arviii/sql-create-context · Datasets at Hugging Face

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