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int64
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800
coffee_shop
CREATE TABLE "shop" ( "Shop_ID" int, "Address" text, "Num_of_staff" text, "Score" real, "Open_Year" text, PRIMARY KEY ( "Shop_ID" ) ); CREATE TABLE "member" ( "Member_ID" int, "Name" text, "Membership_card" text, "Age" int, "Time_of_purchase" int, "Level_of_membership" int, "Address" text, PRIMARY KEY (...
What are the average score and average staff number of all shops?
SELECT avg(num_of_staff) , avg(score) FROM shop
801
coffee_shop
CREATE TABLE "shop" ( "Shop_ID" int, "Address" text, "Num_of_staff" text, "Score" real, "Open_Year" text, PRIMARY KEY ( "Shop_ID" ) ); CREATE TABLE "member" ( "Member_ID" int, "Name" text, "Membership_card" text, "Age" int, "Time_of_purchase" int, "Level_of_membership" int, "Address" text, PRIMARY KEY (...
Find the id and address of the shops whose score is below the average score.
SELECT shop_id , address FROM shop WHERE score < (SELECT avg(score) FROM shop)
802
coffee_shop
CREATE TABLE "shop" ( "Shop_ID" int, "Address" text, "Num_of_staff" text, "Score" real, "Open_Year" text, PRIMARY KEY ( "Shop_ID" ) ); CREATE TABLE "member" ( "Member_ID" int, "Name" text, "Membership_card" text, "Age" int, "Time_of_purchase" int, "Level_of_membership" int, "Address" text, PRIMARY KEY (...
Find the address and staff number of the shops that do not have any happy hour.
SELECT address , num_of_staff FROM shop WHERE shop_id NOT IN (SELECT shop_id FROM happy_hour)
803
coffee_shop
CREATE TABLE "shop" ( "Shop_ID" int, "Address" text, "Num_of_staff" text, "Score" real, "Open_Year" text, PRIMARY KEY ( "Shop_ID" ) ); CREATE TABLE "member" ( "Member_ID" int, "Name" text, "Membership_card" text, "Age" int, "Time_of_purchase" int, "Level_of_membership" int, "Address" text, PRIMARY KEY (...
What are the id and address of the shops which have a happy hour in May?
SELECT t1.address , t1.shop_id FROM shop AS t1 JOIN happy_hour AS t2 ON t1.shop_id = t2.shop_id WHERE MONTH = 'May'
804
coffee_shop
CREATE TABLE "shop" ( "Shop_ID" int, "Address" text, "Num_of_staff" text, "Score" real, "Open_Year" text, PRIMARY KEY ( "Shop_ID" ) ); CREATE TABLE "member" ( "Member_ID" int, "Name" text, "Membership_card" text, "Age" int, "Time_of_purchase" int, "Level_of_membership" int, "Address" text, PRIMARY KEY (...
which shop has happy hour most frequently? List its id and number of happy hours.
SELECT shop_id , count(*) FROM happy_hour GROUP BY shop_id ORDER BY count(*) DESC LIMIT 1
805
coffee_shop
CREATE TABLE "shop" ( "Shop_ID" int, "Address" text, "Num_of_staff" text, "Score" real, "Open_Year" text, PRIMARY KEY ( "Shop_ID" ) ); CREATE TABLE "member" ( "Member_ID" int, "Name" text, "Membership_card" text, "Age" int, "Time_of_purchase" int, "Level_of_membership" int, "Address" text, PRIMARY KEY (...
Which month has the most happy hours?
SELECT MONTH FROM happy_hour GROUP BY MONTH ORDER BY count(*) DESC LIMIT 1
806
coffee_shop
CREATE TABLE "shop" ( "Shop_ID" int, "Address" text, "Num_of_staff" text, "Score" real, "Open_Year" text, PRIMARY KEY ( "Shop_ID" ) ); CREATE TABLE "member" ( "Member_ID" int, "Name" text, "Membership_card" text, "Age" int, "Time_of_purchase" int, "Level_of_membership" int, "Address" text, PRIMARY KEY (...
Which months have more than 2 happy hours?
SELECT MONTH FROM happy_hour GROUP BY MONTH HAVING count(*) > 2
807
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
How many albums are there?
SELECT count(*) FROM ALBUM
808
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the number of albums.
SELECT count(*) FROM ALBUM
809
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
List the names of all music genres.
SELECT Name FROM GENRE
810
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the names of different music genres?
SELECT Name FROM GENRE
811
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find all the customer information in state NY.
SELECT * FROM CUSTOMER WHERE State = "NY"
812
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is all the customer information for customers in NY state?
SELECT * FROM CUSTOMER WHERE State = "NY"
813
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the first names and last names of the employees who live in Calgary city.
SELECT FirstName , LastName FROM EMPLOYEE WHERE City = "Calgary"
814
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the full names of employees living in the city of Calgary.
SELECT FirstName , LastName FROM EMPLOYEE WHERE City = "Calgary"
815
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the distinct billing countries of the invoices?
SELECT distinct(BillingCountry) FROM INVOICE
816
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the different billing countries for all invoices.
SELECT distinct(BillingCountry) FROM INVOICE
817
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the names of all artists that have "a" in their names.
SELECT Name FROM ARTIST WHERE Name LIKE "%a%"
818
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the names of artist who have the letter 'a' in their names?
SELECT Name FROM ARTIST WHERE Name LIKE "%a%"
819
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the title of all the albums of the artist "AC/DC".
SELECT Title FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "AC/DC"
820
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the titles of albums by the artist "AC/DC"?
SELECT Title FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "AC/DC"
821
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Hom many albums does the artist "Metallica" have?
SELECT COUNT(*) FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "Metallica"
822
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the number of albums by the artist "Metallica".
SELECT COUNT(*) FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "Metallica"
823
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Which artist does the album "Balls to the Wall" belong to?
SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T1.Title = "Balls to the Wall"
824
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the name of the artist who made the album "Balls to the Wall".
SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T1.Title = "Balls to the Wall"
825
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Which artist has the most albums?
SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId GROUP BY T2.Name ORDER BY COUNT(*) DESC LIMIT 1
826
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is the name of the artist with the greatest number of albums?
SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId GROUP BY T2.Name ORDER BY COUNT(*) DESC LIMIT 1
827
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the names of all the tracks that contain the word "you".
SELECT Name FROM TRACK WHERE Name LIKE '%you%'
828
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the names of tracks that contain the the word you in them?
SELECT Name FROM TRACK WHERE Name LIKE '%you%'
829
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is the average unit price of all the tracks?
SELECT AVG(UnitPrice) FROM TRACK
830
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the average unit price for a track.
SELECT AVG(UnitPrice) FROM TRACK
831
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the durations of the longest and the shortest tracks in milliseconds?
SELECT max(Milliseconds) , min(Milliseconds) FROM TRACK
832
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the maximum and minimum durations of tracks in milliseconds.
SELECT max(Milliseconds) , min(Milliseconds) FROM TRACK
833
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Show the album names, ids and the number of tracks for each album.
SELECT T1.Title , T2.AlbumID , COUNT(*) FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId GROUP BY T2.AlbumID
834
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the names and ids of the different albums, and how many tracks are on each?
SELECT T1.Title , T2.AlbumID , COUNT(*) FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId GROUP BY T2.AlbumID
835
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is the name of the most common genre in all tracks?
SELECT T1.Name FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId GROUP BY T2.GenreId ORDER BY COUNT(*) DESC LIMIT 1
836
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the name of the genre that is most frequent across all tracks.
SELECT T1.Name FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId GROUP BY T2.GenreId ORDER BY COUNT(*) DESC LIMIT 1
837
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is the least common media type in all tracks?
SELECT T1.Name FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId GROUP BY T2.MediaTypeId ORDER BY COUNT(*) ASC LIMIT 1
838
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is the name of the media type that is least common across all tracks?
SELECT T1.Name FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId GROUP BY T2.MediaTypeId ORDER BY COUNT(*) ASC LIMIT 1
839
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Show the album names and ids for albums that contain tracks with unit price bigger than 1.
SELECT T1.Title , T2.AlbumID FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId WHERE T2.UnitPrice > 1 GROUP BY T2.AlbumID
840
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the titles and ids for albums containing tracks with unit price greater than 1?
SELECT T1.Title , T2.AlbumID FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId WHERE T2.UnitPrice > 1 GROUP BY T2.AlbumID
841
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
How many tracks belong to rock genre?
SELECT COUNT(*) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock"
842
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Count the number of tracks that are part of the rock genre.
SELECT COUNT(*) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock"
843
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is the average unit price of tracks that belong to Jazz genre?
SELECT AVG(UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Jazz"
844
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the average unit price of jazz tracks.
SELECT AVG(UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Jazz"
845
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is the first name and last name of the customer that has email "luisg@embraer.com.br"?
SELECT FirstName , LastName FROM CUSTOMER WHERE Email = "luisg@embraer.com.br"
846
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the full name of the customer with the email "luisg@embraer.com.br".
SELECT FirstName , LastName FROM CUSTOMER WHERE Email = "luisg@embraer.com.br"
847
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
How many customers have email that contains "gmail.com"?
SELECT COUNT(*) FROM CUSTOMER WHERE Email LIKE "%gmail.com%"
848
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Count the number of customers that have an email containing "gmail.com".
SELECT COUNT(*) FROM CUSTOMER WHERE Email LIKE "%gmail.com%"
849
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is the first name and last name employee helps the customer with first name Leonie?
SELECT T2.FirstName , T2.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.FirstName = "Leonie"
850
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the full names of employees who help customers with the first name Leonie.
SELECT T2.FirstName , T2.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.FirstName = "Leonie"
851
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What city does the employee who helps the customer with postal code 70174 live in?
SELECT T2.City FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.PostalCode = "70174"
852
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the cities corresponding to employees who help customers with the postal code 70174.
SELECT T2.City FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.PostalCode = "70174"
853
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
How many distinct cities does the employees live in?
SELECT COUNT(DISTINCT city) FROM EMPLOYEE
854
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the number of different cities that employees live in.
SELECT COUNT(DISTINCT city) FROM EMPLOYEE
855
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find all invoice dates corresponding to customers with first name Astrid and last name Gruber.
SELECT T2.InvoiceDate FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.FirstName = "Astrid" AND LastName = "Gruber"
856
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the invoice dates for customers with the first name Astrid and the last name Gruber?
SELECT T2.InvoiceDate FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.FirstName = "Astrid" AND LastName = "Gruber"
857
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find all the customer last names that do not have invoice totals larger than 20.
SELECT LastName FROM CUSTOMER EXCEPT SELECT T1.LastName FROM CUSTOMER AS T1 JOIN Invoice AS T2 ON T1.CustomerId = T2.CustomerId WHERE T2.total > 20
858
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the last names of customers without invoice totals exceeding 20?
SELECT LastName FROM CUSTOMER EXCEPT SELECT T1.LastName FROM CUSTOMER AS T1 JOIN Invoice AS T2 ON T1.CustomerId = T2.CustomerId WHERE T2.total > 20
859
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the first names of all customers that live in Brazil and have an invoice.
SELECT DISTINCT T1.FirstName FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Brazil"
860
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the different first names for customers from Brazil who have also had an invoice?
SELECT DISTINCT T1.FirstName FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Brazil"
861
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the address of all customers that live in Germany and have invoice.
SELECT DISTINCT T1.Address FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Germany"
862
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the addresses of customers living in Germany who have had an invoice?
SELECT DISTINCT T1.Address FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Germany"
863
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
List the phone numbers of all employees.
SELECT Phone FROM EMPLOYEE
864
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the phone numbers for each employee?
SELECT Phone FROM EMPLOYEE
865
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
How many tracks are in the AAC audio file media type?
SELECT COUNT(*) FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId WHERE T1.Name = "AAC audio file"
866
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Count the number of tracks that are of the media type "AAC audio file".
SELECT COUNT(*) FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId WHERE T1.Name = "AAC audio file"
867
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is the average duration in milliseconds of tracks that belong to Latin or Pop genre?
SELECT AVG(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Latin" OR T1.Name = "Pop"
868
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the average millisecond length of Latin and Pop tracks.
SELECT AVG(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Latin" OR T1.Name = "Pop"
869
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Please show the employee first names and ids of employees who serve at least 10 customers.
SELECT T1.FirstName , T1.SupportRepId FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) >= 10
870
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the first names and support rep ids for employees serving 10 or more customers?
SELECT T1.FirstName , T1.SupportRepId FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) >= 10
871
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Please show the employee last names that serves no more than 20 customers.
SELECT T1.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) <= 20
872
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the last names of employees who serve at most 20 customers?
SELECT T1.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) <= 20
873
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Please list all album titles in alphabetical order.
SELECT Title FROM ALBUM ORDER BY Title
874
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are all the album titles, in alphabetical order?
SELECT Title FROM ALBUM ORDER BY Title
875
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Please list the name and id of all artists that have at least 3 albums in alphabetical order.
SELECT T2.Name , T1.ArtistId FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistID GROUP BY T1.ArtistId HAVING COUNT(*) >= 3 ORDER BY T2.Name
876
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the names and ids of artists with 3 or more albums, listed in alphabetical order?
SELECT T2.Name , T1.ArtistId FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistID GROUP BY T1.ArtistId HAVING COUNT(*) >= 3 ORDER BY T2.Name
877
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the names of artists that do not have any albums.
SELECT Name FROM ARTIST EXCEPT SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId
878
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the names of artists who have not released any albums?
SELECT Name FROM ARTIST EXCEPT SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId
879
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What is the average unit price of rock tracks?
SELECT AVG(T2.UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock"
880
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the average unit price of tracks from the Rock genre.
SELECT AVG(T2.UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock"
881
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the duration of the longest and shortest pop tracks in milliseconds?
SELECT max(Milliseconds) , min(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Pop"
882
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the maximum and minimum millisecond lengths of pop tracks.
SELECT max(Milliseconds) , min(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Pop"
883
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the birth dates of employees living in Edmonton?
SELECT BirthDate FROM EMPLOYEE WHERE City = "Edmonton"
884
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the birth dates corresponding to employees who live in the city of Edmonton.
SELECT BirthDate FROM EMPLOYEE WHERE City = "Edmonton"
885
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the distinct unit prices of all tracks?
SELECT distinct(UnitPrice) FROM TRACK
886
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the distinct unit prices for tracks.
SELECT distinct(UnitPrice) FROM TRACK
887
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
How many artists do not have any album?
SELECT count(*) FROM ARTIST WHERE artistid NOT IN(SELECT artistid FROM ALBUM)
888
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Cound the number of artists who have not released an album.
SELECT count(*) FROM ARTIST WHERE artistid NOT IN(SELECT artistid FROM ALBUM)
889
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
What are the album titles for albums containing both 'Reggae' and 'Rock' genre tracks?
SELECT T1.Title FROM Album AS T1 JOIN Track AS T2 ON T1.AlbumId = T2.AlbumId JOIN Genre AS T3 ON T2.GenreID = T3.GenreID WHERE T3.Name = 'Reggae' INTERSECT SELECT T1.Title FROM Album AS T1 JOIN Track AS T2 ON T1.AlbumId = T2.AlbumId JOIN Genre AS T3 ON T2.GenreID = T3.GenreID WHERE T3.Name = 'Rock'
890
chinook_1
CREATE TABLE `Album` ( `AlbumId` integer NOT NULL , `Title` varchar( 160 ) NOT NULL , `ArtistId` integer NOT NULL , PRIMARY KEY ( `AlbumId` ) , CONSTRAINT `FK_AlbumArtistId` FOREIGN KEY ( `ArtistId` ) REFERENCES `Artist` ( `ArtistId` ) ON DELETE NO ACTION ON UPDATE NO ACTION ...
Find the titles of albums that contain tracks of both the Reggae and Rock genres.
SELECT T1.Title FROM Album AS T1 JOIN Track AS T2 ON T1.AlbumId = T2.AlbumId JOIN Genre AS T3 ON T2.GenreID = T3.GenreID WHERE T3.Name = 'Reggae' INTERSECT SELECT T1.Title FROM Album AS T1 JOIN Track AS T2 ON T1.AlbumId = T2.AlbumId JOIN Genre AS T3 ON T2.GenreID = T3.GenreID WHERE T3.Name = 'Rock'
891
insurance_fnol
CREATE TABLE Customers ( Customer_ID INTEGER NOT NULL, Customer_name VARCHAR( 40 ), PRIMARY KEY ( Customer_ID ) ); CREATE TABLE Services ( Service_ID INTEGER NOT NULL, Service_name VARCHAR( 40 ), PRIMARY KEY ( Service_ID ) ); CREATE TABLE Available_Policies ( Policy_ID INTEGER NOT NUL...
Find all the phone numbers.
SELECT customer_phone FROM available_policies
892
insurance_fnol
CREATE TABLE Customers ( Customer_ID INTEGER NOT NULL, Customer_name VARCHAR( 40 ), PRIMARY KEY ( Customer_ID ) ); CREATE TABLE Services ( Service_ID INTEGER NOT NULL, Service_name VARCHAR( 40 ), PRIMARY KEY ( Service_ID ) ); CREATE TABLE Available_Policies ( Policy_ID INTEGER NOT NUL...
What are all the phone numbers?
SELECT customer_phone FROM available_policies
893
insurance_fnol
CREATE TABLE Customers ( Customer_ID INTEGER NOT NULL, Customer_name VARCHAR( 40 ), PRIMARY KEY ( Customer_ID ) ); CREATE TABLE Services ( Service_ID INTEGER NOT NULL, Service_name VARCHAR( 40 ), PRIMARY KEY ( Service_ID ) ); CREATE TABLE Available_Policies ( Policy_ID INTEGER NOT NUL...
What are the customer phone numbers under the policy "Life Insurance"?
SELECT customer_phone FROM available_policies WHERE policy_type_code = "Life Insurance"
894
insurance_fnol
CREATE TABLE Customers ( Customer_ID INTEGER NOT NULL, Customer_name VARCHAR( 40 ), PRIMARY KEY ( Customer_ID ) ); CREATE TABLE Services ( Service_ID INTEGER NOT NULL, Service_name VARCHAR( 40 ), PRIMARY KEY ( Service_ID ) ); CREATE TABLE Available_Policies ( Policy_ID INTEGER NOT NUL...
What are the phone numbers of customers using the policy with the code "Life Insurance"?
SELECT customer_phone FROM available_policies WHERE policy_type_code = "Life Insurance"
895
insurance_fnol
CREATE TABLE Customers ( Customer_ID INTEGER NOT NULL, Customer_name VARCHAR( 40 ), PRIMARY KEY ( Customer_ID ) ); CREATE TABLE Services ( Service_ID INTEGER NOT NULL, Service_name VARCHAR( 40 ), PRIMARY KEY ( Service_ID ) ); CREATE TABLE Available_Policies ( Policy_ID INTEGER NOT NUL...
Which policy type has the most records in the database?
SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY count(*) DESC LIMIT 1
896
insurance_fnol
CREATE TABLE Customers ( Customer_ID INTEGER NOT NULL, Customer_name VARCHAR( 40 ), PRIMARY KEY ( Customer_ID ) ); CREATE TABLE Services ( Service_ID INTEGER NOT NULL, Service_name VARCHAR( 40 ), PRIMARY KEY ( Service_ID ) ); CREATE TABLE Available_Policies ( Policy_ID INTEGER NOT NUL...
Which policy type appears most frequently in the available policies?
SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY count(*) DESC LIMIT 1
897
insurance_fnol
CREATE TABLE Customers ( Customer_ID INTEGER NOT NULL, Customer_name VARCHAR( 40 ), PRIMARY KEY ( Customer_ID ) ); CREATE TABLE Services ( Service_ID INTEGER NOT NULL, Service_name VARCHAR( 40 ), PRIMARY KEY ( Service_ID ) ); CREATE TABLE Available_Policies ( Policy_ID INTEGER NOT NUL...
What are all the customer phone numbers under the most popular policy type?
SELECT customer_phone FROM available_policies WHERE policy_type_code = (SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY count(*) DESC LIMIT 1)
898
insurance_fnol
CREATE TABLE Customers ( Customer_ID INTEGER NOT NULL, Customer_name VARCHAR( 40 ), PRIMARY KEY ( Customer_ID ) ); CREATE TABLE Services ( Service_ID INTEGER NOT NULL, Service_name VARCHAR( 40 ), PRIMARY KEY ( Service_ID ) ); CREATE TABLE Available_Policies ( Policy_ID INTEGER NOT NUL...
Find the phone numbers of customers using the most common policy type among the available policies.
SELECT customer_phone FROM available_policies WHERE policy_type_code = (SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY count(*) DESC LIMIT 1)
899
insurance_fnol
CREATE TABLE Customers ( Customer_ID INTEGER NOT NULL, Customer_name VARCHAR( 40 ), PRIMARY KEY ( Customer_ID ) ); CREATE TABLE Services ( Service_ID INTEGER NOT NULL, Service_name VARCHAR( 40 ), PRIMARY KEY ( Service_ID ) ); CREATE TABLE Available_Policies ( Policy_ID INTEGER NOT NUL...
Find the policy type used by more than 4 customers.
SELECT policy_type_code FROM available_policies GROUP BY policy_type_code HAVING count(*) > 4