question
string
answer
string
context
string
dialect
string
spider_hardness
string
extended_hardness
string
classification_status
string
syntax_status
string
ddl_syntax_status
string
schema_status
string
syntax_errors
list
ddl_syntax_errors
list
schema_errors
list
classification_errors
list
status_history
list
correction_status
string
component1
int64
component2
int64
others
int64
validation_source
dict
For each classroom, report the classroom number and the number of grades using it.
SELECT classroom, COUNT(DISTINCT grade) FROM list GROUP BY classroom
CREATE TABLE list (classroom VARCHAR(255), grade VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2368 }
Which classroom has the most students?
SELECT classroom FROM list GROUP BY classroom ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE list (classroom VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2369 }
Report the number of students in each classroom.
SELECT classroom, COUNT(*) FROM list GROUP BY classroom
CREATE TABLE list (classroom VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2370 }
For each grade 0 classroom, report the total number of students.
SELECT classroom, COUNT(*) FROM list WHERE grade = '0' GROUP BY classroom
CREATE TABLE list (classroom VARCHAR(255), grade VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 2371 }
Report the total number of students for each fourth-grade classroom.
SELECT classroom, COUNT(*) FROM list WHERE grade = '4' GROUP BY classroom
CREATE TABLE list (classroom VARCHAR(255), grade VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 2372 }
Find the name of the teacher who teaches the largest number of students.
SELECT T2.firstname, T2.lastname FROM list AS T1 JOIN teachers AS T2 ON T1.classroom = T2.classroom GROUP BY T2.firstname, T2.lastname ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE list (classroom VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE teachers (firstname VARCHAR(255), lastname VARCHAR(255), classroom VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
2
{ "split": "train", "row_index": 2373 }
Find the number of students in one classroom.
SELECT COUNT(*), classroom FROM list GROUP BY classroom
CREATE TABLE list (classroom VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2374 }
How many companies are headquartered in the US?
SELECT COUNT(*) FROM company WHERE Headquarters = 'USA'
CREATE TABLE company (Headquarters VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2375 }
List the names of companies by ascending number of sales.
SELECT Name FROM company ORDER BY Sales_in_Billion
CREATE TABLE company (Name VARCHAR(255), Sales_in_Billion VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2376 }
What are the headquarters and industries of all companies?
SELECT Headquarters, Industry FROM company
CREATE TABLE company (Headquarters VARCHAR(255), Industry VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 2377 }
Show the names of companies in the banking or retailing industry?
SELECT Name FROM company WHERE Industry = 'Banking' OR Industry = 'Retailing'
CREATE TABLE company (Name VARCHAR(255), Industry VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 2378 }
What is the maximum and minimum market value of companies?
SELECT MAX(Market_Value_in_Billion), MIN(Market_Value_in_Billion) FROM company
CREATE TABLE company (Market_Value_in_Billion INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
2
{ "split": "train", "row_index": 2379 }
What is the headquarter of the company with the largest sales?
SELECT Headquarters FROM company ORDER BY Sales_in_Billion DESC LIMIT 1
CREATE TABLE company (Headquarters VARCHAR(255), Sales_in_Billion VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 2380 }
Show the different headquarters and number of companies at each headquarter.
SELECT Headquarters, COUNT(*) FROM company GROUP BY Headquarters
CREATE TABLE company (Headquarters VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2381 }
Show the most common headquarter for companies.
SELECT Headquarters FROM company GROUP BY Headquarters ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE company (Headquarters VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2382 }
Show the headquarters that have at least two companies.
SELECT Headquarters FROM company GROUP BY Headquarters HAVING COUNT(*) >= 2
CREATE TABLE company (Headquarters VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2383 }
Show the headquarters that have both companies in banking industry and companies in oil and gas industry.
SELECT Headquarters FROM company WHERE Industry = 'Banking' INTERSECT SELECT Headquarters FROM company WHERE Industry = 'Oil and gas'
CREATE TABLE company (Headquarters VARCHAR(255), Industry VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 2384 }
Show the names of companies and of employees.
SELECT T3.Name, T2.Name FROM employment AS T1 JOIN people AS T2 ON T1.People_ID = T2.People_ID JOIN company AS T3 ON T1.Company_ID = T3.Company_ID
CREATE TABLE company (Name VARCHAR(255), Company_ID VARCHAR(255)); CREATE TABLE employment (People_ID VARCHAR(255), Company_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE people (Name VARCHAR(255), People_ID VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 2385 }
Show names of companies and that of employees in descending order of number of years working for that employee.
SELECT T3.Name, T2.Name FROM employment AS T1 JOIN people AS T2 ON T1.People_ID = T2.People_ID JOIN company AS T3 ON T1.Company_ID = T3.Company_ID ORDER BY T1.Year_working
CREATE TABLE people (Name VARCHAR(255), People_ID VARCHAR(255)); CREATE TABLE company (Name VARCHAR(255), Company_ID VARCHAR(255)); CREATE TABLE employment (People_ID VARCHAR(255), Company_ID VARCHAR(255), Year_working VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 2386 }
Show the names of employees that work for companies with sales bigger than 200.
SELECT T2.Name FROM employment AS T1 JOIN people AS T2 ON T1.People_ID = T2.People_ID JOIN company AS T3 ON T1.Company_ID = T3.Company_ID WHERE T3.Sales_in_Billion > 200
CREATE TABLE employment (People_ID VARCHAR(255), Company_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE company (Company_ID VARCHAR(255), Sales_in_Billion INTEGER); CREATE TABLE people (Name VARCHAR(255), People_ID VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2387 }
Show the names of companies and the number of employees they have
SELECT T3.Name, COUNT(*) FROM employment AS T1 JOIN people AS T2 ON T1.People_ID = T2.People_ID JOIN company AS T3 ON T1.Company_ID = T3.Company_ID GROUP BY T3.Name
CREATE TABLE company (Name VARCHAR(255), Company_ID VARCHAR(255)); CREATE TABLE people (People_ID VARCHAR(255), Name VARCHAR(255)); CREATE TABLE employment (People_ID VARCHAR(255), Company_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 2388 }
List the names of people that are not employed by any company
SELECT Name FROM people WHERE NOT People_ID IN (SELECT People_ID FROM employment)
CREATE TABLE employment (Name VARCHAR(255), People_ID VARCHAR(255)); CREATE TABLE people (Name VARCHAR(255), People_ID VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 2389 }
list the names of the companies with more than 200 sales in the descending order of sales and profits.
SELECT name FROM company WHERE Sales_in_Billion > 200 ORDER BY Sales_in_Billion, Profits_in_Billion DESC
CREATE TABLE company (name VARCHAR(255), Sales_in_Billion INTEGER, Profits_in_Billion VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
null
medium
extended
valid
valid
valid
[]
[]
[]
[ "Mixed ORDER BY directions" ]
[ "validated", "extended", "extended" ]
null
null
null
null
{ "split": "train", "row_index": 2390 }
How many film are there?
SELECT COUNT(*) FROM film
CREATE TABLE film (Id VARCHAR(255), Title VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2391 }
List the distinct director of all films.
SELECT DISTINCT Director FROM film
CREATE TABLE film (Director VARCHAR(255), Title VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2392 }
What is the average ticket sales gross in dollars of films?
SELECT AVG(Gross_in_dollar) FROM film
CREATE TABLE film (Gross_in_dollar INTEGER, Title VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2393 }
What are the low and high estimates of film markets?
SELECT Low_Estimate, High_Estimate FROM film_market_estimation
CREATE TABLE film_market_estimation (Low_Estimate VARCHAR(255), High_Estimate VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 2394 }
What are the types of film market estimations in year 1995?
SELECT TYPE FROM film_market_estimation WHERE YEAR = 1995
CREATE TABLE film_market_estimation (TYPE VARCHAR(255), YEAR VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2395 }
What are the maximum and minimum number of cities in all markets.
SELECT MAX(Number_cities), MIN(Number_cities) FROM market
CREATE TABLE market (Number_cities INTEGER, Country VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
2
{ "split": "train", "row_index": 2396 }
How many markets have number of cities smaller than 300?
SELECT COUNT(*) FROM market WHERE Number_cities < 300
CREATE TABLE market (Number_cities INTEGER, Country VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2397 }
List all countries of markets in ascending alphabetical order.
SELECT Country FROM market ORDER BY Country
CREATE TABLE market (Country VARCHAR(255), Number_cities INTEGER)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2398 }
List all countries of markets in descending order of number of cities.
SELECT Country FROM market ORDER BY Number_cities DESC
CREATE TABLE market (Country VARCHAR(255), Number_cities VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2399 }
Please show the titles of films and the types of market estimations.
SELECT T1.Title, T2.Type FROM film AS T1 JOIN film_market_estimation AS T2 ON T1.Film_ID = T2.Film_ID
CREATE TABLE film (Title VARCHAR(255), Film_ID VARCHAR(255)); CREATE TABLE film_market_estimation (Type VARCHAR(255), Film_ID VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2400 }
Show the distinct director of films with market estimation in the year of 1995.
SELECT DISTINCT T1.Director FROM film AS T1 JOIN film_market_estimation AS T2 ON T1.Film_ID = T2.Film_ID WHERE T2.Year = 1995
CREATE TABLE film (Director VARCHAR(255), Film_ID VARCHAR(255)); CREATE TABLE film_market_estimation (Film_ID VARCHAR(255), Year VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 2401 }
What is the average number of cities of markets with low film market estimate bigger than 10000?
SELECT AVG(T2.Number_cities) FROM film_market_estimation AS T1 JOIN market AS T2 ON T1.Market_ID = T2.Market_ID WHERE T1.Low_Estimate > 10000
CREATE TABLE market (Number_cities INTEGER, Market_ID VARCHAR(255)); CREATE TABLE film_market_estimation (Market_ID VARCHAR(255), Low_Estimate INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 2402 }
Please list the countries and years of film market estimations.
SELECT T2.Country, T1.Year FROM film_market_estimation AS T1 JOIN market AS T2 ON T1.Market_ID = T2.Market_ID
CREATE TABLE film_market_estimation (Year VARCHAR(255), Market_ID VARCHAR(255)); CREATE TABLE market (Country VARCHAR(255), Market_ID VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2403 }
Please list the years of film market estimations when the market is in country "Japan" in descending order.
SELECT T1.Year FROM film_market_estimation AS T1 JOIN market AS T2 ON T1.Market_ID = T2.Market_ID WHERE T2.Country = 'Japan' ORDER BY T1.Year DESC
CREATE TABLE film_market_estimation (Year VARCHAR(255), Market_ID VARCHAR(255)); CREATE TABLE market (Market_ID VARCHAR(255), Country VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2404 }
List the studios of each film and the number of films produced by that studio.
SELECT Studio, COUNT(*) FROM film GROUP BY Studio
CREATE TABLE film (Studio VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2405 }
List the name of film studio that have the most number of films.
SELECT Studio FROM film GROUP BY Studio ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE film (Studio VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2406 }
List the names of studios that have at least two films.
SELECT Studio FROM film GROUP BY Studio HAVING COUNT(*) >= 2
CREATE TABLE film (Studio VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2407 }
List the title of films that do not have any market estimation.
SELECT Title FROM film WHERE NOT Film_ID IN (SELECT Film_ID FROM film_market_estimation)
CREATE TABLE film_market_estimation (Title VARCHAR(255), Film_ID VARCHAR(255)); CREATE TABLE film (Title VARCHAR(255), Film_ID VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 2408 }
Show the studios that have produced films with director "Nicholas Meyer" and "Walter Hill".
SELECT Studio FROM film WHERE Director = 'Nicholas Meyer' INTERSECT SELECT Studio FROM film WHERE Director = 'Walter Hill'
CREATE TABLE film (Studio VARCHAR(255), Director VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 2409 }
Find the titles and studios of the films that are produced by some film studios that contained the word "Universal".
SELECT title, Studio FROM film WHERE Studio LIKE '%Universal%'
CREATE TABLE film (title VARCHAR(255), Studio VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 2410 }
Show the studios that have not produced films with director "Walter Hill".
SELECT Studio FROM film EXCEPT SELECT Studio FROM film WHERE Director = 'Walter Hill'
CREATE TABLE film (Studio VARCHAR(255), Director VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
1
0
{ "split": "train", "row_index": 2411 }
List the studios which average gross is above 4500000.
SELECT Studio FROM film GROUP BY Studio HAVING AVG(Gross_in_dollar) >= 4500000
CREATE TABLE film (Studio VARCHAR(255), Gross_in_dollar INTEGER)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2412 }
What are the titles and directors of the films were never presented in China?
SELECT title, director FROM film WHERE NOT film_id IN (SELECT film_id FROM film_market_estimation AS T1 JOIN market AS T2 ON T1.market_id = T2.Market_ID WHERE country = 'China')
CREATE TABLE film (title VARCHAR(255), director VARCHAR(255), film_id VARCHAR(255), country VARCHAR(255)); CREATE TABLE market (Market_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE film_market_estimation (market_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 2414 }
How many calendar items do we have?
SELECT COUNT(*) FROM Ref_calendar
CREATE TABLE Ref_calendar (Id VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2415 }
Show all calendar dates and day Numbers.
SELECT calendar_date, day_Number FROM Ref_calendar
CREATE TABLE Ref_calendar (calendar_date VARCHAR(255), day_Number VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 2416 }
Show the number of document types.
SELECT COUNT(*) FROM Ref_document_types
CREATE TABLE Ref_document_types (Id VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2417 }
List all document type codes and document type names.
SELECT document_type_code, document_type_name FROM Ref_document_types
CREATE TABLE Ref_document_types (document_type_code VARCHAR(255), document_type_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 2418 }
What is the name and description for document type code RV?
SELECT document_type_name, document_type_description FROM Ref_document_types WHERE document_type_code = 'RV'
CREATE TABLE Ref_document_types (document_type_name VARCHAR(255), document_type_description VARCHAR(255), document_type_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2419 }
What is the document type code for document type "Paper"?
SELECT document_type_code FROM Ref_document_types WHERE document_type_name = 'Paper'
CREATE TABLE Ref_document_types (document_type_code VARCHAR(255), document_type_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2420 }
Show the number of documents with document type code CV or BK.
SELECT COUNT(*) FROM All_documents WHERE document_type_code = 'CV' OR document_type_code = 'BK'
CREATE TABLE All_documents (document_type_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 2421 }
What is the date when the document "Marry CV" was stored?
SELECT date_stored FROM All_documents WHERE Document_name = 'Marry CV'
CREATE TABLE All_documents (date_stored VARCHAR(255), Document_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2422 }
What is the document type name for the document with name "How to read a book"?
SELECT T2.document_type_name FROM All_documents AS T1 JOIN Ref_document_types AS T2 ON T1.document_type_code = T2.document_type_code WHERE T1.document_name = 'How to read a book'
CREATE TABLE Ref_document_types (document_type_name VARCHAR(255), document_type_code VARCHAR(255)); CREATE TABLE All_documents (document_type_code VARCHAR(255), document_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 2424 }
Show the number of locations.
SELECT COUNT(*) FROM Ref_locations
CREATE TABLE Ref_locations (Id VARCHAR(255), location_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2425 }
List all location codes and location names.
SELECT location_code, location_name FROM Ref_locations
CREATE TABLE Ref_locations (location_code VARCHAR(255), location_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 2426 }
What are the name and description for location code x?
SELECT location_name, location_description FROM Ref_locations WHERE location_code = 'x'
CREATE TABLE Ref_locations (location_name VARCHAR(255), location_description VARCHAR(255), location_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2427 }
What is the location code for the country "Canada"?
SELECT location_code FROM Ref_locations WHERE location_name = 'Canada'
CREATE TABLE Ref_locations (location_code VARCHAR(255), location_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2428 }
How many roles are there?
SELECT COUNT(*) FROM ROLES
CREATE TABLE ROLES (Id VARCHAR(255), role_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2429 }
List all role codes, role names, and role descriptions.
SELECT role_code, role_name, role_description FROM ROLES
CREATE TABLE ROLES (role_code VARCHAR(255), role_name VARCHAR(255), role_description VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 2430 }
What are the name and description for role code "MG"?
SELECT role_name, role_description FROM ROLES WHERE role_code = 'MG'
CREATE TABLE ROLES (role_name VARCHAR(255), role_description VARCHAR(255), role_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2431 }
Show the description for role name "Proof Reader".
SELECT role_description FROM ROLES WHERE role_name = 'Proof Reader'
CREATE TABLE ROLES (role_description VARCHAR(255), role_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2432 }
What is the id for the employee called Ebba?
SELECT employee_ID FROM Employees WHERE employee_name = 'Ebba'
CREATE TABLE Employees (employee_ID VARCHAR(255), employee_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2434 }
Show the names of all the employees with role "HR".
SELECT employee_name FROM Employees WHERE role_code = 'HR'
CREATE TABLE Employees (employee_name VARCHAR(255), role_code VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2435 }
Show all role codes and the number of employees in each role.
SELECT role_code, COUNT(*) FROM Employees GROUP BY role_code
CREATE TABLE Employees (role_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2436 }
What is the role code with the largest number of employees?
SELECT role_code FROM Employees GROUP BY role_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Employees (role_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2437 }
Show all role codes with at least 3 employees.
SELECT role_code FROM Employees GROUP BY role_code HAVING COUNT(*) >= 3
CREATE TABLE Employees (role_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2438 }
Show the role code with the least employees.
SELECT role_code FROM Employees GROUP BY role_code ORDER BY COUNT(*) LIMIT 1
CREATE TABLE Employees (role_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2439 }
What is the role name and role description for employee called Ebba?
SELECT T2.role_name, T2.role_description FROM Employees AS T1 JOIN ROLES AS T2 ON T1.role_code = T2.role_code WHERE T1.employee_name = 'Ebba'
CREATE TABLE Employees (role_code VARCHAR(255), employee_name VARCHAR(255)); CREATE TABLE ROLES (role_name VARCHAR(255), role_description VARCHAR(255), role_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 2440 }
Show the names of employees with role name Editor.
SELECT T1.employee_name FROM Employees AS T1 JOIN ROLES AS T2 ON T1.role_code = T2.role_code WHERE T2.role_name = 'Editor'
CREATE TABLE ROLES (role_code VARCHAR(255), role_name VARCHAR(255)); CREATE TABLE Employees (employee_name VARCHAR(255), role_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 2441 }
Show the employee ids for all employees with role name "Human Resource" or "Manager".
SELECT T1.employee_id FROM Employees AS T1 JOIN ROLES AS T2 ON T1.role_code = T2.role_code WHERE T2.role_name = 'Human Resource' OR T2.role_name = 'Manager'
CREATE TABLE Employees (employee_id VARCHAR(255), role_code VARCHAR(255)); CREATE TABLE ROLES (role_code VARCHAR(255), role_name VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 2442 }
What are the different location codes for documents?
SELECT DISTINCT location_code FROM Document_locations
CREATE TABLE Document_locations (location_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2443 }
Show the location name for document "Robin CV".
SELECT T3.location_name FROM All_documents AS T1 JOIN Document_locations AS T2 ON T1.document_id = T2.document_id JOIN Ref_locations AS T3 ON T2.location_code = T3.location_code WHERE T1.document_name = 'Robin CV'
CREATE TABLE Ref_locations (location_name VARCHAR(255), location_code VARCHAR(255)); CREATE TABLE All_documents (document_id VARCHAR(255), document_name VARCHAR(255)); CREATE TABLE Document_locations (document_id VARCHAR(255), location_code VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2444 }
Show the location code, the starting date and ending data in that location for all the documents.
SELECT location_code, date_in_location_from, date_in_locaton_to FROM Document_locations
CREATE TABLE Document_locations (location_code VARCHAR(255), date_in_location_from VARCHAR(255), date_in_locaton_to VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 2445 }
What is "the date in location from" and "the date in location to" for the document with name "Robin CV"?
SELECT T1.date_in_location_from, T1.date_in_locaton_to FROM Document_locations AS T1 JOIN All_documents AS T2 ON T1.document_id = T2.document_id WHERE T2.document_name = 'Robin CV'
CREATE TABLE All_documents (document_id VARCHAR(255), document_name VARCHAR(255)); CREATE TABLE Document_locations (date_in_location_from VARCHAR(255), date_in_locaton_to VARCHAR(255), document_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 2446 }
Show the location codes and the number of documents in each location.
SELECT location_code, COUNT(*) FROM Document_locations GROUP BY location_code
CREATE TABLE Document_locations (location_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2447 }
What is the location code with the most documents?
SELECT location_code FROM Document_locations GROUP BY location_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Document_locations (location_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2448 }
Show the location codes with at least 3 documents.
SELECT location_code FROM Document_locations GROUP BY location_code HAVING COUNT(*) >= 3
CREATE TABLE Document_locations (location_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 2449 }
Show the location name and code with the least documents.
SELECT T2.location_name, T1.location_code FROM Document_locations AS T1 JOIN Ref_locations AS T2 ON T1.location_code = T2.location_code GROUP BY T1.location_code ORDER BY COUNT(*) LIMIT 1
CREATE TABLE Document_locations (location_code VARCHAR(255)); CREATE TABLE Ref_locations (location_name VARCHAR(255), location_code VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 2450 }
What are the names of the employees who authorised the destruction and the employees who destroyed the corresponding documents?
SELECT T2.employee_name, T3.employee_name FROM Documents_to_be_destroyed AS T1 JOIN Employees AS T2 ON T1.Destruction_Authorised_by_Employee_ID = T2.employee_id JOIN Employees AS T3 ON T1.Destroyed_by_Employee_ID = T3.employee_id
CREATE TABLE Employees (employee_name VARCHAR(255), employee_id VARCHAR(255)); CREATE TABLE Documents_to_be_destroyed (Destruction_Authorised_by_Employee_ID VARCHAR(255), Destroyed_by_Employee_ID VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 2451 }
Show the id of each employee and the number of document destruction authorised by that employee.
SELECT Destruction_Authorised_by_Employee_ID, COUNT(*) FROM Documents_to_be_destroyed GROUP BY Destruction_Authorised_by_Employee_ID
CREATE TABLE Documents_to_be_destroyed (Destruction_Authorised_by_Employee_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2452 }
Show the employee ids and the number of documents destroyed by each employee.
SELECT Destroyed_by_Employee_ID, COUNT(*) FROM Documents_to_be_destroyed GROUP BY Destroyed_by_Employee_ID
CREATE TABLE Documents_to_be_destroyed (Destroyed_by_Employee_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 2453 }
Show the ids of the employees who don't authorize destruction for any document.
SELECT employee_id FROM Employees EXCEPT SELECT Destruction_Authorised_by_Employee_ID FROM Documents_to_be_destroyed
CREATE TABLE Documents_to_be_destroyed (employee_id VARCHAR(255), Destruction_Authorised_by_Employee_ID VARCHAR(255)); CREATE TABLE Employees (employee_id VARCHAR(255), Destruction_Authorised_by_Employee_ID VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
1
0
{ "split": "train", "row_index": 2454 }
Show the ids of all employees who have authorized destruction.
SELECT DISTINCT Destruction_Authorised_by_Employee_ID FROM Documents_to_be_destroyed
CREATE TABLE Documents_to_be_destroyed (Destruction_Authorised_by_Employee_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2455 }
Show the ids of all employees who have destroyed a document.
SELECT DISTINCT Destroyed_by_Employee_ID FROM Documents_to_be_destroyed
CREATE TABLE Documents_to_be_destroyed (Destroyed_by_Employee_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2456 }
Show the ids of all employees who don't destroy any document.
SELECT employee_id FROM Employees EXCEPT SELECT Destroyed_by_Employee_ID FROM Documents_to_be_destroyed
CREATE TABLE Employees (employee_id VARCHAR(255), Destroyed_by_Employee_ID VARCHAR(255)); CREATE TABLE Documents_to_be_destroyed (employee_id VARCHAR(255), Destroyed_by_Employee_ID VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
1
0
{ "split": "train", "row_index": 2457 }
Show the ids of all employees who have either destroyed a document or made an authorization to do this.
SELECT Destroyed_by_Employee_ID FROM Documents_to_be_destroyed UNION SELECT Destruction_Authorised_by_Employee_ID FROM Documents_to_be_destroyed
CREATE TABLE Documents_to_be_destroyed (Destroyed_by_Employee_ID VARCHAR(255), Destruction_Authorised_by_Employee_ID VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
1
0
{ "split": "train", "row_index": 2458 }
What are the names of all clubs?
SELECT clubname FROM club
CREATE TABLE club (clubname VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2459 }
How many students are there?
SELECT COUNT(*) FROM student
CREATE TABLE student (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2460 }
What are the first names of all the students?
SELECT DISTINCT fname FROM student
CREATE TABLE student (fname VARCHAR(255), lname VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 2461 }
Find the last names of the members of the club "Bootup Baltimore".
SELECT t3.lname FROM club AS t1 JOIN member_of_club AS t2 ON t1.clubid = t2.clubid JOIN student AS t3 ON t2.stuid = t3.stuid WHERE t1.clubname = 'Bootup Baltimore'
CREATE TABLE club (clubid VARCHAR(255), clubname VARCHAR(255)); CREATE TABLE student (lname VARCHAR(255), stuid VARCHAR(255)); CREATE TABLE member_of_club (clubid VARCHAR(255), stuid VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2462 }
Who are the members of the club named "Hopkins Student Enterprises"? Show the last name.
SELECT t3.lname FROM club AS t1 JOIN member_of_club AS t2 ON t1.clubid = t2.clubid JOIN student AS t3 ON t2.stuid = t3.stuid WHERE t1.clubname = 'Hopkins Student Enterprises'
CREATE TABLE club (clubid VARCHAR(255), clubname VARCHAR(255)); CREATE TABLE student (lname VARCHAR(255), stuid VARCHAR(255)); CREATE TABLE member_of_club (clubid VARCHAR(255), stuid VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2463 }
How many members does the club "Tennis Club" has?
SELECT COUNT(*) FROM club AS t1 JOIN member_of_club AS t2 ON t1.clubid = t2.clubid JOIN student AS t3 ON t2.stuid = t3.stuid WHERE t1.clubname = 'Tennis Club'
CREATE TABLE club (clubid VARCHAR(255), clubname VARCHAR(255)); CREATE TABLE member_of_club (clubid VARCHAR(255), stuid VARCHAR(255)); CREATE TABLE student (stuid VARCHAR(255), fname VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2464 }
Find the number of members of club "Pen and Paper Gaming".
SELECT COUNT(*) FROM club AS t1 JOIN member_of_club AS t2 ON t1.clubid = t2.clubid JOIN student AS t3 ON t2.stuid = t3.stuid WHERE t1.clubname = 'Pen and Paper Gaming'
CREATE TABLE club (clubid VARCHAR(255), clubname VARCHAR(255)); CREATE TABLE member_of_club (clubid VARCHAR(255), stuid VARCHAR(255)); CREATE TABLE student (stuid VARCHAR(255), fname VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 2465 }
How many clubs does "Linda Smith" belong to?
SELECT COUNT(*) FROM club AS t1 JOIN member_of_club AS t2 ON t1.clubid = t2.clubid JOIN student AS t3 ON t2.stuid = t3.stuid WHERE t3.fname = 'Linda' AND t3.lname = 'Smith'
CREATE TABLE student (stuid VARCHAR(255), fname VARCHAR(255), lname VARCHAR(255)); CREATE TABLE member_of_club (clubid VARCHAR(255), stuid VARCHAR(255)); CREATE TABLE club (clubid VARCHAR(255), clubname VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 2466 }
Find the number of clubs where "Tracy Kim" is a member.
SELECT COUNT(*) FROM club AS t1 JOIN member_of_club AS t2 ON t1.clubid = t2.clubid JOIN student AS t3 ON t2.stuid = t3.stuid WHERE t3.fname = 'Tracy' AND t3.lname = 'Kim'
CREATE TABLE student (stuid VARCHAR(255), fname VARCHAR(255), lname VARCHAR(255)); CREATE TABLE member_of_club (clubid VARCHAR(255), stuid VARCHAR(255)); CREATE TABLE club (clubid VARCHAR(255), clubname VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 2467 }
Find all the female members of club "Bootup Baltimore". Show the first name and last name.
SELECT t3.fname, t3.lname FROM club AS t1 JOIN member_of_club AS t2 ON t1.clubid = t2.clubid JOIN student AS t3 ON t2.stuid = t3.stuid WHERE t1.clubname = 'Bootup Baltimore' AND t3.sex = 'F'
CREATE TABLE club (clubid VARCHAR(255), clubname VARCHAR(255)); CREATE TABLE student (fname VARCHAR(255), lname VARCHAR(255), stuid VARCHAR(255), sex VARCHAR(255)); CREATE TABLE member_of_club (clubid VARCHAR(255), stuid VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
2
{ "split": "train", "row_index": 2468 }
Find all the male members of club "Hopkins Student Enterprises". Show the first name and last name.
SELECT t3.fname, t3.lname FROM club AS t1 JOIN member_of_club AS t2 ON t1.clubid = t2.clubid JOIN student AS t3 ON t2.stuid = t3.stuid WHERE t1.clubname = 'Hopkins Student Enterprises' AND t3.sex = 'M'
CREATE TABLE club (clubid VARCHAR(255), clubname VARCHAR(255)); CREATE TABLE student (fname VARCHAR(255), lname VARCHAR(255), stuid VARCHAR(255), sex VARCHAR(255)); CREATE TABLE member_of_club (clubid VARCHAR(255), stuid VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
2
{ "split": "train", "row_index": 2469 }
Find all members of "Bootup Baltimore" whose major is "600". Show the first name and last name.
SELECT t3.fname, t3.lname FROM club AS t1 JOIN member_of_club AS t2 ON t1.clubid = t2.clubid JOIN student AS t3 ON t2.stuid = t3.stuid WHERE t1.clubname = 'Bootup Baltimore' AND t3.major = '600'
CREATE TABLE club (clubid VARCHAR(255), clubname VARCHAR(255)); CREATE TABLE member_of_club (clubid VARCHAR(255), stuid VARCHAR(255)); CREATE TABLE student (fname VARCHAR(255), lname VARCHAR(255), stuid VARCHAR(255), major VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
2
{ "split": "train", "row_index": 2470 }