this behavior isn't acceptable, change the second line to RANDOMIZE TIMER for +6 bytes.) Finally, it draws the pipes to the screen.
The maximum dimensions that can be entered are 80 (width) by 25 (height). Giving a height of 25 will result in the bottom row getting cut off when QBasic says "Press any key to continue."
How?
TL;DR: A lot of math.
The current row and column are r and c; the current direction is d and the previous direction is b. Direction values 0-3 are down, right, up, left. Arithmetic translates those into the correct step values for r and c, as well as the correct edge coordinates to start on.
The box drawing characters │┐└─┘┌ are code points 179, 191, 192, 196, 217, and 218 in QBasic. Those appear pretty random, but it still used fewer characters to generate the numbers with some (pretty convoluted, I'm-not-sure-even-I-understand-it) math than to do a bunch of conditional statements.
The code for changing direction generates a random number between -0.125 and 1.125 and takes its floor. This gives -1 10% of the time, 0 80% of the time, and 1 10% of the time. We then add this to the current value of d, mod 4. Adding 0 keeps the current direction; adding +/-1 makes a turn.
As for control flow, the WHILE t ... WEND is the main loop; the section before it, starting with line number 1 (1b=INT(RND*4)), restarts the pipe at a random edge. Whenever r and c are outside the window, we GOTO 1.
Show me the GIF!
Here you go:
This was generated by a somewhat ungolfed version with animation, color, and an automatic random seed:
INPUT t, w, h
RANDOMIZE TIMER
CLS
restart:
' Calculate an edge to start from
b = INT(RND * 4)
'0: top edge (moving down)
'1: left edge (moving right)
'2: bottom edge (moving up)
'3: right edge (moving left)
d = b
' Calculate column and row for a random point on that edge
IF b MOD 2 THEN
c = (b - 1) / 2 * (w - 1) + 1
r = 1 + INT(RND * h)
ELSE
c = 1 + INT(RND * w)
r = b / 2 * (h - 1) + 1
END IF
COLOR INT(RND * 15) + 1
WHILE t
' Mathemagic to generate the correct box-drawing character
m = (b + d) MOD 4
IF b = d THEN
x = 17 * m / 2
ELSE
x = 13 * m + (1 < ((b MOD m * 3) + m) MOD 5)
END IF
LOCATE r, c
PRINT CHR$(179 + x);
' Update row and column
r = r - (d - 1) MOD 2
c = c - (d - 2) MOD 2
' Generate new direction (10% turn one way, 10% turn the other way,
' 80% go straight)
b = d
d = (4 + d + INT(RND * 1.25 - .125)) MOD 4
' Pause
z = TIMER
WHILE TIMER < z + 0.01
IF z > TIMER THEN z = z - 86400
WEND
t = t - 1
IF r > h OR c > w OR r = 0 OR c = 0 THEN GOTO restart
WEND
I've typed this into my MS-DOS v6.22 VM :-)
Python 2.7, 624 616 569 548 552 bytes
from random import*
from time import*
i=randint
z=lambda a,b:dict(zip(a,b))
c={'u':z('lur',u'┐│┌'),'d':z('ldr',u'┘│└'),'l':z('uld',u'└─┌'),'r':z('urd',u'┘─┐')}
m=z('udlr',[[0,-1],[0,1],[-1,0],[1,0]])
def f(e,t,w,h):
seed(e);s=[w*[' ',]for _ in' '*h]
while t>0:
_=i(0,1);x,y=((i(0,w-1),i(0,1)*(h-1)),(i(0,1)*(w-1),i(0,h-1)))[_];o=('du'[y>0],'rl'[x>0])[_]
while t>0:
d=c[o].keys()[i(7,16)//8];s[y][x]=c[o][d];x+=m[d][0];y+=m[d][1];t-=1;sleep(.5);print'\n'.join([''.join(k)for k in s]);o=d
if(x*y<0)+(x>=w)+(y>=h):break
The first parameter is a seed, same seeds will generate the same output, printing each step with a 500 ms delay.
-10 bytes thanks to @TuukkaX
repl it
Example run
f(5,6,3,3)
will output
┐
─┐
──┐
┘─┐
┐
┘─┐
verbose version
import random as r
from time import *
char={
'u':{'u':'│','l':'┐','r':'┌'},
'd':{'d':'│','l':'┘','r':'└'},
'l':{'u':'└','d':'┌','l':'─'},
'r':{'u':'┘','d':'┐','r':'─'}
}
move={'u':[0,-1],'d':[0,1],'l':[-1,0],'r':[1,0]}
def f(seed,steps,w,h):
r.seed(seed)
screen=[[' ',]*w for _ in ' '*h]
while steps > 0:
if r.randint(0,1):
x,y=r.randint(0,w-1),r.randint(0,1)*(h-1)
origin='du'[y>0]
else:
x,y=r.randint(0,1)*(w-1),r.randint(0,h-1)
origin = 'rl'[x>0]
while steps > 0:
direction = char[origin].keys()[r.randint(0,2)]
screen[y][x]=char[origin][direction]
x+=move[direction][0]
y+=move[direction][1]
steps-=1
sleep(0.5)
print '\n'.join([''.join(k) for k in screen]),''
if x<0 or y<0 or x>=w or y>=h:
break
origin=direction
There's an useless whitespace at if x*y<0 or. 0.5 can be reduced to .5. import * could be import*. ''.join(k) for has an useless whitespace. You should also be able to keep dict in a variable and call it each time you use it. Haven't tested how much this saves, but by saving the dict(zip(a,b)) in a lambda that does the job for two strings (a, b), it should chop some. +1.
C (GCC/linux), 402 353 352 302 300 298 296 288 bytes
#define R rand()%
x,y,w,h,r;main(c){srand(time(0));scanf(
"%d%d",&w,&h);for(printf("\e[2J");x%~w*
(y%~h)||(c=R 8,(r=R 4)&1?x=1+R w,y=r&2
?1:h:(y=1+R h,x=r&2?1:w));usleep('??'))
printf("\e[%dm\e[%d;%dH\342\224%c\e[H\n",
30+c,y,x,2*"@J_FHAF__L@HL_JA"[r*4|(r^=R 5
?0:1|R 4)]),x+=--r%2,y+=~-r++%2;}
Credit to edc65 for storing the direction in a single 4-bit number.
Reads a width/height on stdin before looping the screensaver forever. E.g.:
gcc -w golf.c && echo "25 25" | ./a.out
Or for a full-screen screensaver:
gcc -w golf.c && resize | sed 's/[^0-9]*//g' | ./a.out
For readability I added newlines. Requires a linux machine with a terminal respecting ANSI codes. Has colors! If you remove color support it costs 17 bytes less.
Ruby, 413 403 396 bytes
A function that takes a number of ticks and a width as input and returns the final screen as a string. Could no doubt be golfed more.
->t,w{k=[-1,0,1,0,-1]
b=(" "*w+$/)*w
f=->t,a=[[0,m=rand(w),2],[w-1,m,0],[m,0,1],[m,w-1,3]].sample{n,m,i=a
d=k[i,2]
q=->n,m,i{_,g,j=rand>0.2?[[1,0],[3,0],[0,1],[2,1]].assoc(i):"021322033132243140251350".chars.map(&:to_i).each_slice(3).select{|c,|c==i}.sample
v,u=k[j||=i,2]
y=n+v
x=m+u
[g,y,x,j]}
g,y,x,j=q[n,m,i]
b[n*w+n+m]="─│┌┐┘└"[g]
y>=0&&y<w&&x>=0&&x<w ?t>1?f[t-1,[y,x,j]]:b:f[t]}
f[t]}
See it on repl.it: https://repl.it/Db5h/4
In order to see it in action, insert the following after the line that begins b[n*w+n+m]=:
puts b; sleep 0.2
...then assign the lambda to a variable e.g. pipes=->... and call it like pipes[100,20] (for 100 ticks and a 20x20 screen).
Ungolfed & explanation
# Anonymous function
# t - Number of ticks
# w - Screen width
->t,w{
# The cardinal directions ([y,x] vectors)
# Up = k[0..1], Right = k[1..2] etc.
k = [-1, 0, 1, 0, -1]
# An empty screen as a string
b = (" " * w + $/) * w
# Main tick function (recursive)
# t - The number of ticks remaining
# a - The current position and vector index; if not given is generated randomly
f = ->t,a=[[0,m=rand(w),2], [w-1,m,0], [m,0,1], [m,w-1,3]].sample{
# Current row, column, and vector index
n, m, i = a
d = k[i,2] # Get vector by index
# Function to get the next move based on the previous position (n,m) and direction (d)
q = ->n,m,i{
# Choose the next pipe (`g` for glyph) and get the subsequent vector index (j)
_, g, j = (
rand > 0.2 ?
[[1,0], [3,0], [0,1], [2,1]].assoc(i) : # 80% of the time go straight
"021322033132243140251350".chars.map(&:to_i).each_slice(3)
.select{|c,|c==i}.sample
)
# Next vector (`v` for vertical, `u` for horizontal)
# If straight, `j` will be nil so previous index `i` is used
v, u = k[j||=i, 2]
# Calculate next position
y = n + v
x = m + u
# Return next glyph, position and vector index
[g, y, x, j]
}
# Get next glyph, and subsequent position and vector index
g, y, x, j = q[n, m, i]
# Draw the glyph
b[n * w + n + m] = "─│┌┐┘└"[g]
# Check for out-of-bounds
y >= 0 && y < w && x >=0 && x < w ?
# In bounds; check number of ticks remaining
t > 1 ?
f[t-1, [y,x,j]] : # Ticks remain; start next iteration
b : # No more ticks; return final screen
# Out of bounds; repeat tick with new random start position
f[t]
}
f[t]
}
| 21,320 |
https://stackoverflow.com/questions/53213053 | StackExchange | Open Web | CC-By-SA | 2,018 | Stack Exchange | Edmund Lee, https://stackoverflow.com/users/4164651 | English | Spoken | 338 | 816 | Creating a new Phabricator task with javascript
I am trying to connect to Phabricator conduit API and create a task via a javascript bonded to a google sheet.
The Conduit API Docs linked here doesn't really explain as much. I have seen better API documentations!
Below is what I have in mind but this is a cURL and I have no idea how to make it Javascript or wither this would work or not? I appreciate the help
curl https://secure.phabricator.com/api/maniphest.edit \
-d api.token=api-token \
-d param= [
{
"type": "title",
"value": "A value from a cell on the googlesheet"
},
{
"type": "description",
"value": "A value from a cell on the googlesheet"
},
{
"type": "subscribers.add",
"value": "A value from a cell on the googlesheet"
}
] \
Did you figure this out?
Generally speaking the steps are:
First, generate an API token in:
https://phabricator.yourdomain.com/settings/user/username/page/apitokens/
where phabricator.yourdomain.com must be changed by the subdomain you have Phabricator installed and username must be changed by your administration user name.
Then, let's say you have installed Phabricator in phabricator.yourdomain.com, you can request the API methods with URLs of the following type
https://phabricator.yourdomain.com/api/method_name?parameter1=value1¶meter2=value2...
where method_name must be replaced by the descriptor of a real method from this catalog:
https://secure.phabricator.com/conduit/
For example, if you want to read the contents of task number 125, with a generated API token of value api-svhcp2a3qmgkkjfa5f6sh7cm4joz, use the method maniphest.info to complete a URL like this:
http://phabricator.yourdomain.com/api/maniphest.info?api.token=api-svhcp2a3qmgkkjfa5f6sh7cm4joz&task_id=125&output=json
This URL can be directly tested in your preferred browser to obtain a JSON response with the information about task number 125 (make sure that task ID exists). Firefox will even show the returned JSON in a human-readable fashion.
These working URLs can be then inserted in Javascript as
window.location.href=http://phabricator.yourdomain.com/api/maniphest.info?api.token=api-svhcp2a3qmgkkjfa5f6sh7cm4joz&task_id=125&output=json
or as an asynchronous Ajax call.
I had a similar problem as you (I used HTTParty with Ruby).
To solve it I used the following body (using your example):
"transactions[0][type]=title&transactions[0][value][0]=A value from a cell on the googlesheet&transactions[1][type]=description&transactions[1][value]=A value from a cell on the googlesheet&transactions[2][type]=subscribers.add&transactions[2][value][0]=A value from a cell on the googlesheet"
| 13,189 |
https://stackoverflow.com/questions/20935036 | StackExchange | Open Web | CC-By-SA | 2,014 | Stack Exchange | English | Spoken | 155 | 282 | Simplifying small piece of code
Right now I have
var index = link.attr('href');
index =index.replace("#",'');
and later in my code I use the "index" again
It feels like I could write this easier, but I don't know how..
Why not in a single line
var index = link.attr("href").replace("#", "");
You can do both operations on the same line:
var index = link.attr('href').replace("#", "");
The concept is called method chaining, check out a detailed post on the subject: http://schier.co/post/method-chaining-in-javascript
You could use function chaining here. Something like this:
var index = link.attr('href').replace("#",'');
The attr() function actually returns the resulting attribute string itself. So you just call the replace() function on the returned string value of attr().
For example, if the href attribute's value is http://google.com, your code would evaluate to something like this:
var index = "http://google.com".replace("#",'');
To the JavaScript code, once it has extracted your href attribute, you're just executing a replace() on a string.
| 292 | |
https://stackoverflow.com/questions/11987313 | StackExchange | Open Web | CC-By-SA | 2,012 | Stack Exchange | English | Spoken | 124 | 235 | Change elements of matrix based on condition
Suppose I have Matrix A:
A =
1 2 3
4 5 6
For every element x<2, add 10,
for every element x>5, add 20, and
for every element 2<=x<=5, add 30.
So for my example matrix A, I need to end up with the following matrix B:
B =
11 32 33
34 35 26
I need to be able to do this in a general way, since the actual matrix will be quite large. Any suggestions? Can I use the IF statement?
You don't need an IF you need logical indexing:
IndexOfLessThan2 = A < 2;
IndexOfGreaterThan5 = A > 5;
IndexBtw2and5 = ~(IndexOfLessThan2 | IndexOfGreaterThan5);
A = A + IndexOfLessThan2*10 + IndexOfGreaterThan5*20 + IndexBtw2and5*30;
| 13,967 | |
https://ca.wikipedia.org/wiki/Federico%20Jusid | Wikipedia | Open Web | CC-By-SA | 2,023 | Federico Jusid | https://ca.wikipedia.org/w/index.php?title=Federico Jusid&action=history | Catalan | Spoken | 1,415 | 2,724 | Federico Jusid (Buenos Aires, 23 d'abril de 1973) és un pianista i compositor argentí que actualment resideix i treballa entre Madrid i Los Angeles. Ha compost la banda sonora original de més de 40 llargmetratges de ficció i documentals, i nombroses partitures i arranjaments per a publicitat i més de 20 sèries de televisió. Entre les seves composicions originals per a cinema figura El secreto de sus ojos (2009, dirigida per Juan José Campanella), guanyadora de l'Oscar a la millor pel·lícula de parla no anglesa, partitura per la que fou nominat al Goya a la millor música original.
Recentment, ha participat en la banda sonora de la pel·lícula de Ridley Scott Exodus: Gods and Kings, amb composicions addicionals per a Alberto Iglesias i ha compost la música de Kidnap (Luis Prieto), amb Halle Berry i Felices 140 (Gracia Querejeta). Altres bandes sonores destacades de la seva autoria són: Francisco, El Padre Jorge (Beda Docampo Feijóo), Todos tenemos un plan (Anna Piterbarg), La fuga (Eduardo Mignogna), La cara oculta (Andrés Baiz), I Want To Be A Soldier (Christian Molina), ¡Atraco! (Eduard Cortés), i més recentment La vida inesperada (Jorge Torregrossa, co-composta amb Lucio Godoy), Magallanes (Salvador del Solar) Getúlio (Joao Jardim), Betibú (Miguel Cohan) i La ignorancia de la sangre (Manuel Gómez Pereira).
A televisió, el seu treball més reconegut és la banda sonora per a la sèrie de Televisió Espanyola Isabel, amb la que va guanyar diversos premis, com ara el de la International Film Music Critics Association (IFMCA) i el Reel Music Award. El 2015 va reatlizar les bandes sonores per les sèries Bajo sospecha, Refugiados (coproducció amb BBC Internacional) i Carlos, rey emperador, nou drama històric, seqüela d' Isabel.
En el camp de la música per a sales de concert, ha compost obres per a diferents agrupacions de cambra, cor i solistes. Com pianista de concert ha girat per prestigiosos teatres d'Europa, Àsia i Amèrica. Ha gravat per als segells BMG, IRCO, Magenta i Melopea Discos.
Començaments i formació
Va iniciar els seus estudis de piano i composició als set anys. Des de llavors, la seva trajectòria s'ha repartit entre la formació (titulat “Professor Superior de Música” Conservatori de Buenos Aires, “Master of Music” The Manhattan School of Music, Nova York; “New England CSS”, Boston; i “Diplôme de Exécution Musicale”, Brussel·les), la interpretació (amb una llarga llista de premis internacionals) i la composició, tant de música acadèmica com pel cinema.
Fill del reconegut director de cinema argentí Juan José Jusid i de l'actriu Luisina Brando, Federico va créixer entre escenaris i sets de filmació. Aviat la seva passió per la música i el cinema es van fondre en un interès únic i va començar la seva carrera com a compositor de música per a cinema. En 1994 va signar la seva primera banda sonora original i des de llavors ha compost les partitures de més de 30 llargmetratges, a més de la música per a més de 15 sèries de televisió.
Carrera
Els seus últims treballs inclouen composicions addicionals per a la BSO d'Alberto Iglesias a Exodus: Gods and Kings de Ridley Scott, i la banda sonora de Kidnap (Luis Prieto), amb Halle Berry, Francisco, El Padre Jorge (Beda Docampo Feijóo), Felices 140 (Gracia Querejeta), Magallanes (Salvador del Solar), La vida inesperada (Jorge Torregrossa), Getúlio (Joao Jardim), Betibú (Miguel Cohan) i La ignorancia de la sangre (Manuel Gómez Pereira).
Entre els seus crèdits en música per a cinema es compten els llargmetratges Todos tenemos un plan (Twentieth Century Fox - Dir. Anna Piterbarg, protagonitzada per Viggo Mortensen), Di que sí (Columbia Pictures - Dir. Juan Calvo); La fuga (Dir.: Eduardo Mignogna), El Custodio (Dir. Rodrigo Moreno), L'habitació de Fermat (Luis Piedrahíta i Rodrigo Sopeña), La cara oculta (Twentieth Century Fox - Dir. Andrés Baiz), I Want To Be A Soldier (Warner - Dir. Christian Molina), Che: Un hombre nuevo (documental, Dir. Tristán Bauer), ¡Atraco! (Dir. Eduard Cortés), i la internacionalment aclamada El secreto de sus ojos (Dir. Juan José Campanella), guanyadora de l'Oscar a Millor Pel·lícula Estrangera, i nominada per als XXIV Premis Goya a Millor Banda Sonora Original.
Els seus treballs en composició de música original per a sèries de televisió inclouen Isabel, Carlos, rey emperador, Bajo sospecha, Refugiados, Gran Reserva, La Señora, 14 de abril. La República, Los Simuladores, Hermanos y detectives i Los misterios de Laura, entre otras.
Altres distincions per al seu treball en cinema i televisió inclouen Millor Música Original al Festival Internacional de Cinema de l'Havana 2010, Millor Música Original en els Premis Clarín 2010 a Buenos Aires, Primer Premi Còndor de Plata de l'Asociación de Cronistas del Espectáculo 2010, Premi Sur de l'Academia de Artes y Ciencias Cinematográficas de la Argentina a Millor Música Original, tots per la pel·lícula El secreto de sus ojos; Premi a Millor Banda Sonora Original per a TV de la International Film Music Critics Association (IFMCA) 2013 i 2014 per la banda sonora d' Isabel, mi reina, que també va guanyar els Reel Music Award 2013 i 2014 a Millor Banda Sonora de Sèrie de TV; Millor Banda Sonora al 12è Festival de Cinema Llatinoamericà de Trieste (Itàlia) en 2006 per la pel·lícula Olga, Victoria Olga; Premi Llanterna del Públic del Programa "El acomodador" per la banda sonora de la pel·lícula Rodrigo en 2001; Primer Premi Pentagrama d'Or 2001 per la banda sonora original de la pel·lícula La fuga al Festival Internacional de Cinema de Mar de Plata; Primer Premi Còndor de Plata per Millor Banda Sonora de l'Any per Bajo Bandera, entregat per l'Asociación Argentina de Cronistas del Espectáculo en 1998.
Concerts i recitals
Federico Jusid compatibilitza la seva activitat en la composició per a cinema amb la composició per a sales de concert i actuacions com a pianista. Les seves últimes composicions inclouen la peça Tango Rhapsody, per a dos pianos i orquestra simfònica, comissionada pel Martha Argerich Project per l'International Music Festival de Lugano; Enigmas, una peça per a piano comissionada per la Universitat d'Alcalá de Henares (Madrid) al seu V Centenari; Finding Sarasate, comissionada per la Universitat de Navarra per estrenar en el Concert Tribut a Pablo Sarasate; i La Librería del Ingenioso Hidalgo, comissionada per les celebracions del IV Centenari de Don Quixote.
Com a intèrpret de piano, Jusid ha tocat com a solista en nombroses gires en alguns dels més prestigiosos teatres d'Amèrica, Àsia i Europa, incloent Carnegie Weill Hall, New York; Teatro Colón, Buenos Aires; Theater Platz, Frankfurt; Israel Philarmonic Orchestra House, Tel Aviv; Saint-Severine, París; Piazza Pola, Sicília; Musikmuseet, Estocolm; Centralen Voenen, Sofia; Temppeliaukion Kirkko, Hèlsinki; Hubbard Hall, Nova York; Biblioteca Nacional, Buenos Aires; Teatro Municipal General San Martín, Buenos Aires; Auditorio Conde Duque, Madrid; Paraninfo de la Universitat Complutense, Madrid; Roma; Shanghai, entre altres.
A més, amb el Sonor Ensemble, integrat per solistes de l'Orquestra Nacional d'Espanya dirigits per Luis Aguirre, ha realitzat gires de concert per Espanya i Europa.
Com a solista ha actuat al capdavant de les següents orquestres: Sonor Ensemble, Sinfonietta de París, Orquestra Simfònica Nacional de l'Argentina, Orquestra Simfònica de Bratislava, Orquestra Simfònica de Còrdova, Orquestra Simfònica de Bahía Blanca, Orquestra Simfònica de Mar del Plata, Orquestra Simfònica de Santa Fe.
Tant la seva música per a sala de concert com per a cinema i televisió ha estat interpretada per orquestres i solistes internacionals.
Entre les distincions rebudes pel seu treball d'intèrpret i compositor de música acadèmica figuren: Primer premi "Beethoven Piano Competition" (1995), Primer premi "A. Williams Piano Competition" (1995), Primer premi "Radio Nacional Argentina" (1996), Primer premi "C.E.P" (Centro de Estudios Pianísticos) (1996), Primer premi Concurs Internacional de Música de Buenos Aires (1998), Primer premi "Friends of Israel Philharmonic Orchestra" (1998), Primer premi Acadèmia Nacional de Belles Arts - Beca Drago Mitre (1999), Beca de "Fundación Antorchas" (1995 i 1996), Beca de "Germaine Pinault School of Music" (1995), Beca Estudis a l'Estranger de Fundación Antorchas (2001), Distinció Especial “IBLA piano and composition contest” (Italia) per l'obra "concertant pera a piano sol Fleeting Fantasies; Menció Artista Distingit “IBLA contest”, per la interpretació del Concert per a piano i orquestra Op. 15 de J. Brahms; Premi XII Festival Internacional de Música de La Mancha per l'estrena de l'obra La librería del ingenioso hidalgo, comissionada per la celebració del IV Centenari d'El Quijote (2005).
Treballs
Música per cinema
Música per televisió
Premis
Referències
Enllaços externs
Web Oficial
Web oficial de l'agrupació musical Sonor Ensemble
Web oficial de "Isabel, mi reina" RTVE
Web oficial d' "El secreto de sus ojos"
Compositors argentins
Músics de Buenos Aires
Los misterios de Laura | 4,567 |
https://lv.wikipedia.org/wiki/Kerijs%20Praiss | Wikipedia | Open Web | CC-By-SA | 2,023 | Kerijs Praiss | https://lv.wikipedia.org/w/index.php?title=Kerijs Praiss&action=history | Latvian | Spoken | 223 | 647 | Kerijs Praiss (; dzimis ) ir profesionāls kanādiešu hokejists, vārtsargs. Pašlaik (2023) Praiss pārstāv Nacionālās hokeja līgas klubu Monreālas "Canadiens".
Spēlētāja karjera
Pēc vairākās Rietumu hokeja līgā aizvadītām sezonā, profesionāļa karjeru Praiss sāka 2006.—07. gada sezonā Amerikas hokeja līgas kluba Hamiltonas "Bulldogs" rindās. Lai gan regulārajā sezonā komandas rindās viņš laukumā devās vien divās spēlēs, izslēgšanas kārtā Praiss aizvadīja 22 spēles, kopā ar komandu izcīnot Kaldera kausu.
2007.—08. gada sezonā Praiss jau spēlēja NHL kluba Monreālas "Canadiens" rindās, dalot spēlet laiku ar Kristobalu Iē. Nākamajās divās sezonās Praiss spēles laiku dalīja ar Jaroslavu Halāku, taču 2010.—11. gada sezonā komandas sastāvā laukumā devās jau 72 regulārās sezonas spēlēs. 2014.—15. gada sezonu Praiss noslēdza kā līderis visos trīs vārtsargu statistikas pamatrādītājos - vidēji spēlē ielaistie vārti (1,96), atvairīto metienu procents (93,3) un uzvaras (44). Kā līgas vērtīgākais spēlētājs ieguva Harta balvu. Kā līgas labākais vārtsargs Praiss ieguva arī Vezinas balvu.
Kanādas hokeja izlases rindās kļuvis par 2014. gada olimpisko čempionu, iegūstot arī turnīra labāka vārtsarga titulu. Praiss kļuvis arī par 2007. gada Pasaules čempionu junioriem. Praiss tika atzīts arī par turnīra vērtīgāko spēlētāju, kā arī labāko vārtsargu.
Ārējās saites
1987. gadā dzimušie
Kanādas hokejisti
Monreālas "Canadiens" spēlētāji
Hokejisti 2014. gada ziemas olimpiskajās spēlēs
NHL drafta pirmās kārtas izvēles
Britu Kolumbijā dzimušie
2014. gada ziemas olimpisko spēļu medaļnieki
Olimpiskie zelta medaļnieki hokejā
Kanādas olimpiskie zelta medaļnieki | 20,278 |
https://ceb.wikipedia.org/wiki/Tukad%20Yangapi | Wikipedia | Open Web | CC-By-SA | 2,023 | Tukad Yangapi | https://ceb.wikipedia.org/w/index.php?title=Tukad Yangapi&action=history | Cebuano | Spoken | 46 | 74 | Suba ang Tukad Yangapi sa Indonesya. Nahimutang ni sa lalawigan sa Provinsi Bali, sa habagatang bahin sa nasod, km sa sidlakan sa Jakarta ang ulohan sa nasod. Ang Tukad Yangapi nahimutang sa pulo sa Bali.
Saysay
Ang mga gi basihan niini
Mga suba sa Provinsi Bali | 1,368 |
https://ceb.wikipedia.org/wiki/Anisozyga%20diversifimbria | Wikipedia | Open Web | CC-By-SA | 2,023 | Anisozyga diversifimbria | https://ceb.wikipedia.org/w/index.php?title=Anisozyga diversifimbria&action=history | Cebuano | Spoken | 42 | 83 | Kaliwatan sa alibangbang ang Anisozyga diversifimbria. Una ning gihulagway ni Louis Beethoven Prout ni adtong 1922. Ang Anisozyga diversifimbria sakop sa kahenera nga Anisozyga, ug kabanay nga Geometridae. Walay nalista nga matang nga sama niini.
Ang mga gi basihan niini
Insekto
Anisozyga | 12,120 |
https://sv.wikipedia.org/wiki/Platypalpus%20puerinus | Wikipedia | Open Web | CC-By-SA | 2,023 | Platypalpus puerinus | https://sv.wikipedia.org/w/index.php?title=Platypalpus puerinus&action=history | Swedish | Spoken | 32 | 70 | Platypalpus puerinus är en tvåvingeart som beskrevs av Axel Leonard Melander 1927. Platypalpus puerinus ingår i släktet Platypalpus och familjen puckeldansflugor. Inga underarter finns listade i Catalogue of Life.
Källor
Puckeldansflugor
puerinus | 9,311 |
https://wo.wikipedia.org/wiki/Jur%C3%B3om%20%C3%B1ett | Wikipedia | Open Web | CC-By-SA | 2,023 | Juróom ñett | https://wo.wikipedia.org/w/index.php?title=Juróom ñett&action=history | Wolof | Spoken | 5 | 14 | Juróom ñett : 8
Xayma | 48,939 |
https://ug.wikipedia.org/wiki/Windows%20%D9%86%D9%89%DA%AD%20%D8%AA%DB%95%D8%B1%DB%95%D9%82%D9%82%D9%89%D9%8A%D8%A7%D8%AA%20%D8%AA%D8%A7%D8%B1%D9%89%D8%AE%D9%89 | Wikipedia | Open Web | CC-By-SA | 2,023 | Windows نىڭ تەرەققىيات تارىخى | https://ug.wikipedia.org/w/index.php?title=Windows نىڭ تەرەققىيات تارىخى&action=history | Uyghur | Spoken | 348 | 1,954 | 1985-يىلى، تۇنجى Windows ئېلان قىلىنغان بولۇپ، ھازىرغا قەدەر 27 يىللىق تارىخىنى بېسىپ بولدى.
1985-يىلى، Window 1.0 دۇنياغا كەلگەن، بۇ MS-DOS نى يادرو قىلغان ھەم گىئومېتىرىيە شەكىللىك مەشغۇلات يۈزى ئىدى. ئەينى ۋاقىتتا MS-DOS كومپيۇتېرنىڭ ئاساسىي ئېقىمدىكى مەشغۇلات سىستېمىسى بولۇپ قالغان.
1987-يىلى، Windows 2.0 دۇنياغا كەلگەن. Word ۋە Excel ئەڭ دەسلەپكى مۇشۇ نەشردىن باشلاپ ۋۇجۇدقا كەلگەن. چۈنكى بۇ نەشردە مىكروسوفت ۋە ئالما شىركىتى تۇنجى مەيدان دەۋالاشقان. ئالما شىركىتى مىكروسوفتنى Macintosh ۋە Lisa دىكى قىسمەن ئامىللارنى ئوغرىلىغان دەپ ئەرز قىلغان، ئەمما ئالما شىركىتى بۇ مەيدان دەۋادا ئۇتتۇرۇپ قويغان.
1990-يىلى، Windows 3.0 دۇنياغا كەلگەن، بۇ نەشردىن باشلاپ سىنبەلگە شەكلىدە MS-DOS يۇشۇرۇلغان بولغانلىقى ئۈچۈن، بۇ تۇنجى مودا بولۇپ قالغان Windows مەشغۇلات سىستېمىسىدۇر.
1991-يىلى، Windows 3.1 دۇنياغا كەلگەن، بۇ نەشرى 90-يىللاردىكى IBM كومپيۇتېر دەۋردىكى ئۆلچەملىك سەپلىمىسى بولۇپ قالغان. بۇ نەشرىمۇ ئەڭ ئاخىرىقى ھەقىقىي مەنىدىكى MS-DOS غا ئوخشايدىغان كومپيۇتېر بولۇپ قالغان.
1995-يىلى، Windows 95 كۈچلۈك زىلزىلە بىلەن دۇنياغا كەلدى. بۇ نەشردە بۇرۇن بولۇپ باقمىغان ۋە كېيىنمۇ بولۇپ باقمايدىغان ئۆزگىرىشلەر بولغان. بۇ نەشردىن باشلاپ ئۈستەل يۈزى، سىنبەلگە، IE، سېۋەت (“垃圾箱”或者“回收站” ئاتالغۇسى Windows 8 نىڭ تەرجىمىسىدە سېۋەت دەپ ئېلىنغان بولغاچقا، بۇ يەردىمۇ مۇشۇ ئاتالغۇ ئىشلىتىلدى.) قاتارلىق ئۇقۇملار مەيدانغا كەلدى. ھازىرغا قەدەر، بۇ ئۇقۇملاردا ئاساسىي جەھەتتىن چوڭ ئۆزگىرىش بولمىغان.
1998-يىلى، Windows 98 دۇنياغا كەلگەن. بۇ نەشرى 95 نەشرىگە قارىغاندا، پەقەت بەزىبىر يېڭى ئىقتىدارلىرى قېتىلغاندىن سىرت چوڭ ئۆزگىرىش بولمىغان.
2000-يىلى، Windows ME دۇنياغا كەلگەن. ME بولسا Millennium (مىڭ يىل) Edition نىڭ قىسقارتىلىپ يېزىلىشى بولۇپ، كۆپچىلىكنىڭ ھەممىسى ئۈمىدنى XP گە باغلىغان بولغانلىقتىن بۇ نەشرنىڭ سېتىلىش ئەھۋالى ئانچە ياخشى بولمىغان.
2001-يىلى، Windows XP دۇنياغا كەلگەن. بۇ قېتىملىق لايىھە ئۆتمۈشتىكى 6 يىللىق Windows نىڭ پۈتكۈل ئارايۈز ئۇسلۇبىنى ئاستىن-ئۈستۈن قىلىۋەتكەن. بۇ كىشىگە ناھايىتى راھەت تۇيغۇ بەرگەن (ھېلىھەم شۇنداققۇ دەيمەن، شۇڭا نۇرغۇن كىشىلەر تا ھازىرغىچە بۇ نەشردىن ۋاز كېچىشكە قىيالماي كەلمەكتە)
2006-يىلى، Windows Vista دۇنياغا كەلگەن، ئەمما بۇنىڭ ئۈنۈمى ئانچە ياخشى بولمىغان (يەنى، مىكروسوفتنىڭ ئەڭ چوڭ مەغلۇبىيىتى بولسا كېرەك) ئەمما يېرىم سۈزۈك بولغان ئارايۈز يەنىلا قالتىس بولغان.
2009-يىلى، Windows 7 دۇنياغا كەلگەن، بۇ نەشرى Vista نىڭ ئاساسىدا ئەلالاشتۇرۇلغان. ئەلالاشتۇرۇلغاندىن كېيىن سىستېما زور دەرىجىدە تۇراقلاشقان.
2012-يىلى، Windows 8 دۇنياغا كەلگەن.
بۇ نەشرى قوللىنىشچانلىق (قوللىنىشچان سىستېمىلىرى) جەھەتتە يەنە بىر قېتىملىق يېڭىلىق يارىتىشتا زىلزىلە قوزغىغان نەشرىدۇر.
2013-يىلى Windows 8نىڭ تولۇقلانغان نەشىرى Windows 8.1 دۇنياغا كەلگەن.
Microsoft Windows | 25,933 |
https://uk.wikipedia.org/wiki/%D0%9B%D1%83%D1%82%D0%B8%D0%B3%D0%B0%20%D1%80%D0%BE%D0%B7%D1%85%D0%B8%D0%BB%D0%B8%D1%81%D1%82%D0%B0 | Wikipedia | Open Web | CC-By-SA | 2,023 | Лутига розхилиста | https://uk.wikipedia.org/w/index.php?title=Лутига розхилиста&action=history | Ukrainian | Spoken | 81 | 298 | Лутига розхилиста (Atriplex patens, можливо синонім до Atriplex laevis) — вид рослин з родини щирицевих (Amaranthaceae).
Морфологічна характеристика
Однорічна рослина 20–70 см заввишки. Листки м'ясисті, соковиті (сухі — товстуваті, вкриті дрібними пухирчастими зморшками), з розсіченим борошнистим нальотом, від довгасто-яйцюватих до ланцетних.
Поширення
Вид поширений у Євразії від України до Казахстану; інтродукований у Білорусі й країнах Балтії.
В Україні вид зростає у посівах, засмічених місцях, біля доріг — на всій території.
Примітки
розхилиста
Флора України
Флора Європи
Флора Казахстану
Рослини, описані 1927 | 34,463 |
https://stackoverflow.com/questions/40061120 | StackExchange | Open Web | CC-By-SA | 2,016 | Stack Exchange | Cheers and hth. - Alf, Code-Apprentice, DavidW, Evgeniy, Lightness Races in Orbit, StoryTeller - Unslander Monica, https://stackoverflow.com/users/1080064, https://stackoverflow.com/users/1440565, https://stackoverflow.com/users/3209505, https://stackoverflow.com/users/464581, https://stackoverflow.com/users/4657412, https://stackoverflow.com/users/493106, https://stackoverflow.com/users/560648, https://stackoverflow.com/users/817643, tkausl, xaxxon | English | Spoken | 785 | 1,266 | Restoring the exact angle from std::cos(angle) using std::acos
Is it guaranteed by the C++ standard that angle == std::acos(std::cos(angle)) if angle is in the range [0, Pi], or in other words is it possible to restore the exact original value of angle from the result of std::cos using std::acos given the mentioned range limit?
The marginal cases when angle is infinity or NaN are omitted.
I don't think so. floating point precision prevents that.
You are limited by precision and rounding
The standard cannot make that guarantee. Simply because the result of std::cos may not be representable exactly by a double, so you get a truncation error, which will affect the result of std::acos.
Can I get an exact answer for floating point operations on arbitrary data? No.
Ignoring the rounding issue, there are multiple angles that give the same cos(angle) (one every 2pi radians), so it would not be possible for acos to return to the same one.
Are there any limits on angle?
Even if the angle is in the range [-2 * M_PI, 2 * M_PI], it isn't a question of mathematics, but of hardware and numeric constraints. This is a good read on the subject https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html
Answer by StoryTeller:
The standard cannot make that guarantee, simply because the result of std::cos may not be representable exactly by a double, so you get a truncation error, which will affect the result of std::acos.
Is that actually the reason? Is it impossible to make a conforming implementation that would map the value back to the initial value?
@StoryTeller: If it's a dupe then okay don't answer, but if you are going to answer, please do so in the proper place!
@xaxxon, I'm afraid not. When the function accepts a value, it must treat every digit as significant, and the value as exactly what the caller intended to pass.
as defined here in 7.3.5.1 http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1124.pdf
@LightnessRacesinOrbit, I really hate having to type an answer just to have the question closed in an instant. This comment is just my two cents.
@Cheersandhth.-Alf, Well it's relevant if the OP wants to rely on angle == std::acos(std::cos(angle)) always being true, which is of course impossible.
From cppreference.com:
” If no errors occur, [acos returns] the arc cosine of arg (arccos(arg)) in the range [0 ; π]
In degrees, that's 0 to 180, inclusive, corresponding to cosine values 1 down through -1, inclusive.
Outside that range you can't even get an approximate correspondence. Computing the cosine discards information about which angle you had outside of that range. There's no way to get that information back.
How information is discarded:
First, in degrees, cos(x) = cos(K*360 + x), for arbitrary integer K. Secondly, cos(x) = cos(-x). This adds up to an awful lot of angle values that produce the same cosine value.
Also, even though all readers likely know this, but for completeness: since sines are cosines are very irrational numbers, generally not simple fractions, you can't expect exact results except for maybe cosine 1, which corresponds to 0 degrees.
@xaxxon: pi is little bit irrational, but not much, because it becomes a very systematic, simple sequence of digits when expressed in the factorial numeral system. But the sine of 1.234 degrees, uh oh, that's very irrational. :-p
According to the standard:
This International Standard imposes no requirements on the accuracy
of floating-point operations; see also 18.3.2. — end note ]
http://open-std.org/JTC1/SC22/WG21/docs/papers/2016/n4606.pdf
True, but the reason is to avoid constraining the implementation from using the full power of the hardware. And I think it's irrelevant. Even if a machine existed that could do exact computations every time, the standard cannot make that promise. From a practical standpoint, one should always be careful when dealing with numeric code.
@StoryTeller The question specifically asked if the standard made the guarantee and I think this bit in the standard very specifically precludes the guarantee which is why I added it as an answer -- because I think approaches the answer from the direction requested in the question.
I think it's a bit of stretch reading that from this quote. The standard usually tries to be as little limiting as possible. But yeah, I can see why it could preclude it.
@StoryTeller It precludes the standard making the guarantee because they would be mutually exclusive. I wasn't trying to say it precludes an implementation that does so -- but that's not the question posed.
I see. A rare occasion when a negative claim is provable :)
@StoryTeller Well, you can just rephrase it. Does the standard allow for sin(asin(x)) != x. Yes.
Even mathematically this is impossible. For example, cos(2*PI) is 0, but so is cos(4*PI).
Yeah, I forgot the period the function.
| 9,986 |
https://stackoverflow.com/questions/55672842 | StackExchange | Open Web | CC-By-SA | 2,019 | Stack Exchange | English | Spoken | 386 | 1,095 | serverless-localstack fail to deploy lambda
When trying to deploy lambda with serverless-localstack on localstack, the code is uploaded to s3 bucket but the lambda is not deploy though their are not errors in log.
I run the command
sls deploy --stage local --verbose
and don't see any errors.
the output is:
Serverless: config
.options_stage: local
Serverless: serverless.service.custom.stage: undefined
Serverless: serverless.service.provider.stage: dev
Serverless: config.stage: local
Serverless: Using serverless-localstack
Serverless: Reconfiguring service apigateway to use http://localhost:4567
Serverless: Reconfiguring service cloudformation to use
http://localhost:4581
Serverless: Reconfiguring service cloudwatch to use http://localhost:4582
Serverless: Reconfiguring service lambda to use http://localhost:4574
Serverless: Reconfiguring service dynamodb to use http://localhost:4569
Serverless: Reconfiguring service kinesis to use http://localhost:4568
Serverless: Reconfiguring service route53 to use http://localhost:4580
Serverless: Reconfiguring service firehose to use http://localhost:4573
Serverless: Reconfiguring service stepfunctions to use http://localhost:4585
Serverless: Reconfiguring service es to use http://localhost:4578
Serverless: Reconfiguring service s3 to use http://localhost:4572
Serverless: Reconfiguring service ses to use http://localhost:4579
Serverless: Reconfiguring service sns to use http://localhost:4575
Serverless: Reconfiguring service sqs to use http://localhost:4576
Serverless: Reconfiguring service sts to use http://localhost:4592
Serverless: Reconfiguring service iam to use http://localhost:4593
Serverless: Generated requirements from /root/lambda_detect/requirements.txt in /root/lambda_detect/.serverless/requirements.txt...
Serverless: Installing requirements from /root/lambda_detect/.serverless/requirements/requirements.txt ...
Serverless: Running ...
Serverless: Packaging service...
Serverless: Excluding development dependencies...
Serverless: Injecting required Python packages to package...
Serverless: Uploading CloudFormation file to S3...
Serverless: Uploading artifacts...
Serverless: Uploading service detector.zip file to S3 (971.78 KB)...
Serverless: Validating template...
Serverless: Skipping template validation: Unsupported in Localstack
Serverless: Updating Stack...
Serverless: Checking Stack update progress...
CloudFormation - UPDATE_IN_PROGRESS - AWS::CloudFormation::Stack - detector-local
CloudFormation - UPDATE_COMPLETE - AWS::CloudFormation::Stack - detector-local
Serverless: Stack update finished...
Service Information
service: detector
stage: local
region: us-east-1
stack: detector-local
resources: 3
api keys:
None
endpoints:
None
functions:
detectfile: detector-local-detectfile
layers:
None
Stack Outputs
ServerlessDeploymentBucketName: detector-local-ServerlessDeploymentBucket-A-Z978A-Z005A-Z489
my serverless.yaml file is:
service: detector
provider:
name: aws
runtime: python3.6
region: us-east-1
plugins:
- serverless-python-requirements
- serverless-localstack
custom:
pythonRequirements:
dockerizePip: false
localstack:
host: http://localhost
debug: true
stages:
- local
autostart: true
functions:
detectfile:
handler: lambda_function.lambda_handler
Role: 'arn:aws:iam::000000000001:role/e2e_lambda_role'
Events:
-s3:
Bucket:
Ref: Bucket1
event: 's3:ObjectCreated:*'
filter:
S3Key:
Rules:
- Name: prefix
Value: fmts/
Bucket1:
Type: 'AWS::S3::Bucket'
environment:
bucket: e2e-test
after deploy, I run awslocal lambda list-functions but don't see the lambda their
If you run
docker logs -f <container_name>
you can get the logs inside the docker/localstack container.
docker ps -a
will return you the runnning containers.
| 40,556 | |
https://nl.wikipedia.org/wiki/Tsukumi | Wikipedia | Open Web | CC-By-SA | 2,023 | Tsukumi | https://nl.wikipedia.org/w/index.php?title=Tsukumi&action=history | Dutch | Spoken | 57 | 122 | is een stad in de prefectuur Oita, Japan.
De stad heeft 22,336 inwoners (2003) en een bevolkingsdichtheid van 281,38 inwoners per km². Het beslaat een gebied van 79,38 km².
De stad is op 1 april 1951 gesticht.
Geboren
Ippei Watanabe (18 maart 1997), zwemmer
Externe links
Tsukumi officiële website in het Japans
Stad in de prefectuur Oita | 38,370 |
https://lv.wikipedia.org/wiki/Leslija%20Manna | Wikipedia | Open Web | CC-By-SA | 2,023 | Leslija Manna | https://lv.wikipedia.org/w/index.php?title=Leslija Manna&action=history | Latvian | Spoken | 92 | 294 | Leslija Manna (Leslie Mann; dzimusi ) ir amerikāņu aktrise.
Karjeru sākusi 18 gadu vecumā, kad filmējās dažādās televīzijas reklāmās. Mannas pirmā nozīmīgā loma bija 1996. gada filmā "Montieris". Arī turpmākajā aktrises karjerā filmējusies komēdiju žanra filmās, tai skaitā "Džordžs no džungļiem" (1997), "Nevainīgs 40 gadu vecumā" (2005), "Pēkšņi stāvoklī" (2007), "Atkal 17" (2009), "Ērmotie ļaudis" (2009), "Rio" (2011), "Samainīti vietām" (2011), "Mīlestība pieaugušajiem" (2012). Vairākās filmās sadarbojusies ar savu vīru Džūdu Apatovu. Žurnāls Elle 2012. gadā viņu nosauca par "Holivudas komēdiju karalieni".
Atsauces
Ārējās saites
1972. gadā dzimušie
Kalifornijā dzimušie
Amerikāņu kinoaktrises | 1,297 |
https://stackoverflow.com/questions/17384142 | StackExchange | Open Web | CC-By-SA | 2,013 | Stack Exchange | Antonio Foglia, BalusC, JB Nizet, https://stackoverflow.com/users/157882, https://stackoverflow.com/users/2498037, https://stackoverflow.com/users/571407 | English | Spoken | 237 | 538 | Stateful session bean created each time
I have a stateful session bean that initializes a Collections.synchronizedList, I add products to the list and check the list, it works(all during the same session).
But when I restart the browser it doesn't show me the list.
After watching the console I have discovered that the bean is created again.
Here is the code in my Servlet:
@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
carrelloLocal carrello=null;
HttpSession session = request.getSession();
//try to recover bean
String username= (String)session.getAttribute("username");
carrello = (carrelloLocal)session.getAttribute("carrello"+username);
System.out.println("try to recover bean for user "+username);
if(carrello == null){
//new instance of the bean
carrello = lookupcarrelloLocal();
session.setAttribute("carrello"+username, carrello);
System.out.println("created new cart for user "+username);
}
List<Prodotto> lista=carrello.getCarrello();
session.setAttribute("listacarrello", lista);
getServletContext().getRequestDispatcher("/visualizzaCarrello.jsp?listacarrello=listacarrello").forward(request, response);
}
If you restart the browser, you get a new session. That's expected. The session cookie doesn't survive browser restarts.
Not exactly a duplicate question, but the answer to that should be helpful: http://stackoverflow.com/questions/8887140/jsf-request-scoped-bean-keeps-recreating-new-stateful-session-beans-on-every-req/8889612#8889612 You namely seem to misunderstand the meaning of "stateful" in EJBs. The solution to your concrete functional requirement for which you incorrectly thought that this (as in your question) is the right solution, should be solved differently. Start learning about SQL databases and cookies.
thanks for the answers.
with the lookup i create a brand new bean each time and saving it into the http session is useless... i want to know how i can access an existing bean
| 42,372 |
https://ai.stackexchange.com/questions/20303 | StackExchange | Open Web | CC-By-SA | 2,020 | Stack Exchange | Aditya Prakash, Akash G, BG10, Barandon Francien P., Felox9876, Hatim Suwasrawala, Kamal Mehra, Martins Bash, dvdblk, hifz, https://ai.stackexchange.com/users/36072, https://ai.stackexchange.com/users/49917, https://ai.stackexchange.com/users/67628, https://ai.stackexchange.com/users/67629, https://ai.stackexchange.com/users/67630, https://ai.stackexchange.com/users/67632, https://ai.stackexchange.com/users/67633, https://ai.stackexchange.com/users/67634, https://ai.stackexchange.com/users/67645, https://ai.stackexchange.com/users/67683, https://ai.stackexchange.com/users/67691, luqo33 | English | Spoken | 1,127 | 1,587 | Is my understanding of the value function, Q function, policy, reward and return correct?
I'm a beginner in the RL field, and I would like to check that my understanding of certain RL concepts.
Value function: How good it is to be in a state S following policy π.
So, the value functions here are 0.3 and 0.9
Q function(also called state-action value, or just action value): How good it is to be in a state S and perform action A while following policy π. It uses reward to measure the state-action value
So, the state-action values here are 0.03,0.02,0.5 and 0.9
Q value: The overall expected rewards after performing action A in state S, and continuing with policy π until the end of the episode. So, essentially I can only calculate the Q value if I know all the state-action values of the actions I will be taking in the single episode.(Because the Q value takes into account the actions after the current action A, till the end of the episode, following policy π)
Reward: The metric used to tell the agent how good/bad it's action was. It is a constant value.
For e.g
1. Fall in pond --> -1
2. On stone path --> +1
3. Reach home--> +10
Return: The sum of rewards in a single episode
Policy π: A set of specific instructions an agent will follow in an episode. For example, the policy will look like:
In state 1, take action 3 ( which takes me to state 2)
In state 2, take action 2 ( which takes me to state 3)
In state 3, take action 1 ( Which takes me to state 4)
In state 4, take action 2 ( Which takes me to terminal state)
1 episode completed
And my policy will keep updating each episode to get the best return
What is the name of this book that you are using?
Value function: How good it is to be in a state $s$ following policy $\pi$.
There are different value functions. There's the state value function, often denoted as $v(s)$ (or $V(s)$), so it's a function of only one variable, i.e. $s$ (a state). There's the state-action value function $q(s, a)$ (or $Q(s, a$)). A value function is a function, so it's not a number or a vector, or whatever. It's a function, so it maps inputs to outputs. In the first case, it maps states to real numbers. In the second case, it maps states and actions to real numbers. So, we could denote the state value function as $v : \mathcal{S} \rightarrow \mathbb{R}$ (where $\mathcal{S}$ is the set of states in your environment) and state-action value function as $q : \mathcal{S} \times \mathcal{A}\rightarrow \mathbb{R}$ (where $\mathcal{A}$ is the set of actions and $\times$ means "combination of").
So, your definition of a value function is not quite correct. The value function $v(s)$ doesn't represent "how good it is to be in a state $s$ following a policy $\pi$", but "how good it is to be in a state $s$ AND THEN following policy $\pi$". To emphasize this, you often use the notation $v_{\pi}(s)$ rather than simply $v(s)$.
See What are the value functions used in reinforcement learning? for more details about existing value functions in reinforcement learning. And to see the full definition of the value functions, I suggest you read Sutton and Barto's book.
Q function (also called state-action value, or just action value): How good it is to be in a state $s$ and perform action $a$ while following policy $\pi$. It uses reward to measure the state-action value
As I said above, the $q$ function is a "value function" too. It's just a different value function than $v$.
Again, the same thing I said for $v$ also applies here, so "how good it is to be in a state $s$ and perform action $a$ while following policy $\pi$" is incorrect for the same reason your definition for $v$ was incorrect. The $q$ function can be defined as "how good it is to be in a state $s$ and take action $a$, AND, AFTER THAT, follow a given policy $\pi$. Again, to emphasize that $q$ is defined in terms of $\pi$, we often use the notation $q_\pi$.
Reward: The metric used to tell the agent how good/bad it's action was. It is a constant value.
This is roughly correct, but the reward doesn't have to be constant and it depends on your problem. Also, there's also the related notion of "reward function", which is the function that assigns rewards to each action. So, when defining your problem as a Markov decision process, you need to define this reward function. Actually, this is probably the most important function in reinforcement learning (because this is the way you teach the agent to behave).
Return: The sum of rewards in a single episode
This is roughly correct. However, note that the sum can also be a "weighted sum".
Policy: A set of specific instructions an agent will follow in an episode.
This is roughly correct, but a policy can also have some randomness in it. For example, if you are in state $s$, your policy could say "always take action $a_i$", but another policy could say "take action $a_i$ with probability $p$ and action $a_j$ with probably $1 - p$. Also, note that the policy is not restricted to an episode. It's a general function that tells the agent how to behave independently of the episode.
(Sorry, I didn't look at your examples. Maybe I will review this answer later to look at your examples too, but the information in this answer should already tell you if your examples are correct or not).
I think most of it is correct.
Q function(also called state-action value, or just action value): How good it is to be in a state S and perform action A while following policy π. It uses reward to measure the state-action value
This is a bit off. Q function basically tells you how good it is to be in state S and perform action A, and follow policy $\pi$ from the next state onwards. The action A that you take can be any action from the action space and need not be according to the policy $\pi$.
Also, I think Q-function and Q-value are mostly used interchangeably to mean the same thing.
Ah, so my action A can be different , which will affect the subsequent effectiveness policy π? (e.g my policy π requires the state to be 1 to be effective, but when I take action A it takes me to state 3 instead which means my policy isn't as effective. Hence, the Q value of this action(to state 3) will be relatively low)
| 31,956 |
https://hy.wikipedia.org/wiki/%D5%84%D5%B8%D5%BD%D5%AF%D5%B8%D5%BE%D5%B5%D5%A1%D5%B6%20%D6%83%D5%B8%D5%B2%D5%B8%D6%81%20%28%D4%BF%D6%80%D5%A1%D5%BD%D5%B6%D5%AB%20%D5%8D%D5%B8%D6%82%D5%AC%D5%AB%D5%B6%29 | Wikipedia | Open Web | CC-By-SA | 2,023 | Մոսկովյան փողոց (Կրասնի Սուլին) | https://hy.wikipedia.org/w/index.php?title=Մոսկովյան փողոց (Կրասնի Սուլին)&action=history | Armenian | Spoken | 187 | 966 | Մոսկովյան փողոց (), փողոց Ռուսաստանի Ռոստովի մարզի Կրասնի Սուլինի շրջանի և Կրասնոսուլինսկի քաղաքային շրջանի վարչական կենտրոն Կրասնի Սուլին քաղաքում։ Անվանվել է Ռուսաստանի մայրաքաղաք Մոսկվայի պատվին։ Քաղաքի հնագույն փողոցներից է։ Սկիզբ է առնում Բոլշայա Պրոլետարսկայա և 1-ին Զագոֆմանսկայա փողոցների հանգույցից և ավարտվում՝ միանալով Վոկզալնայա փողոցին։ Երկարությունը 1,4 կիլոմետր է։
Նկարագրություն
Կրասնի Սուլին քաղաքի Մոսկովյան փողոցը սկսվում է Բոլշայա Պրոլետարսկայա և 1-ին Զագոֆմանսկայա փողոցների հանգույցից, հատում է Գնիլուշա գետը, Մոսկովսկո-Մեժևոյ, Գրիբոյեդովի, Դոբրոլյուբովի, Վոդոպոյնի նրբանցքները, Խորհրդային բանակի փողոցը և ավարտվում՝ միանալով Վոկզալնայա փողոցին։ Երկարությունը 1400 մետր է։ Փողոցում կա համարակալված 65 շինություն։
Մոսկովյան փողոցի փոստային դասիչն է՝ 346357։
Փողոցն անվանվել է Ռուսաստանի մայրաքաղաք Մոսկվայի պատվին։ Քաղաքի ամենահին փողոցներից է։ Սկսել է ձևավորվել 19-րդ դարի 80-ական թվականներին և «Սուլին» կայարանը կապում էր գործարանային բնակավայրի հետ։ Այնտեղ ապրում էին առևտրով զբաղվող հարուստ կազակները՝ Դենիսովները, Դուլինները, Իլյիչովները, Սվիդովսկիները, Չուգունովները։
Ծանոթագրություններ
Աղբյուրներ
Улица Московская в Красном Сулине
Подробная карта Улица Московская, город Красный Сулин со спутника с номерами домов, индекс и маршруты, Яндекс, Гугл карты
Արտաքին հղումներ
Почтовый индекс 346356 — Красный сулин, Ростовская область
Գրականություն
Мякинченко, В.А. Сулин и сулинцы: словарь-справочник / В.А.Мякинченко. – Красный Сулин, 2002. – 180 с.
Կրասնի Սուլինի փողոցներ | 31,286 |
https://physics.stackexchange.com/questions/327013 | StackExchange | Open Web | CC-By-SA | 2,017 | Stack Exchange | English | Spoken | 366 | 764 | When do we consider states under a $U(1)$ transformations to be physically different?
Consider the Goldstone model of a complex scalar field $\Phi$. It has $U(1)$ global symmetry, so if
we apply the transformation $\Phi \to e^{i\alpha} \Phi$ the Lagrangian is left invariant.
Furthermore, we have an infinite set of possible vacua all with the same non-zero vacuum expectation value.
But the vacuum changes under a $U(1)$ transformation, so $U(1)$ symmetry is spontaneously broken.
In this case we assume that for each value $\alpha$,
$e^{i\alpha} |0 \rangle$ corresponds to a different state,
right?
But since we can't measure a phase, wouldn't it be more natural to
consider them the same states? On the other hand, if I think of the
real and imaginary parts as two independent fields, I would say that
they shouldn't be the same states.
Let's now couple $\Phi$ to a gauge field, such that the Lagrangian is invariant under local $U(1)$ transformations.
Then we do consider $e^{i\alpha} | \Psi \rangle$ to be all the same states, right? But then, doesn't this mean that the Higgs vacuum is unique?
You are wrong from this assumption:
We assume that for each value $\alpha$,
$e^{i\alpha}\Phi | 0 \rangle$ corresponds to a different state.
The wave function of the symmetry broken vacuum (the Higgs vacuum) is not $e^{i\alpha}\Phi | 0 \rangle$ as you assumed, but a coherent state $|v\rangle$ (a boson condensate), obtained by applying the shift operator $D(v)$ to the symmetric vacuum state $|0\rangle$:
$$|v\rangle=D(v)|0\rangle=e^{v\Phi^\dagger-v^*\Phi}|0\rangle,$$
where $v=|v| e^{\text{i}\alpha}$ is the vacuum expectation value (VEV) of the boson operator $\Phi=\int\text{d}^dx\Phi(x)$. Note that in the Higgs vacuum state $|v\rangle$, the operators $\Phi$ and $\Phi^\dagger$ are raised to the exponent and the phase factor $e^{\text{i}\alpha}$ is not exposed as an overall factor of the wave function. One can indeed show that $\forall x:\langle v|\Phi(x)|v\rangle=v$ using the boson commutation relation $[\Phi(x),\Phi(x')^\dagger]=\delta(x-x')$ and the definition of the symmetric vacuum state $\Phi(x)|0\rangle=0$ (see this Wikipedia page for more about coherent states). The Higgs vacuum state transforms nontrivially under the U(1) symmetry action: $|v\rangle\to|ve^{\text{i}\theta}\rangle$. The new vacuum is orthogonal to the old one $\langle v|v e^{\text{i}\theta}\rangle=\delta(\theta)$ (in the thermodynamic limit). So the two states are indeed physically different under the U(1) transformation.
| 11,883 | |
https://stackoverflow.com/questions/53191889 | StackExchange | Open Web | CC-By-SA | 2,018 | Stack Exchange | John Rouhana, Mike Robinson, https://stackoverflow.com/users/10526441, https://stackoverflow.com/users/4811873 | English | Spoken | 423 | 627 | How do i check if two strings are on the same line in a file?
I have a program where I ask the user their username and password, and it checks a file to see if they have the correct username and password. Each username and password is stored on its own line.
this is what I have
if username and password in open("logininfo").read():
print("logged in succesfully")
else:
print("Incorrect username or password")
the problem i'm having with this is that any username can be used with any password since it check the whole file. Is there any way to check if they're on the same line in the file?
You'll simply have to read the file a line at a time, check the value, and break out of the loop when you find a match. (Python code left as an exercise to the reader.)
I recommend making a Username => Password dict. Check if the username given by the user is in the dict. If it is, check if the password matches the dict value.
You can easily make this function checking the file line by line
def check_password(username, password, lines):
for line in lines:
if username in line and password in line:
return True
return False
And you can use this function this way:
check_password(username, password, open(file_name).readlines())
Note that your code does not do what you think it does. It doesn't check the username at all it only checks if the username is not "False".
This does what you want:
def checkpw():
for line in open("foobar"):
if line == (username + " " + password):
print("logged in succesfully")
return
print("Incorrect username or password")
but I strongly recommend that you use some sort of library for a task like this. You should probably at least be hashing your passwords. This code is a terrible idea.
Iterate over open('logininfo').readlines() and check whether the username is in a line AND the password is in a line.
In if username and password in open("logininfo"), it checks whether string username is not empty or None, which is not intended, so you need to check both the username and the password separately like this:
if (username in line) and (password in line):
...
Resources:
Python 3 docs
You can check it reading line by line. Don't forget to close the file or open it in a with statement to free resources when you finish.
def check_user(user, password):
with open('loginfo', 'r') as file:
for line in file:
if user in line and password in line:
return True
return False
| 5,973 |
https://diy.stackexchange.com/questions/139276 | StackExchange | Open Web | CC-By-SA | 2,018 | Stack Exchange | fixer1234, https://diy.stackexchange.com/users/28390 | English | Spoken | 266 | 335 | Laying used laminate floating floor
We had a water leak in our basement. We removed the laminate floating planks. We have let them dry out for weeks while we redid the walls. We are now beginning to relay the flooring. It isn’t going as well as the videos show laying a new floor. We are having to cut the lip on the ends a little so the planks will fit back together. Is there something we are doing wrong? Is there a video anywhere to watch? Will cutting the lip a little hurt in the long run? We are laying the floor on the concrete and only use the room for storage at the moment but still want it to look nice. Thanks you
Can you add a picture of the planks and the lip you are referring to?
The fact that the laminate was soaked with water caused it to swell. Drying it won't allow it to contract completely. Add to that the flooring was laid on concrete. Was there a vapor barrier between the concrete and the laminate? If not then the flooring had been drawing moisture from the concrete for a long time before the water leak. You are lucky you were able to disassemble the flooring without destroying it. As for trimming the lip. This will help reassembly. However, it probably won't hold together as well. My suggestion would be to scrap replacing the laminate and go purchase some LVT (Luxury Vinyl Tile, also known as LVF, Luxury Vinyl Flooring). It's more durable over time and also water proof. Hope this helps.
| 3,392 |
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