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Please show the date of ceremony of the volumes that last more than 2 weeks on top. | CREATE TABLE music_festival (Date_of_ceremony VARCHAR, Volume VARCHAR); CREATE TABLE volume (Volume_ID VARCHAR, Weeks_on_Top INTEGER) | SELECT T1.Date_of_ceremony FROM music_festival AS T1 JOIN volume AS T2 ON T1.Volume = T2.Volume_ID WHERE T2.Weeks_on_Top > 2 | ### Context: CREATE TABLE music_festival (Date_of_ceremony VARCHAR, Volume VARCHAR); CREATE TABLE volume (Volume_ID VARCHAR, Weeks_on_Top INTEGER) ### Question: Please show the date of ceremony of the volumes that last more than 2 weeks on top. ### Answer: SELECT T1.Date_of_ceremony FROM music_festival AS T1 JOIN volume AS T2 ON T1.Volume = T2.Volume_ID WHERE T2.Weeks_on_Top > 2 |
Please show the songs that have result "nominated" at music festivals. | CREATE TABLE volume (Song VARCHAR, Volume_ID VARCHAR); CREATE TABLE music_festival (Volume VARCHAR, Result VARCHAR) | SELECT T2.Song FROM music_festival AS T1 JOIN volume AS T2 ON T1.Volume = T2.Volume_ID WHERE T1.Result = "Nominated" | ### Context: CREATE TABLE volume (Song VARCHAR, Volume_ID VARCHAR); CREATE TABLE music_festival (Volume VARCHAR, Result VARCHAR) ### Question: Please show the songs that have result "nominated" at music festivals. ### Answer: SELECT T2.Song FROM music_festival AS T1 JOIN volume AS T2 ON T1.Volume = T2.Volume_ID WHERE T1.Result = "Nominated" |
What are the issue dates of volumes associated with the artist "Gorgoroth"? | CREATE TABLE artist (Artist_ID VARCHAR, Artist VARCHAR); CREATE TABLE volume (Issue_Date VARCHAR, Artist_ID VARCHAR) | SELECT T2.Issue_Date FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T1.Artist = "Gorgoroth" | ### Context: CREATE TABLE artist (Artist_ID VARCHAR, Artist VARCHAR); CREATE TABLE volume (Issue_Date VARCHAR, Artist_ID VARCHAR) ### Question: What are the issue dates of volumes associated with the artist "Gorgoroth"? ### Answer: SELECT T2.Issue_Date FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T1.Artist = "Gorgoroth" |
What are the songs in volumes associated with the artist aged 32 or older? | CREATE TABLE volume (Song VARCHAR, Artist_ID VARCHAR); CREATE TABLE artist (Artist_ID VARCHAR, age VARCHAR) | SELECT T2.Song FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T1.age >= 32 | ### Context: CREATE TABLE volume (Song VARCHAR, Artist_ID VARCHAR); CREATE TABLE artist (Artist_ID VARCHAR, age VARCHAR) ### Question: What are the songs in volumes associated with the artist aged 32 or older? ### Answer: SELECT T2.Song FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T1.age >= 32 |
What is the average weeks on top of volumes associated with the artist aged 25 or younger? | CREATE TABLE volume (Weeks_on_Top INTEGER, Artist_ID VARCHAR); CREATE TABLE artist (Artist_ID VARCHAR, age VARCHAR) | SELECT AVG(T2.Weeks_on_Top) FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T1.age <= 25 | ### Context: CREATE TABLE volume (Weeks_on_Top INTEGER, Artist_ID VARCHAR); CREATE TABLE artist (Artist_ID VARCHAR, age VARCHAR) ### Question: What is the average weeks on top of volumes associated with the artist aged 25 or younger? ### Answer: SELECT AVG(T2.Weeks_on_Top) FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T1.age <= 25 |
What are the famous title of the artists associated with volumes with more than 2 weeks on top? | CREATE TABLE artist (Famous_Title VARCHAR, Artist_ID VARCHAR); CREATE TABLE volume (Artist_ID VARCHAR, Weeks_on_Top INTEGER) | SELECT T1.Famous_Title FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T2.Weeks_on_Top > 2 | ### Context: CREATE TABLE artist (Famous_Title VARCHAR, Artist_ID VARCHAR); CREATE TABLE volume (Artist_ID VARCHAR, Weeks_on_Top INTEGER) ### Question: What are the famous title of the artists associated with volumes with more than 2 weeks on top? ### Answer: SELECT T1.Famous_Title FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T2.Weeks_on_Top > 2 |
Please list the age and famous title of artists in descending order of age. | CREATE TABLE artist (Famous_Title VARCHAR, Age VARCHAR) | SELECT Famous_Title, Age FROM artist ORDER BY Age DESC | ### Context: CREATE TABLE artist (Famous_Title VARCHAR, Age VARCHAR) ### Question: Please list the age and famous title of artists in descending order of age. ### Answer: SELECT Famous_Title, Age FROM artist ORDER BY Age DESC |
What is the famous release date of the artist with the oldest age? | CREATE TABLE artist (Famous_Release_date VARCHAR, Age VARCHAR) | SELECT Famous_Release_date FROM artist ORDER BY Age DESC LIMIT 1 | ### Context: CREATE TABLE artist (Famous_Release_date VARCHAR, Age VARCHAR) ### Question: What is the famous release date of the artist with the oldest age? ### Answer: SELECT Famous_Release_date FROM artist ORDER BY Age DESC LIMIT 1 |
Please show the categories of the music festivals and the count. | CREATE TABLE music_festival (Category VARCHAR) | SELECT Category, COUNT(*) FROM music_festival GROUP BY Category | ### Context: CREATE TABLE music_festival (Category VARCHAR) ### Question: Please show the categories of the music festivals and the count. ### Answer: SELECT Category, COUNT(*) FROM music_festival GROUP BY Category |
What is the most common result of the music festival? | CREATE TABLE music_festival (RESULT VARCHAR) | SELECT RESULT FROM music_festival GROUP BY RESULT ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE music_festival (RESULT VARCHAR) ### Question: What is the most common result of the music festival? ### Answer: SELECT RESULT FROM music_festival GROUP BY RESULT ORDER BY COUNT(*) DESC LIMIT 1 |
Please show the categories of the music festivals with count more than 1. | CREATE TABLE music_festival (Category VARCHAR) | SELECT Category FROM music_festival GROUP BY Category HAVING COUNT(*) > 1 | ### Context: CREATE TABLE music_festival (Category VARCHAR) ### Question: Please show the categories of the music festivals with count more than 1. ### Answer: SELECT Category FROM music_festival GROUP BY Category HAVING COUNT(*) > 1 |
What is the song in the volume with the maximum weeks on top? | CREATE TABLE volume (Song VARCHAR, Weeks_on_Top VARCHAR) | SELECT Song FROM volume ORDER BY Weeks_on_Top DESC LIMIT 1 | ### Context: CREATE TABLE volume (Song VARCHAR, Weeks_on_Top VARCHAR) ### Question: What is the song in the volume with the maximum weeks on top? ### Answer: SELECT Song FROM volume ORDER BY Weeks_on_Top DESC LIMIT 1 |
Find the famous titles of artists that do not have any volume. | CREATE TABLE volume (Famous_Title VARCHAR, Artist_ID VARCHAR); CREATE TABLE artist (Famous_Title VARCHAR, Artist_ID VARCHAR) | SELECT Famous_Title FROM artist WHERE NOT Artist_ID IN (SELECT Artist_ID FROM volume) | ### Context: CREATE TABLE volume (Famous_Title VARCHAR, Artist_ID VARCHAR); CREATE TABLE artist (Famous_Title VARCHAR, Artist_ID VARCHAR) ### Question: Find the famous titles of artists that do not have any volume. ### Answer: SELECT Famous_Title FROM artist WHERE NOT Artist_ID IN (SELECT Artist_ID FROM volume) |
Show the famous titles of the artists with both volumes that lasted more than 2 weeks on top and volumes that lasted less than 2 weeks on top. | CREATE TABLE artist (Famous_Title VARCHAR, Artist_ID VARCHAR); CREATE TABLE volume (Artist_ID VARCHAR, Weeks_on_Top INTEGER) | SELECT T1.Famous_Title FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T2.Weeks_on_Top > 2 INTERSECT SELECT T1.Famous_Title FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T2.Weeks_on_Top < 2 | ### Context: CREATE TABLE artist (Famous_Title VARCHAR, Artist_ID VARCHAR); CREATE TABLE volume (Artist_ID VARCHAR, Weeks_on_Top INTEGER) ### Question: Show the famous titles of the artists with both volumes that lasted more than 2 weeks on top and volumes that lasted less than 2 weeks on top. ### Answer: SELECT T1.Famous_Title FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T2.Weeks_on_Top > 2 INTERSECT SELECT T1.Famous_Title FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T2.Weeks_on_Top < 2 |
What are the date of ceremony of music festivals with category "Best Song" and result "Awarded"? | CREATE TABLE music_festival (Date_of_ceremony VARCHAR, Category VARCHAR, RESULT VARCHAR) | SELECT Date_of_ceremony FROM music_festival WHERE Category = "Best Song" AND RESULT = "Awarded" | ### Context: CREATE TABLE music_festival (Date_of_ceremony VARCHAR, Category VARCHAR, RESULT VARCHAR) ### Question: What are the date of ceremony of music festivals with category "Best Song" and result "Awarded"? ### Answer: SELECT Date_of_ceremony FROM music_festival WHERE Category = "Best Song" AND RESULT = "Awarded" |
What is the issue date of the volume with the minimum weeks on top? | CREATE TABLE volume (Issue_Date VARCHAR, Weeks_on_Top VARCHAR) | SELECT Issue_Date FROM volume ORDER BY Weeks_on_Top LIMIT 1 | ### Context: CREATE TABLE volume (Issue_Date VARCHAR, Weeks_on_Top VARCHAR) ### Question: What is the issue date of the volume with the minimum weeks on top? ### Answer: SELECT Issue_Date FROM volume ORDER BY Weeks_on_Top LIMIT 1 |
Please show the results of music festivals and the number of music festivals that have had each, ordered by this count. | CREATE TABLE music_festival (RESULT VARCHAR) | SELECT RESULT, COUNT(*) FROM music_festival GROUP BY RESULT ORDER BY COUNT(*) DESC | ### Context: CREATE TABLE music_festival (RESULT VARCHAR) ### Question: Please show the results of music festivals and the number of music festivals that have had each, ordered by this count. ### Answer: SELECT RESULT, COUNT(*) FROM music_festival GROUP BY RESULT ORDER BY COUNT(*) DESC |
What are the issue dates of volumes associated with the artist aged 23 or younger? | CREATE TABLE volume (Artist_ID VARCHAR); CREATE TABLE artist (Artist_ID VARCHAR, age VARCHAR) | SELECT Issue_Date FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T1.age <= 23 | ### Context: CREATE TABLE volume (Artist_ID VARCHAR); CREATE TABLE artist (Artist_ID VARCHAR, age VARCHAR) ### Question: What are the issue dates of volumes associated with the artist aged 23 or younger? ### Answer: SELECT Issue_Date FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T1.age <= 23 |
How many roller coasters are there? | CREATE TABLE roller_coaster (Id VARCHAR) | SELECT COUNT(*) FROM roller_coaster | ### Context: CREATE TABLE roller_coaster (Id VARCHAR) ### Question: How many roller coasters are there? ### Answer: SELECT COUNT(*) FROM roller_coaster |
List the names of roller coasters by ascending order of length. | CREATE TABLE roller_coaster (Name VARCHAR, LENGTH VARCHAR) | SELECT Name FROM roller_coaster ORDER BY LENGTH | ### Context: CREATE TABLE roller_coaster (Name VARCHAR, LENGTH VARCHAR) ### Question: List the names of roller coasters by ascending order of length. ### Answer: SELECT Name FROM roller_coaster ORDER BY LENGTH |
What are the lengths and heights of roller coasters? | CREATE TABLE roller_coaster (LENGTH VARCHAR, Height VARCHAR) | SELECT LENGTH, Height FROM roller_coaster | ### Context: CREATE TABLE roller_coaster (LENGTH VARCHAR, Height VARCHAR) ### Question: What are the lengths and heights of roller coasters? ### Answer: SELECT LENGTH, Height FROM roller_coaster |
List the names of countries whose language is not "German". | CREATE TABLE country (Name VARCHAR, Languages VARCHAR) | SELECT Name FROM country WHERE Languages <> "German" | ### Context: CREATE TABLE country (Name VARCHAR, Languages VARCHAR) ### Question: List the names of countries whose language is not "German". ### Answer: SELECT Name FROM country WHERE Languages <> "German" |
Show the statuses of roller coasters longer than 3300 or higher than 100. | CREATE TABLE roller_coaster (Status VARCHAR, LENGTH VARCHAR, Height VARCHAR) | SELECT Status FROM roller_coaster WHERE LENGTH > 3300 OR Height > 100 | ### Context: CREATE TABLE roller_coaster (Status VARCHAR, LENGTH VARCHAR, Height VARCHAR) ### Question: Show the statuses of roller coasters longer than 3300 or higher than 100. ### Answer: SELECT Status FROM roller_coaster WHERE LENGTH > 3300 OR Height > 100 |
What are the speeds of the longest roller coaster? | CREATE TABLE roller_coaster (Speed VARCHAR, LENGTH VARCHAR) | SELECT Speed FROM roller_coaster ORDER BY LENGTH DESC LIMIT 1 | ### Context: CREATE TABLE roller_coaster (Speed VARCHAR, LENGTH VARCHAR) ### Question: What are the speeds of the longest roller coaster? ### Answer: SELECT Speed FROM roller_coaster ORDER BY LENGTH DESC LIMIT 1 |
What is the average speed of roller coasters? | CREATE TABLE roller_coaster (Speed INTEGER) | SELECT AVG(Speed) FROM roller_coaster | ### Context: CREATE TABLE roller_coaster (Speed INTEGER) ### Question: What is the average speed of roller coasters? ### Answer: SELECT AVG(Speed) FROM roller_coaster |
Show the different statuses and the numbers of roller coasters for each status. | CREATE TABLE roller_coaster (Status VARCHAR) | SELECT Status, COUNT(*) FROM roller_coaster GROUP BY Status | ### Context: CREATE TABLE roller_coaster (Status VARCHAR) ### Question: Show the different statuses and the numbers of roller coasters for each status. ### Answer: SELECT Status, COUNT(*) FROM roller_coaster GROUP BY Status |
Please show the most common status of roller coasters. | CREATE TABLE roller_coaster (Status VARCHAR) | SELECT Status FROM roller_coaster GROUP BY Status ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE roller_coaster (Status VARCHAR) ### Question: Please show the most common status of roller coasters. ### Answer: SELECT Status FROM roller_coaster GROUP BY Status ORDER BY COUNT(*) DESC LIMIT 1 |
List the status shared by more than two roller coaster. | CREATE TABLE roller_coaster (Status VARCHAR) | SELECT Status FROM roller_coaster GROUP BY Status HAVING COUNT(*) > 2 | ### Context: CREATE TABLE roller_coaster (Status VARCHAR) ### Question: List the status shared by more than two roller coaster. ### Answer: SELECT Status FROM roller_coaster GROUP BY Status HAVING COUNT(*) > 2 |
Show the park of the roller coaster with the highest speed. | CREATE TABLE roller_coaster (Park VARCHAR, Speed VARCHAR) | SELECT Park FROM roller_coaster ORDER BY Speed DESC LIMIT 1 | ### Context: CREATE TABLE roller_coaster (Park VARCHAR, Speed VARCHAR) ### Question: Show the park of the roller coaster with the highest speed. ### Answer: SELECT Park FROM roller_coaster ORDER BY Speed DESC LIMIT 1 |
Show the names of roller coasters and names of country they are in. | CREATE TABLE country (Name VARCHAR, Country_ID VARCHAR); CREATE TABLE roller_coaster (Name VARCHAR, Country_ID VARCHAR) | SELECT T2.Name, T1.Name FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID | ### Context: CREATE TABLE country (Name VARCHAR, Country_ID VARCHAR); CREATE TABLE roller_coaster (Name VARCHAR, Country_ID VARCHAR) ### Question: Show the names of roller coasters and names of country they are in. ### Answer: SELECT T2.Name, T1.Name FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID |
Show the names of countries that have more than one roller coaster. | CREATE TABLE roller_coaster (Country_ID VARCHAR); CREATE TABLE country (Name VARCHAR, Country_ID VARCHAR) | SELECT T1.Name FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID GROUP BY T1.Name HAVING COUNT(*) > 1 | ### Context: CREATE TABLE roller_coaster (Country_ID VARCHAR); CREATE TABLE country (Name VARCHAR, Country_ID VARCHAR) ### Question: Show the names of countries that have more than one roller coaster. ### Answer: SELECT T1.Name FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID GROUP BY T1.Name HAVING COUNT(*) > 1 |
Show the name and population of the country that has the highest roller coaster. | CREATE TABLE roller_coaster (Country_ID VARCHAR, Height VARCHAR); CREATE TABLE country (Name VARCHAR, population VARCHAR, Country_ID VARCHAR) | SELECT T1.Name, T1.population FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID ORDER BY T2.Height DESC LIMIT 1 | ### Context: CREATE TABLE roller_coaster (Country_ID VARCHAR, Height VARCHAR); CREATE TABLE country (Name VARCHAR, population VARCHAR, Country_ID VARCHAR) ### Question: Show the name and population of the country that has the highest roller coaster. ### Answer: SELECT T1.Name, T1.population FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID ORDER BY T2.Height DESC LIMIT 1 |
Show the names of countries and the average speed of roller coasters from each country. | CREATE TABLE roller_coaster (Speed INTEGER, Country_ID VARCHAR); CREATE TABLE country (Name VARCHAR, Country_ID VARCHAR) | SELECT T1.Name, AVG(T2.Speed) FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID GROUP BY T1.Name | ### Context: CREATE TABLE roller_coaster (Speed INTEGER, Country_ID VARCHAR); CREATE TABLE country (Name VARCHAR, Country_ID VARCHAR) ### Question: Show the names of countries and the average speed of roller coasters from each country. ### Answer: SELECT T1.Name, AVG(T2.Speed) FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID GROUP BY T1.Name |
How many countries do not have an roller coaster longer than 3000? | CREATE TABLE country (country_id VARCHAR, LENGTH INTEGER); CREATE TABLE roller_coaster (country_id VARCHAR, LENGTH INTEGER) | SELECT COUNT(*) FROM country WHERE NOT country_id IN (SELECT country_id FROM roller_coaster WHERE LENGTH > 3000) | ### Context: CREATE TABLE country (country_id VARCHAR, LENGTH INTEGER); CREATE TABLE roller_coaster (country_id VARCHAR, LENGTH INTEGER) ### Question: How many countries do not have an roller coaster longer than 3000? ### Answer: SELECT COUNT(*) FROM country WHERE NOT country_id IN (SELECT country_id FROM roller_coaster WHERE LENGTH > 3000) |
What are the country names, area and population which has both roller coasters with speed higher | CREATE TABLE country (name VARCHAR, area VARCHAR, population VARCHAR, Country_ID VARCHAR); CREATE TABLE roller_coaster (Country_ID VARCHAR, speed INTEGER) | SELECT T1.name, T1.area, T1.population FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID WHERE T2.speed > 60 INTERSECT SELECT T1.name, T1.area, T1.population FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID WHERE T2.speed < 55 | ### Context: CREATE TABLE country (name VARCHAR, area VARCHAR, population VARCHAR, Country_ID VARCHAR); CREATE TABLE roller_coaster (Country_ID VARCHAR, speed INTEGER) ### Question: What are the country names, area and population which has both roller coasters with speed higher ### Answer: SELECT T1.name, T1.area, T1.population FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID WHERE T2.speed > 60 INTERSECT SELECT T1.name, T1.area, T1.population FROM country AS T1 JOIN roller_coaster AS T2 ON T1.Country_ID = T2.Country_ID WHERE T2.speed < 55 |
How many different captain ranks are there? | CREATE TABLE captain (rank VARCHAR) | SELECT COUNT(DISTINCT rank) FROM captain | ### Context: CREATE TABLE captain (rank VARCHAR) ### Question: How many different captain ranks are there? ### Answer: SELECT COUNT(DISTINCT rank) FROM captain |
How many captains are in each rank? | CREATE TABLE captain (rank VARCHAR) | SELECT COUNT(*), rank FROM captain GROUP BY rank | ### Context: CREATE TABLE captain (rank VARCHAR) ### Question: How many captains are in each rank? ### Answer: SELECT COUNT(*), rank FROM captain GROUP BY rank |
How many captains with younger than 50 are in each rank? | CREATE TABLE captain (rank VARCHAR, age INTEGER) | SELECT COUNT(*), rank FROM captain WHERE age < 50 GROUP BY rank | ### Context: CREATE TABLE captain (rank VARCHAR, age INTEGER) ### Question: How many captains with younger than 50 are in each rank? ### Answer: SELECT COUNT(*), rank FROM captain WHERE age < 50 GROUP BY rank |
Sort all captain names by their ages from old to young. | CREATE TABLE captain (name VARCHAR, age VARCHAR) | SELECT name FROM captain ORDER BY age DESC | ### Context: CREATE TABLE captain (name VARCHAR, age VARCHAR) ### Question: Sort all captain names by their ages from old to young. ### Answer: SELECT name FROM captain ORDER BY age DESC |
Find the name, class and rank of all captains. | CREATE TABLE captain (name VARCHAR, CLASS VARCHAR, rank VARCHAR) | SELECT name, CLASS, rank FROM captain | ### Context: CREATE TABLE captain (name VARCHAR, CLASS VARCHAR, rank VARCHAR) ### Question: Find the name, class and rank of all captains. ### Answer: SELECT name, CLASS, rank FROM captain |
Which rank is the most common among captains? | CREATE TABLE captain (rank VARCHAR) | SELECT rank FROM captain GROUP BY rank ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE captain (rank VARCHAR) ### Question: Which rank is the most common among captains? ### Answer: SELECT rank FROM captain GROUP BY rank ORDER BY COUNT(*) DESC LIMIT 1 |
Which classes have more than two captains? | CREATE TABLE captain (CLASS VARCHAR) | SELECT CLASS FROM captain GROUP BY CLASS HAVING COUNT(*) > 2 | ### Context: CREATE TABLE captain (CLASS VARCHAR) ### Question: Which classes have more than two captains? ### Answer: SELECT CLASS FROM captain GROUP BY CLASS HAVING COUNT(*) > 2 |
Find the name of captains whose rank are either Midshipman or Lieutenant. | CREATE TABLE captain (name VARCHAR, rank VARCHAR) | SELECT name FROM captain WHERE rank = 'Midshipman' OR rank = 'Lieutenant' | ### Context: CREATE TABLE captain (name VARCHAR, rank VARCHAR) ### Question: Find the name of captains whose rank are either Midshipman or Lieutenant. ### Answer: SELECT name FROM captain WHERE rank = 'Midshipman' OR rank = 'Lieutenant' |
What are the average and minimum age of captains in different class? | CREATE TABLE captain (CLASS VARCHAR, age INTEGER) | SELECT AVG(age), MIN(age), CLASS FROM captain GROUP BY CLASS | ### Context: CREATE TABLE captain (CLASS VARCHAR, age INTEGER) ### Question: What are the average and minimum age of captains in different class? ### Answer: SELECT AVG(age), MIN(age), CLASS FROM captain GROUP BY CLASS |
Find the captain rank that has some captains in both Cutter and Armed schooner classes. | CREATE TABLE captain (rank VARCHAR, CLASS VARCHAR) | SELECT rank FROM captain WHERE CLASS = 'Cutter' INTERSECT SELECT rank FROM captain WHERE CLASS = 'Armed schooner' | ### Context: CREATE TABLE captain (rank VARCHAR, CLASS VARCHAR) ### Question: Find the captain rank that has some captains in both Cutter and Armed schooner classes. ### Answer: SELECT rank FROM captain WHERE CLASS = 'Cutter' INTERSECT SELECT rank FROM captain WHERE CLASS = 'Armed schooner' |
Find the captain rank that has no captain in Third-rate ship of the line class. | CREATE TABLE captain (rank VARCHAR, CLASS VARCHAR) | SELECT rank FROM captain EXCEPT SELECT rank FROM captain WHERE CLASS = 'Third-rate ship of the line' | ### Context: CREATE TABLE captain (rank VARCHAR, CLASS VARCHAR) ### Question: Find the captain rank that has no captain in Third-rate ship of the line class. ### Answer: SELECT rank FROM captain EXCEPT SELECT rank FROM captain WHERE CLASS = 'Third-rate ship of the line' |
What is the name of the youngest captain? | CREATE TABLE captain (name VARCHAR, age VARCHAR) | SELECT name FROM captain ORDER BY age LIMIT 1 | ### Context: CREATE TABLE captain (name VARCHAR, age VARCHAR) ### Question: What is the name of the youngest captain? ### Answer: SELECT name FROM captain ORDER BY age LIMIT 1 |
Find the name, type, and flag of the ship that is built in the most recent year. | CREATE TABLE ship (name VARCHAR, TYPE VARCHAR, flag VARCHAR, built_year VARCHAR) | SELECT name, TYPE, flag FROM ship ORDER BY built_year DESC LIMIT 1 | ### Context: CREATE TABLE ship (name VARCHAR, TYPE VARCHAR, flag VARCHAR, built_year VARCHAR) ### Question: Find the name, type, and flag of the ship that is built in the most recent year. ### Answer: SELECT name, TYPE, flag FROM ship ORDER BY built_year DESC LIMIT 1 |
Group by ships by flag, and return number of ships that have each flag. | CREATE TABLE ship (flag VARCHAR) | SELECT COUNT(*), flag FROM ship GROUP BY flag | ### Context: CREATE TABLE ship (flag VARCHAR) ### Question: Group by ships by flag, and return number of ships that have each flag. ### Answer: SELECT COUNT(*), flag FROM ship GROUP BY flag |
Which flag is most widely used among all ships? | CREATE TABLE ship (flag VARCHAR) | SELECT flag FROM ship GROUP BY flag ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE ship (flag VARCHAR) ### Question: Which flag is most widely used among all ships? ### Answer: SELECT flag FROM ship GROUP BY flag ORDER BY COUNT(*) DESC LIMIT 1 |
List all ship names in the order of built year and class. | CREATE TABLE ship (name VARCHAR, built_year VARCHAR, CLASS VARCHAR) | SELECT name FROM ship ORDER BY built_year, CLASS | ### Context: CREATE TABLE ship (name VARCHAR, built_year VARCHAR, CLASS VARCHAR) ### Question: List all ship names in the order of built year and class. ### Answer: SELECT name FROM ship ORDER BY built_year, CLASS |
Find the ship type that are used by both ships with Panama and Malta flags. | CREATE TABLE ship (TYPE VARCHAR, flag VARCHAR) | SELECT TYPE FROM ship WHERE flag = 'Panama' INTERSECT SELECT TYPE FROM ship WHERE flag = 'Malta' | ### Context: CREATE TABLE ship (TYPE VARCHAR, flag VARCHAR) ### Question: Find the ship type that are used by both ships with Panama and Malta flags. ### Answer: SELECT TYPE FROM ship WHERE flag = 'Panama' INTERSECT SELECT TYPE FROM ship WHERE flag = 'Malta' |
In which year were most of ships built? | CREATE TABLE ship (built_year VARCHAR) | SELECT built_year FROM ship GROUP BY built_year ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE ship (built_year VARCHAR) ### Question: In which year were most of ships built? ### Answer: SELECT built_year FROM ship GROUP BY built_year ORDER BY COUNT(*) DESC LIMIT 1 |
Find the name of the ships that have more than one captain. | CREATE TABLE captain (ship_id VARCHAR); CREATE TABLE ship (name VARCHAR, ship_id VARCHAR) | SELECT t1.name FROM ship AS t1 JOIN captain AS t2 ON t1.ship_id = t2.ship_id GROUP BY t2.ship_id HAVING COUNT(*) > 1 | ### Context: CREATE TABLE captain (ship_id VARCHAR); CREATE TABLE ship (name VARCHAR, ship_id VARCHAR) ### Question: Find the name of the ships that have more than one captain. ### Answer: SELECT t1.name FROM ship AS t1 JOIN captain AS t2 ON t1.ship_id = t2.ship_id GROUP BY t2.ship_id HAVING COUNT(*) > 1 |
what are the names and classes of the ships that do not have any captain yet? | CREATE TABLE ship (name VARCHAR, CLASS VARCHAR, ship_id VARCHAR); CREATE TABLE captain (name VARCHAR, CLASS VARCHAR, ship_id VARCHAR) | SELECT name, CLASS FROM ship WHERE NOT ship_id IN (SELECT ship_id FROM captain) | ### Context: CREATE TABLE ship (name VARCHAR, CLASS VARCHAR, ship_id VARCHAR); CREATE TABLE captain (name VARCHAR, CLASS VARCHAR, ship_id VARCHAR) ### Question: what are the names and classes of the ships that do not have any captain yet? ### Answer: SELECT name, CLASS FROM ship WHERE NOT ship_id IN (SELECT ship_id FROM captain) |
Find the name of the ship that is steered by the youngest captain. | CREATE TABLE ship (name VARCHAR, ship_id VARCHAR); CREATE TABLE captain (ship_id VARCHAR, age VARCHAR) | SELECT t1.name FROM ship AS t1 JOIN captain AS t2 ON t1.ship_id = t2.ship_id ORDER BY t2.age LIMIT 1 | ### Context: CREATE TABLE ship (name VARCHAR, ship_id VARCHAR); CREATE TABLE captain (ship_id VARCHAR, age VARCHAR) ### Question: Find the name of the ship that is steered by the youngest captain. ### Answer: SELECT t1.name FROM ship AS t1 JOIN captain AS t2 ON t1.ship_id = t2.ship_id ORDER BY t2.age LIMIT 1 |
Find the name and flag of ships that are not steered by any captain with Midshipman rank. | CREATE TABLE captain (name VARCHAR, flag VARCHAR, ship_id VARCHAR, rank VARCHAR); CREATE TABLE ship (name VARCHAR, flag VARCHAR, ship_id VARCHAR, rank VARCHAR) | SELECT name, flag FROM ship WHERE NOT ship_id IN (SELECT ship_id FROM captain WHERE rank = 'Midshipman') | ### Context: CREATE TABLE captain (name VARCHAR, flag VARCHAR, ship_id VARCHAR, rank VARCHAR); CREATE TABLE ship (name VARCHAR, flag VARCHAR, ship_id VARCHAR, rank VARCHAR) ### Question: Find the name and flag of ships that are not steered by any captain with Midshipman rank. ### Answer: SELECT name, flag FROM ship WHERE NOT ship_id IN (SELECT ship_id FROM captain WHERE rank = 'Midshipman') |
Find the name of the ships that are steered by both a captain with Midshipman rank and a captain with Lieutenant rank. | CREATE TABLE ship (name VARCHAR, ship_id VARCHAR); CREATE TABLE captain (ship_id VARCHAR, rank VARCHAR) | SELECT t1.name FROM ship AS t1 JOIN captain AS t2 ON t1.ship_id = t2.ship_id WHERE t2.rank = 'Midshipman' INTERSECT SELECT t1.name FROM ship AS t1 JOIN captain AS t2 ON t1.ship_id = t2.ship_id WHERE t2.rank = 'Lieutenant' | ### Context: CREATE TABLE ship (name VARCHAR, ship_id VARCHAR); CREATE TABLE captain (ship_id VARCHAR, rank VARCHAR) ### Question: Find the name of the ships that are steered by both a captain with Midshipman rank and a captain with Lieutenant rank. ### Answer: SELECT t1.name FROM ship AS t1 JOIN captain AS t2 ON t1.ship_id = t2.ship_id WHERE t2.rank = 'Midshipman' INTERSECT SELECT t1.name FROM ship AS t1 JOIN captain AS t2 ON t1.ship_id = t2.ship_id WHERE t2.rank = 'Lieutenant' |
What is id of the city that hosted events in the most recent year? | CREATE TABLE hosting_city (host_city VARCHAR, YEAR VARCHAR) | SELECT host_city FROM hosting_city ORDER BY YEAR DESC LIMIT 1 | ### Context: CREATE TABLE hosting_city (host_city VARCHAR, YEAR VARCHAR) ### Question: What is id of the city that hosted events in the most recent year? ### Answer: SELECT host_city FROM hosting_city ORDER BY YEAR DESC LIMIT 1 |
Find the match ids of the cities that hosted competition "1994 FIFA World Cup qualification"? | CREATE TABLE MATCH (match_id VARCHAR, competition VARCHAR) | SELECT match_id FROM MATCH WHERE competition = "1994 FIFA World Cup qualification" | ### Context: CREATE TABLE MATCH (match_id VARCHAR, competition VARCHAR) ### Question: Find the match ids of the cities that hosted competition "1994 FIFA World Cup qualification"? ### Answer: SELECT match_id FROM MATCH WHERE competition = "1994 FIFA World Cup qualification" |
Find the cities which were once a host city after 2010? | CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR, year INTEGER) | SELECT T1.city FROM city AS T1 JOIN hosting_city AS T2 ON T1.city_id = T2.host_city WHERE T2.year > 2010 | ### Context: CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR, year INTEGER) ### Question: Find the cities which were once a host city after 2010? ### Answer: SELECT T1.city FROM city AS T1 JOIN hosting_city AS T2 ON T1.city_id = T2.host_city WHERE T2.year > 2010 |
Which city has hosted the most events? | CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR) | SELECT T1.city FROM city AS T1 JOIN hosting_city AS T2 ON T1.city_id = T2.host_city GROUP BY T2.host_city ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR) ### Question: Which city has hosted the most events? ### Answer: SELECT T1.city FROM city AS T1 JOIN hosting_city AS T2 ON T1.city_id = T2.host_city GROUP BY T2.host_city ORDER BY COUNT(*) DESC LIMIT 1 |
What is the venue of the competition "1994 FIFA World Cup qualification" hosted by "Nanjing ( Jiangsu )"? | CREATE TABLE city (city_id VARCHAR, city VARCHAR); CREATE TABLE MATCH (venue VARCHAR, match_id VARCHAR, competition VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR, match_id VARCHAR) | SELECT T3.venue FROM city AS T1 JOIN hosting_city AS T2 ON T1.city_id = T2.host_city JOIN MATCH AS T3 ON T2.match_id = T3.match_id WHERE T1.city = "Nanjing ( Jiangsu )" AND T3.competition = "1994 FIFA World Cup qualification" | ### Context: CREATE TABLE city (city_id VARCHAR, city VARCHAR); CREATE TABLE MATCH (venue VARCHAR, match_id VARCHAR, competition VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR, match_id VARCHAR) ### Question: What is the venue of the competition "1994 FIFA World Cup qualification" hosted by "Nanjing ( Jiangsu )"? ### Answer: SELECT T3.venue FROM city AS T1 JOIN hosting_city AS T2 ON T1.city_id = T2.host_city JOIN MATCH AS T3 ON T2.match_id = T3.match_id WHERE T1.city = "Nanjing ( Jiangsu )" AND T3.competition = "1994 FIFA World Cup qualification" |
Give me the temperature of Shanghai in January. | CREATE TABLE city (city_id VARCHAR, city VARCHAR); CREATE TABLE temperature (Jan VARCHAR, city_id VARCHAR) | SELECT T2.Jan FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id WHERE T1.city = "Shanghai" | ### Context: CREATE TABLE city (city_id VARCHAR, city VARCHAR); CREATE TABLE temperature (Jan VARCHAR, city_id VARCHAR) ### Question: Give me the temperature of Shanghai in January. ### Answer: SELECT T2.Jan FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id WHERE T1.city = "Shanghai" |
What is the host year of city "Taizhou ( Zhejiang )"? | CREATE TABLE city (city_id VARCHAR, city VARCHAR); CREATE TABLE hosting_city (year VARCHAR, host_city VARCHAR) | SELECT T2.year FROM city AS T1 JOIN hosting_city AS T2 ON T1.city_id = T2.host_city WHERE T1.city = "Taizhou ( Zhejiang )" | ### Context: CREATE TABLE city (city_id VARCHAR, city VARCHAR); CREATE TABLE hosting_city (year VARCHAR, host_city VARCHAR) ### Question: What is the host year of city "Taizhou ( Zhejiang )"? ### Answer: SELECT T2.year FROM city AS T1 JOIN hosting_city AS T2 ON T1.city_id = T2.host_city WHERE T1.city = "Taizhou ( Zhejiang )" |
Which three cities have the largest regional population? | CREATE TABLE city (city VARCHAR, regional_population VARCHAR) | SELECT city FROM city ORDER BY regional_population DESC LIMIT 3 | ### Context: CREATE TABLE city (city VARCHAR, regional_population VARCHAR) ### Question: Which three cities have the largest regional population? ### Answer: SELECT city FROM city ORDER BY regional_population DESC LIMIT 3 |
Which city has the lowest GDP? Please list the city name and its GDP. | CREATE TABLE city (city VARCHAR, GDP VARCHAR) | SELECT city, GDP FROM city ORDER BY GDP LIMIT 1 | ### Context: CREATE TABLE city (city VARCHAR, GDP VARCHAR) ### Question: Which city has the lowest GDP? Please list the city name and its GDP. ### Answer: SELECT city, GDP FROM city ORDER BY GDP LIMIT 1 |
Which city has the highest temperature in February? | CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE temperature (city_id VARCHAR, Feb VARCHAR) | SELECT T1.city FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id ORDER BY T2.Feb DESC LIMIT 1 | ### Context: CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE temperature (city_id VARCHAR, Feb VARCHAR) ### Question: Which city has the highest temperature in February? ### Answer: SELECT T1.city FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id ORDER BY T2.Feb DESC LIMIT 1 |
Give me a list of cities whose temperature in March is lower than that in July or higher than that in Oct? | CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE temperature (city_id VARCHAR, Mar VARCHAR, Jul VARCHAR, Oct VARCHAR) | SELECT T1.city FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id WHERE T2.Mar < T2.Jul OR T2.Mar > T2.Oct | ### Context: CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE temperature (city_id VARCHAR, Mar VARCHAR, Jul VARCHAR, Oct VARCHAR) ### Question: Give me a list of cities whose temperature in March is lower than that in July or higher than that in Oct? ### Answer: SELECT T1.city FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id WHERE T2.Mar < T2.Jul OR T2.Mar > T2.Oct |
Give me a list of cities whose temperature in Mar is lower than that in July and which have also served as host cities? | CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR); CREATE TABLE temperature (city_id VARCHAR, Mar INTEGER, Jul VARCHAR) | SELECT T1.city FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id WHERE T2.Mar < T2.Jul INTERSECT SELECT T3.city FROM city AS T3 JOIN hosting_city AS T4 ON T3.city_id = T4.host_city | ### Context: CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR); CREATE TABLE temperature (city_id VARCHAR, Mar INTEGER, Jul VARCHAR) ### Question: Give me a list of cities whose temperature in Mar is lower than that in July and which have also served as host cities? ### Answer: SELECT T1.city FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id WHERE T2.Mar < T2.Jul INTERSECT SELECT T3.city FROM city AS T3 JOIN hosting_city AS T4 ON T3.city_id = T4.host_city |
Give me a list of cities whose temperature in Mar is lower than that in Dec and which have never been host cities. | CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR); CREATE TABLE temperature (city_id VARCHAR, Mar INTEGER, Dec VARCHAR) | SELECT T1.city FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id WHERE T2.Mar < T2.Dec EXCEPT SELECT T3.city FROM city AS T3 JOIN hosting_city AS T4 ON T3.city_id = T4.host_city | ### Context: CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR); CREATE TABLE temperature (city_id VARCHAR, Mar INTEGER, Dec VARCHAR) ### Question: Give me a list of cities whose temperature in Mar is lower than that in Dec and which have never been host cities. ### Answer: SELECT T1.city FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id WHERE T2.Mar < T2.Dec EXCEPT SELECT T3.city FROM city AS T3 JOIN hosting_city AS T4 ON T3.city_id = T4.host_city |
Give me a list of cities whose temperature in Feb is higher than that in Jun or cities that were once host cities? | CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR); CREATE TABLE temperature (city_id VARCHAR, Feb INTEGER, Jun VARCHAR) | SELECT T1.city FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id WHERE T2.Feb > T2.Jun UNION SELECT T3.city FROM city AS T3 JOIN hosting_city AS T4 ON T3.city_id = T4.host_city | ### Context: CREATE TABLE city (city VARCHAR, city_id VARCHAR); CREATE TABLE hosting_city (host_city VARCHAR); CREATE TABLE temperature (city_id VARCHAR, Feb INTEGER, Jun VARCHAR) ### Question: Give me a list of cities whose temperature in Feb is higher than that in Jun or cities that were once host cities? ### Answer: SELECT T1.city FROM city AS T1 JOIN temperature AS T2 ON T1.city_id = T2.city_id WHERE T2.Feb > T2.Jun UNION SELECT T3.city FROM city AS T3 JOIN hosting_city AS T4 ON T3.city_id = T4.host_city |
Please give me a list of cities whose regional population is over 10000000. | CREATE TABLE city (city VARCHAR, regional_population INTEGER) | SELECT city FROM city WHERE regional_population > 10000000 | ### Context: CREATE TABLE city (city VARCHAR, regional_population INTEGER) ### Question: Please give me a list of cities whose regional population is over 10000000. ### Answer: SELECT city FROM city WHERE regional_population > 10000000 |
Please give me a list of cities whose regional population is over 8000000 or under 5000000. | CREATE TABLE city (city VARCHAR, regional_population INTEGER) | SELECT city FROM city WHERE regional_population > 10000000 UNION SELECT city FROM city WHERE regional_population < 5000000 | ### Context: CREATE TABLE city (city VARCHAR, regional_population INTEGER) ### Question: Please give me a list of cities whose regional population is over 8000000 or under 5000000. ### Answer: SELECT city FROM city WHERE regional_population > 10000000 UNION SELECT city FROM city WHERE regional_population < 5000000 |
Find the number of matches in different competitions. | CREATE TABLE MATCH (Competition VARCHAR) | SELECT COUNT(*), Competition FROM MATCH GROUP BY Competition | ### Context: CREATE TABLE MATCH (Competition VARCHAR) ### Question: Find the number of matches in different competitions. ### Answer: SELECT COUNT(*), Competition FROM MATCH GROUP BY Competition |
List venues of all matches in the order of their dates starting from the most recent one. | CREATE TABLE MATCH (venue VARCHAR, date VARCHAR) | SELECT venue FROM MATCH ORDER BY date DESC | ### Context: CREATE TABLE MATCH (venue VARCHAR, date VARCHAR) ### Question: List venues of all matches in the order of their dates starting from the most recent one. ### Answer: SELECT venue FROM MATCH ORDER BY date DESC |
what is the GDP of the city with the largest population. | CREATE TABLE city (gdp VARCHAR, Regional_Population VARCHAR) | SELECT gdp FROM city ORDER BY Regional_Population DESC LIMIT 1 | ### Context: CREATE TABLE city (gdp VARCHAR, Regional_Population VARCHAR) ### Question: what is the GDP of the city with the largest population. ### Answer: SELECT gdp FROM city ORDER BY Regional_Population DESC LIMIT 1 |
What are the GDP and population of the city that already served as a host more than once? | CREATE TABLE hosting_city (host_city VARCHAR); CREATE TABLE city (city_id VARCHAR) | SELECT t1.gdp, t1.Regional_Population FROM city AS T1 JOIN hosting_city AS T2 ON T1.city_id = T2.host_city GROUP BY t2.Host_City HAVING COUNT(*) > 1 | ### Context: CREATE TABLE hosting_city (host_city VARCHAR); CREATE TABLE city (city_id VARCHAR) ### Question: What are the GDP and population of the city that already served as a host more than once? ### Answer: SELECT t1.gdp, t1.Regional_Population FROM city AS T1 JOIN hosting_city AS T2 ON T1.city_id = T2.host_city GROUP BY t2.Host_City HAVING COUNT(*) > 1 |
List every individual's first name, middle name and last name in alphabetical order by last name. | CREATE TABLE individuals (individual_first_name VARCHAR, individual_middle_name VARCHAR, individual_last_name VARCHAR) | SELECT individual_first_name, individual_middle_name, individual_last_name FROM individuals ORDER BY individual_last_name | ### Context: CREATE TABLE individuals (individual_first_name VARCHAR, individual_middle_name VARCHAR, individual_last_name VARCHAR) ### Question: List every individual's first name, middle name and last name in alphabetical order by last name. ### Answer: SELECT individual_first_name, individual_middle_name, individual_last_name FROM individuals ORDER BY individual_last_name |
List all the types of forms. | CREATE TABLE forms (form_type_code VARCHAR) | SELECT DISTINCT form_type_code FROM forms | ### Context: CREATE TABLE forms (form_type_code VARCHAR) ### Question: List all the types of forms. ### Answer: SELECT DISTINCT form_type_code FROM forms |
Find the name of the most popular party form. | CREATE TABLE party_forms (form_id VARCHAR); CREATE TABLE forms (form_name VARCHAR, form_id VARCHAR) | SELECT t1.form_name FROM forms AS t1 JOIN party_forms AS t2 ON t1.form_id = t2.form_id GROUP BY t2.form_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE party_forms (form_id VARCHAR); CREATE TABLE forms (form_name VARCHAR, form_id VARCHAR) ### Question: Find the name of the most popular party form. ### Answer: SELECT t1.form_name FROM forms AS t1 JOIN party_forms AS t2 ON t1.form_id = t2.form_id GROUP BY t2.form_id ORDER BY COUNT(*) DESC LIMIT 1 |
Find the payment method and phone of the party with email "[email protected]". | CREATE TABLE parties (payment_method_code VARCHAR, party_phone VARCHAR, party_email VARCHAR) | SELECT payment_method_code, party_phone FROM parties WHERE party_email = "[email protected]" | ### Context: CREATE TABLE parties (payment_method_code VARCHAR, party_phone VARCHAR, party_email VARCHAR) ### Question: Find the payment method and phone of the party with email "[email protected]". ### Answer: SELECT payment_method_code, party_phone FROM parties WHERE party_email = "[email protected]" |
Find the emails of parties with the most popular party form. | CREATE TABLE party_forms (form_id VARCHAR); CREATE TABLE parties (party_email VARCHAR, party_id VARCHAR); CREATE TABLE party_forms (party_id VARCHAR, form_id VARCHAR) | SELECT t1.party_email FROM parties AS t1 JOIN party_forms AS t2 ON t1.party_id = t2.party_id WHERE t2.form_id = (SELECT form_id FROM party_forms GROUP BY form_id ORDER BY COUNT(*) DESC LIMIT 1) | ### Context: CREATE TABLE party_forms (form_id VARCHAR); CREATE TABLE parties (party_email VARCHAR, party_id VARCHAR); CREATE TABLE party_forms (party_id VARCHAR, form_id VARCHAR) ### Question: Find the emails of parties with the most popular party form. ### Answer: SELECT t1.party_email FROM parties AS t1 JOIN party_forms AS t2 ON t1.party_id = t2.party_id WHERE t2.form_id = (SELECT form_id FROM party_forms GROUP BY form_id ORDER BY COUNT(*) DESC LIMIT 1) |
List all the name of organizations in order of the date formed. | CREATE TABLE organizations (organization_name VARCHAR, date_formed VARCHAR) | SELECT organization_name FROM organizations ORDER BY date_formed | ### Context: CREATE TABLE organizations (organization_name VARCHAR, date_formed VARCHAR) ### Question: List all the name of organizations in order of the date formed. ### Answer: SELECT organization_name FROM organizations ORDER BY date_formed |
Find the name of the youngest organization. | CREATE TABLE organizations (organization_name VARCHAR, date_formed VARCHAR) | SELECT organization_name FROM organizations ORDER BY date_formed DESC LIMIT 1 | ### Context: CREATE TABLE organizations (organization_name VARCHAR, date_formed VARCHAR) ### Question: Find the name of the youngest organization. ### Answer: SELECT organization_name FROM organizations ORDER BY date_formed DESC LIMIT 1 |
Find the last name of the latest contact individual of the organization "Labour Party". | CREATE TABLE organizations (organization_id VARCHAR, organization_name VARCHAR); CREATE TABLE individuals (individual_last_name VARCHAR, individual_id VARCHAR); CREATE TABLE organization_contact_individuals (organization_id VARCHAR, individual_id VARCHAR, date_contact_to VARCHAR) | SELECT t3.individual_last_name FROM organizations AS t1 JOIN organization_contact_individuals AS t2 ON t1.organization_id = t2.organization_id JOIN individuals AS t3 ON t2.individual_id = t3.individual_id WHERE t1.organization_name = "Labour Party" ORDER BY t2.date_contact_to DESC LIMIT 1 | ### Context: CREATE TABLE organizations (organization_id VARCHAR, organization_name VARCHAR); CREATE TABLE individuals (individual_last_name VARCHAR, individual_id VARCHAR); CREATE TABLE organization_contact_individuals (organization_id VARCHAR, individual_id VARCHAR, date_contact_to VARCHAR) ### Question: Find the last name of the latest contact individual of the organization "Labour Party". ### Answer: SELECT t3.individual_last_name FROM organizations AS t1 JOIN organization_contact_individuals AS t2 ON t1.organization_id = t2.organization_id JOIN individuals AS t3 ON t2.individual_id = t3.individual_id WHERE t1.organization_name = "Labour Party" ORDER BY t2.date_contact_to DESC LIMIT 1 |
Find the last name of the first ever contact person of the organization with the highest UK Vat number. | CREATE TABLE organizations (uk_vat_number INTEGER); CREATE TABLE individuals (individual_last_name VARCHAR, individual_id VARCHAR); CREATE TABLE organizations (organization_id VARCHAR, uk_vat_number INTEGER); CREATE TABLE organization_contact_individuals (organization_id VARCHAR, individual_id VARCHAR, date_contact_to VARCHAR) | SELECT t3.individual_last_name FROM organizations AS t1 JOIN organization_contact_individuals AS t2 ON t1.organization_id = t2.organization_id JOIN individuals AS t3 ON t2.individual_id = t3.individual_id WHERE t1.uk_vat_number = (SELECT MAX(uk_vat_number) FROM organizations) ORDER BY t2.date_contact_to LIMIT 1 | ### Context: CREATE TABLE organizations (uk_vat_number INTEGER); CREATE TABLE individuals (individual_last_name VARCHAR, individual_id VARCHAR); CREATE TABLE organizations (organization_id VARCHAR, uk_vat_number INTEGER); CREATE TABLE organization_contact_individuals (organization_id VARCHAR, individual_id VARCHAR, date_contact_to VARCHAR) ### Question: Find the last name of the first ever contact person of the organization with the highest UK Vat number. ### Answer: SELECT t3.individual_last_name FROM organizations AS t1 JOIN organization_contact_individuals AS t2 ON t1.organization_id = t2.organization_id JOIN individuals AS t3 ON t2.individual_id = t3.individual_id WHERE t1.uk_vat_number = (SELECT MAX(uk_vat_number) FROM organizations) ORDER BY t2.date_contact_to LIMIT 1 |
Find name of the services that has never been used. | CREATE TABLE services (service_name VARCHAR); CREATE TABLE party_services (service_id VARCHAR); CREATE TABLE services (service_name VARCHAR, service_id VARCHAR) | SELECT service_name FROM services EXCEPT SELECT t1.service_name FROM services AS t1 JOIN party_services AS t2 ON t1.service_id = t2.service_id | ### Context: CREATE TABLE services (service_name VARCHAR); CREATE TABLE party_services (service_id VARCHAR); CREATE TABLE services (service_name VARCHAR, service_id VARCHAR) ### Question: Find name of the services that has never been used. ### Answer: SELECT service_name FROM services EXCEPT SELECT t1.service_name FROM services AS t1 JOIN party_services AS t2 ON t1.service_id = t2.service_id |
Find the name of all the cities and states. | CREATE TABLE addresses (town_city VARCHAR, state_province_county VARCHAR) | SELECT town_city FROM addresses UNION SELECT state_province_county FROM addresses | ### Context: CREATE TABLE addresses (town_city VARCHAR, state_province_county VARCHAR) ### Question: Find the name of all the cities and states. ### Answer: SELECT town_city FROM addresses UNION SELECT state_province_county FROM addresses |
How many cities are there in state "Colorado"? | CREATE TABLE addresses (state_province_county VARCHAR) | SELECT COUNT(*) FROM addresses WHERE state_province_county = "Colorado" | ### Context: CREATE TABLE addresses (state_province_county VARCHAR) ### Question: How many cities are there in state "Colorado"? ### Answer: SELECT COUNT(*) FROM addresses WHERE state_province_county = "Colorado" |
Find the payment method code used by more than 3 parties. | CREATE TABLE parties (payment_method_code VARCHAR) | SELECT payment_method_code FROM parties GROUP BY payment_method_code HAVING COUNT(*) > 3 | ### Context: CREATE TABLE parties (payment_method_code VARCHAR) ### Question: Find the payment method code used by more than 3 parties. ### Answer: SELECT payment_method_code FROM parties GROUP BY payment_method_code HAVING COUNT(*) > 3 |
Find the name of organizations whose names contain "Party". | CREATE TABLE organizations (organization_name VARCHAR) | SELECT organization_name FROM organizations WHERE organization_name LIKE "%Party%" | ### Context: CREATE TABLE organizations (organization_name VARCHAR) ### Question: Find the name of organizations whose names contain "Party". ### Answer: SELECT organization_name FROM organizations WHERE organization_name LIKE "%Party%" |
How many distinct payment methods are used by parties? | CREATE TABLE parties (payment_method_code VARCHAR) | SELECT COUNT(DISTINCT payment_method_code) FROM parties | ### Context: CREATE TABLE parties (payment_method_code VARCHAR) ### Question: How many distinct payment methods are used by parties? ### Answer: SELECT COUNT(DISTINCT payment_method_code) FROM parties |
Which is the email of the party that has used the services the most number of times? | CREATE TABLE parties (party_email VARCHAR, party_id VARCHAR); CREATE TABLE party_services (customer_id VARCHAR) | SELECT t1.party_email FROM parties AS t1 JOIN party_services AS t2 ON t1.party_id = t2.customer_id GROUP BY t1.party_email ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE parties (party_email VARCHAR, party_id VARCHAR); CREATE TABLE party_services (customer_id VARCHAR) ### Question: Which is the email of the party that has used the services the most number of times? ### Answer: SELECT t1.party_email FROM parties AS t1 JOIN party_services AS t2 ON t1.party_id = t2.customer_id GROUP BY t1.party_email ORDER BY COUNT(*) DESC LIMIT 1 |
Which state can address "6862 Kaitlyn Knolls" possibly be in? | CREATE TABLE addresses (state_province_county VARCHAR, line_1_number_building VARCHAR) | SELECT state_province_county FROM addresses WHERE line_1_number_building LIKE "%6862 Kaitlyn Knolls%" | ### Context: CREATE TABLE addresses (state_province_county VARCHAR, line_1_number_building VARCHAR) ### Question: Which state can address "6862 Kaitlyn Knolls" possibly be in? ### Answer: SELECT state_province_county FROM addresses WHERE line_1_number_building LIKE "%6862 Kaitlyn Knolls%" |
What is the name of organization that has the greatest number of contact individuals? | CREATE TABLE organization_contact_individuals (organization_id VARCHAR); CREATE TABLE organizations (organization_name VARCHAR, organization_id VARCHAR) | SELECT t1.organization_name FROM organizations AS t1 JOIN organization_contact_individuals AS t2 ON t1.organization_id = t2.organization_id GROUP BY t1.organization_name ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE organization_contact_individuals (organization_id VARCHAR); CREATE TABLE organizations (organization_name VARCHAR, organization_id VARCHAR) ### Question: What is the name of organization that has the greatest number of contact individuals? ### Answer: SELECT t1.organization_name FROM organizations AS t1 JOIN organization_contact_individuals AS t2 ON t1.organization_id = t2.organization_id GROUP BY t1.organization_name ORDER BY COUNT(*) DESC LIMIT 1 |
Find the last name of the individuals that have been contact individuals of an organization. | CREATE TABLE individuals (individual_last_name VARCHAR, individual_id VARCHAR); CREATE TABLE organization_contact_individuals (individual_id VARCHAR) | SELECT DISTINCT t1.individual_last_name FROM individuals AS t1 JOIN organization_contact_individuals AS t2 ON t1.individual_id = t2.individual_id | ### Context: CREATE TABLE individuals (individual_last_name VARCHAR, individual_id VARCHAR); CREATE TABLE organization_contact_individuals (individual_id VARCHAR) ### Question: Find the last name of the individuals that have been contact individuals of an organization. ### Answer: SELECT DISTINCT t1.individual_last_name FROM individuals AS t1 JOIN organization_contact_individuals AS t2 ON t1.individual_id = t2.individual_id |
How many drivers are there? | CREATE TABLE driver (Id VARCHAR) | SELECT COUNT(*) FROM driver | ### Context: CREATE TABLE driver (Id VARCHAR) ### Question: How many drivers are there? ### Answer: SELECT COUNT(*) FROM driver |
Show the name, home city, and age for all drivers. | CREATE TABLE driver (name VARCHAR, home_city VARCHAR, age VARCHAR) | SELECT name, home_city, age FROM driver | ### Context: CREATE TABLE driver (name VARCHAR, home_city VARCHAR, age VARCHAR) ### Question: Show the name, home city, and age for all drivers. ### Answer: SELECT name, home_city, age FROM driver |
Show the party and the number of drivers in each party. | CREATE TABLE driver (party VARCHAR) | SELECT party, COUNT(*) FROM driver GROUP BY party | ### Context: CREATE TABLE driver (party VARCHAR) ### Question: Show the party and the number of drivers in each party. ### Answer: SELECT party, COUNT(*) FROM driver GROUP BY party |
Subsets and Splits