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A list of the top 5 countries by number of invoices. List country name and number of invoices.
CREATE TABLE invoices (billing_country VARCHAR)
SELECT billing_country, COUNT(*) FROM invoices GROUP BY billing_country ORDER BY COUNT(*) DESC LIMIT 5
### Context: CREATE TABLE invoices (billing_country VARCHAR) ### Question: A list of the top 5 countries by number of invoices. List country name and number of invoices. ### Answer: SELECT billing_country, COUNT(*) FROM invoices GROUP BY billing_country ORDER BY COUNT(*) DESC LIMIT 5
A list of the top 8 countries by gross/total invoice size. List country name and gross invoice size.
CREATE TABLE invoices (billing_country VARCHAR, total INTEGER)
SELECT billing_country, SUM(total) FROM invoices GROUP BY billing_country ORDER BY SUM(total) DESC LIMIT 8
### Context: CREATE TABLE invoices (billing_country VARCHAR, total INTEGER) ### Question: A list of the top 8 countries by gross/total invoice size. List country name and gross invoice size. ### Answer: SELECT billing_country, SUM(total) FROM invoices GROUP BY billing_country ORDER BY SUM(total) DESC LIMIT 8
A list of the top 10 countries by average invoice size. List country name and average invoice size.
CREATE TABLE invoices (billing_country VARCHAR, total INTEGER)
SELECT billing_country, AVG(total) FROM invoices GROUP BY billing_country ORDER BY AVG(total) DESC LIMIT 10
### Context: CREATE TABLE invoices (billing_country VARCHAR, total INTEGER) ### Question: A list of the top 10 countries by average invoice size. List country name and average invoice size. ### Answer: SELECT billing_country, AVG(total) FROM invoices GROUP BY billing_country ORDER BY AVG(total) DESC LIMIT 10
Find out 5 customers who most recently purchased something. List customers' first and last name.
CREATE TABLE customers (first_name VARCHAR, last_name VARCHAR, id VARCHAR); CREATE TABLE invoices (customer_id VARCHAR, invoice_date VARCHAR)
SELECT T1.first_name, T1.last_name FROM customers AS T1 JOIN invoices AS T2 ON T2.customer_id = T1.id ORDER BY T2.invoice_date DESC LIMIT 5
### Context: CREATE TABLE customers (first_name VARCHAR, last_name VARCHAR, id VARCHAR); CREATE TABLE invoices (customer_id VARCHAR, invoice_date VARCHAR) ### Question: Find out 5 customers who most recently purchased something. List customers' first and last name. ### Answer: SELECT T1.first_name, T1.last_name FROM customers AS T1 JOIN invoices AS T2 ON T2.customer_id = T1.id ORDER BY T2.invoice_date DESC LIMIT 5
Find out the top 10 customers by total number of orders. List customers' first and last name and the number of total orders.
CREATE TABLE customers (first_name VARCHAR, last_name VARCHAR, id VARCHAR); CREATE TABLE invoices (customer_id VARCHAR)
SELECT T1.first_name, T1.last_name, COUNT(*) FROM customers AS T1 JOIN invoices AS T2 ON T2.customer_id = T1.id GROUP BY T1.id ORDER BY COUNT(*) DESC LIMIT 10
### Context: CREATE TABLE customers (first_name VARCHAR, last_name VARCHAR, id VARCHAR); CREATE TABLE invoices (customer_id VARCHAR) ### Question: Find out the top 10 customers by total number of orders. List customers' first and last name and the number of total orders. ### Answer: SELECT T1.first_name, T1.last_name, COUNT(*) FROM customers AS T1 JOIN invoices AS T2 ON T2.customer_id = T1.id GROUP BY T1.id ORDER BY COUNT(*) DESC LIMIT 10
List the top 10 customers by total gross sales. List customers' first and last name and total gross sales.
CREATE TABLE invoices (total INTEGER, customer_id VARCHAR); CREATE TABLE customers (first_name VARCHAR, last_name VARCHAR, id VARCHAR)
SELECT T1.first_name, T1.last_name, SUM(T2.total) FROM customers AS T1 JOIN invoices AS T2 ON T2.customer_id = T1.id GROUP BY T1.id ORDER BY SUM(T2.total) DESC LIMIT 10
### Context: CREATE TABLE invoices (total INTEGER, customer_id VARCHAR); CREATE TABLE customers (first_name VARCHAR, last_name VARCHAR, id VARCHAR) ### Question: List the top 10 customers by total gross sales. List customers' first and last name and total gross sales. ### Answer: SELECT T1.first_name, T1.last_name, SUM(T2.total) FROM customers AS T1 JOIN invoices AS T2 ON T2.customer_id = T1.id GROUP BY T1.id ORDER BY SUM(T2.total) DESC LIMIT 10
List the top 5 genres by number of tracks. List genres name and total tracks.
CREATE TABLE tracks (genre_id VARCHAR); CREATE TABLE genres (name VARCHAR, id VARCHAR)
SELECT T1.name, COUNT(*) FROM genres AS T1 JOIN tracks AS T2 ON T2.genre_id = T1.id GROUP BY T1.id ORDER BY COUNT(*) DESC LIMIT 5
### Context: CREATE TABLE tracks (genre_id VARCHAR); CREATE TABLE genres (name VARCHAR, id VARCHAR) ### Question: List the top 5 genres by number of tracks. List genres name and total tracks. ### Answer: SELECT T1.name, COUNT(*) FROM genres AS T1 JOIN tracks AS T2 ON T2.genre_id = T1.id GROUP BY T1.id ORDER BY COUNT(*) DESC LIMIT 5
List every album's title.
CREATE TABLE albums (title VARCHAR)
SELECT title FROM albums
### Context: CREATE TABLE albums (title VARCHAR) ### Question: List every album's title. ### Answer: SELECT title FROM albums
List every album ordered by album title in ascending order.
CREATE TABLE albums (title VARCHAR)
SELECT title FROM albums ORDER BY title
### Context: CREATE TABLE albums (title VARCHAR) ### Question: List every album ordered by album title in ascending order. ### Answer: SELECT title FROM albums ORDER BY title
List every album whose title starts with A in alphabetical order.
CREATE TABLE albums (title VARCHAR)
SELECT title FROM albums WHERE title LIKE 'A%' ORDER BY title
### Context: CREATE TABLE albums (title VARCHAR) ### Question: List every album whose title starts with A in alphabetical order. ### Answer: SELECT title FROM albums WHERE title LIKE 'A%' ORDER BY title
List the customers first and last name of 10 least expensive invoices.
CREATE TABLE customers (first_name VARCHAR, last_name VARCHAR, id VARCHAR); CREATE TABLE invoices (customer_id VARCHAR)
SELECT T1.first_name, T1.last_name FROM customers AS T1 JOIN invoices AS T2 ON T2.customer_id = T1.id ORDER BY total LIMIT 10
### Context: CREATE TABLE customers (first_name VARCHAR, last_name VARCHAR, id VARCHAR); CREATE TABLE invoices (customer_id VARCHAR) ### Question: List the customers first and last name of 10 least expensive invoices. ### Answer: SELECT T1.first_name, T1.last_name FROM customers AS T1 JOIN invoices AS T2 ON T2.customer_id = T1.id ORDER BY total LIMIT 10
List total amount of invoice from Chicago, IL.
CREATE TABLE invoices (total INTEGER, billing_city VARCHAR, billing_state VARCHAR)
SELECT SUM(total) FROM invoices WHERE billing_city = "Chicago" AND billing_state = "IL"
### Context: CREATE TABLE invoices (total INTEGER, billing_city VARCHAR, billing_state VARCHAR) ### Question: List total amount of invoice from Chicago, IL. ### Answer: SELECT SUM(total) FROM invoices WHERE billing_city = "Chicago" AND billing_state = "IL"
List the number of invoices from Chicago, IL.
CREATE TABLE invoices (billing_city VARCHAR, billing_state VARCHAR)
SELECT COUNT(*) FROM invoices WHERE billing_city = "Chicago" AND billing_state = "IL"
### Context: CREATE TABLE invoices (billing_city VARCHAR, billing_state VARCHAR) ### Question: List the number of invoices from Chicago, IL. ### Answer: SELECT COUNT(*) FROM invoices WHERE billing_city = "Chicago" AND billing_state = "IL"
List the number of invoices from the US, grouped by state.
CREATE TABLE invoices (billing_state VARCHAR, billing_country VARCHAR)
SELECT billing_state, COUNT(*) FROM invoices WHERE billing_country = "USA" GROUP BY billing_state
### Context: CREATE TABLE invoices (billing_state VARCHAR, billing_country VARCHAR) ### Question: List the number of invoices from the US, grouped by state. ### Answer: SELECT billing_state, COUNT(*) FROM invoices WHERE billing_country = "USA" GROUP BY billing_state
List the state in the US with the most invoices.
CREATE TABLE invoices (billing_state VARCHAR, billing_country VARCHAR)
SELECT billing_state, COUNT(*) FROM invoices WHERE billing_country = "USA" GROUP BY billing_state ORDER BY COUNT(*) DESC LIMIT 1
### Context: CREATE TABLE invoices (billing_state VARCHAR, billing_country VARCHAR) ### Question: List the state in the US with the most invoices. ### Answer: SELECT billing_state, COUNT(*) FROM invoices WHERE billing_country = "USA" GROUP BY billing_state ORDER BY COUNT(*) DESC LIMIT 1
List the number of invoices and the invoice total from California.
CREATE TABLE invoices (billing_state VARCHAR, total INTEGER)
SELECT billing_state, COUNT(*), SUM(total) FROM invoices WHERE billing_state = "CA"
### Context: CREATE TABLE invoices (billing_state VARCHAR, total INTEGER) ### Question: List the number of invoices and the invoice total from California. ### Answer: SELECT billing_state, COUNT(*), SUM(total) FROM invoices WHERE billing_state = "CA"
List Aerosmith's albums.
CREATE TABLE artists (id VARCHAR, name VARCHAR); CREATE TABLE albums (title VARCHAR, artist_id VARCHAR)
SELECT T1.title FROM albums AS T1 JOIN artists AS T2 ON T1.artist_id = T2.id WHERE T2.name = "Aerosmith"
### Context: CREATE TABLE artists (id VARCHAR, name VARCHAR); CREATE TABLE albums (title VARCHAR, artist_id VARCHAR) ### Question: List Aerosmith's albums. ### Answer: SELECT T1.title FROM albums AS T1 JOIN artists AS T2 ON T1.artist_id = T2.id WHERE T2.name = "Aerosmith"
How many albums does Billy Cobham has?
CREATE TABLE artists (id VARCHAR, name VARCHAR); CREATE TABLE albums (artist_id VARCHAR)
SELECT COUNT(*) FROM albums AS T1 JOIN artists AS T2 ON T1.artist_id = T2.id WHERE T2.name = "Billy Cobham"
### Context: CREATE TABLE artists (id VARCHAR, name VARCHAR); CREATE TABLE albums (artist_id VARCHAR) ### Question: How many albums does Billy Cobham has? ### Answer: SELECT COUNT(*) FROM albums AS T1 JOIN artists AS T2 ON T1.artist_id = T2.id WHERE T2.name = "Billy Cobham"
Eduardo Martins is a customer at which company?
CREATE TABLE customers (company VARCHAR, first_name VARCHAR, last_name VARCHAR)
SELECT company FROM customers WHERE first_name = "Eduardo" AND last_name = "Martins"
### Context: CREATE TABLE customers (company VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: Eduardo Martins is a customer at which company? ### Answer: SELECT company FROM customers WHERE first_name = "Eduardo" AND last_name = "Martins"
What is Astrid Gruber's email and phone number?
CREATE TABLE customers (email VARCHAR, phone VARCHAR, first_name VARCHAR, last_name VARCHAR)
SELECT email, phone FROM customers WHERE first_name = "Astrid" AND last_name = "Gruber"
### Context: CREATE TABLE customers (email VARCHAR, phone VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: What is Astrid Gruber's email and phone number? ### Answer: SELECT email, phone FROM customers WHERE first_name = "Astrid" AND last_name = "Gruber"
How many customers live in Prague city?
CREATE TABLE customers (city VARCHAR)
SELECT COUNT(*) FROM customers WHERE city = "Prague"
### Context: CREATE TABLE customers (city VARCHAR) ### Question: How many customers live in Prague city? ### Answer: SELECT COUNT(*) FROM customers WHERE city = "Prague"
How many customers in state of CA?
CREATE TABLE customers (state VARCHAR)
SELECT COUNT(*) FROM customers WHERE state = "CA"
### Context: CREATE TABLE customers (state VARCHAR) ### Question: How many customers in state of CA? ### Answer: SELECT COUNT(*) FROM customers WHERE state = "CA"
What country does Roberto Almeida live?
CREATE TABLE customers (country VARCHAR, first_name VARCHAR, last_name VARCHAR)
SELECT country FROM customers WHERE first_name = "Roberto" AND last_name = "Almeida"
### Context: CREATE TABLE customers (country VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: What country does Roberto Almeida live? ### Answer: SELECT country FROM customers WHERE first_name = "Roberto" AND last_name = "Almeida"
List the name of albums that are released by aritist whose name has 'Led'
CREATE TABLE artists (id VARCHAR, name VARCHAR); CREATE TABLE albums (title VARCHAR, artist_id VARCHAR)
SELECT T2.title FROM artists AS T1 JOIN albums AS T2 ON T1.id = T2.artist_id WHERE T1.name LIKE '%Led%'
### Context: CREATE TABLE artists (id VARCHAR, name VARCHAR); CREATE TABLE albums (title VARCHAR, artist_id VARCHAR) ### Question: List the name of albums that are released by aritist whose name has 'Led' ### Answer: SELECT T2.title FROM artists AS T1 JOIN albums AS T2 ON T1.id = T2.artist_id WHERE T1.name LIKE '%Led%'
How many customers does Steve Johnson support?
CREATE TABLE employees (id VARCHAR, first_name VARCHAR, last_name VARCHAR); CREATE TABLE customers (support_rep_id VARCHAR)
SELECT COUNT(*) FROM employees AS T1 JOIN customers AS T2 ON T2.support_rep_id = T1.id WHERE T1.first_name = "Steve" AND T1.last_name = "Johnson"
### Context: CREATE TABLE employees (id VARCHAR, first_name VARCHAR, last_name VARCHAR); CREATE TABLE customers (support_rep_id VARCHAR) ### Question: How many customers does Steve Johnson support? ### Answer: SELECT COUNT(*) FROM employees AS T1 JOIN customers AS T2 ON T2.support_rep_id = T1.id WHERE T1.first_name = "Steve" AND T1.last_name = "Johnson"
What is the title, phone and hire date of Nancy Edwards?
CREATE TABLE employees (title VARCHAR, phone VARCHAR, hire_date VARCHAR, first_name VARCHAR, last_name VARCHAR)
SELECT title, phone, hire_date FROM employees WHERE first_name = "Nancy" AND last_name = "Edwards"
### Context: CREATE TABLE employees (title VARCHAR, phone VARCHAR, hire_date VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: What is the title, phone and hire date of Nancy Edwards? ### Answer: SELECT title, phone, hire_date FROM employees WHERE first_name = "Nancy" AND last_name = "Edwards"
find the full name of employees who report to Nancy Edwards?
CREATE TABLE employees (id VARCHAR, first_name VARCHAR, last_name VARCHAR); CREATE TABLE employees (first_name VARCHAR, last_name VARCHAR, reports_to VARCHAR)
SELECT T2.first_name, T2.last_name FROM employees AS T1 JOIN employees AS T2 ON T1.id = T2.reports_to WHERE T1.first_name = "Nancy" AND T1.last_name = "Edwards"
### Context: CREATE TABLE employees (id VARCHAR, first_name VARCHAR, last_name VARCHAR); CREATE TABLE employees (first_name VARCHAR, last_name VARCHAR, reports_to VARCHAR) ### Question: find the full name of employees who report to Nancy Edwards? ### Answer: SELECT T2.first_name, T2.last_name FROM employees AS T1 JOIN employees AS T2 ON T1.id = T2.reports_to WHERE T1.first_name = "Nancy" AND T1.last_name = "Edwards"
What is the address of employee Nancy Edwards?
CREATE TABLE employees (address VARCHAR, first_name VARCHAR, last_name VARCHAR)
SELECT address FROM employees WHERE first_name = "Nancy" AND last_name = "Edwards"
### Context: CREATE TABLE employees (address VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: What is the address of employee Nancy Edwards? ### Answer: SELECT address FROM employees WHERE first_name = "Nancy" AND last_name = "Edwards"
Find the full name of employee who supported the most number of customers.
CREATE TABLE employees (first_name VARCHAR, last_name VARCHAR, id VARCHAR); CREATE TABLE customers (support_rep_id VARCHAR)
SELECT T1.first_name, T1.last_name FROM employees AS T1 JOIN customers AS T2 ON T1.id = T2.support_rep_id GROUP BY T1.id ORDER BY COUNT(*) DESC LIMIT 1
### Context: CREATE TABLE employees (first_name VARCHAR, last_name VARCHAR, id VARCHAR); CREATE TABLE customers (support_rep_id VARCHAR) ### Question: Find the full name of employee who supported the most number of customers. ### Answer: SELECT T1.first_name, T1.last_name FROM employees AS T1 JOIN customers AS T2 ON T1.id = T2.support_rep_id GROUP BY T1.id ORDER BY COUNT(*) DESC LIMIT 1
How many employees are living in Canada?
CREATE TABLE employees (country VARCHAR)
SELECT COUNT(*) FROM employees WHERE country = "Canada"
### Context: CREATE TABLE employees (country VARCHAR) ### Question: How many employees are living in Canada? ### Answer: SELECT COUNT(*) FROM employees WHERE country = "Canada"
What is employee Nancy Edwards's phone number?
CREATE TABLE employees (phone VARCHAR, first_name VARCHAR, last_name VARCHAR)
SELECT phone FROM employees WHERE first_name = "Nancy" AND last_name = "Edwards"
### Context: CREATE TABLE employees (phone VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: What is employee Nancy Edwards's phone number? ### Answer: SELECT phone FROM employees WHERE first_name = "Nancy" AND last_name = "Edwards"
Who is the youngest employee in the company? List employee's first and last name.
CREATE TABLE employees (first_name VARCHAR, last_name VARCHAR, birth_date VARCHAR)
SELECT first_name, last_name FROM employees ORDER BY birth_date DESC LIMIT 1
### Context: CREATE TABLE employees (first_name VARCHAR, last_name VARCHAR, birth_date VARCHAR) ### Question: Who is the youngest employee in the company? List employee's first and last name. ### Answer: SELECT first_name, last_name FROM employees ORDER BY birth_date DESC LIMIT 1
List top 10 employee work longest in the company. List employee's first and last name.
CREATE TABLE employees (first_name VARCHAR, last_name VARCHAR, hire_date VARCHAR)
SELECT first_name, last_name FROM employees ORDER BY hire_date LIMIT 10
### Context: CREATE TABLE employees (first_name VARCHAR, last_name VARCHAR, hire_date VARCHAR) ### Question: List top 10 employee work longest in the company. List employee's first and last name. ### Answer: SELECT first_name, last_name FROM employees ORDER BY hire_date LIMIT 10
Find the number of employees whose title is IT Staff from each city?
CREATE TABLE employees (city VARCHAR, title VARCHAR)
SELECT COUNT(*), city FROM employees WHERE title = 'IT Staff' GROUP BY city
### Context: CREATE TABLE employees (city VARCHAR, title VARCHAR) ### Question: Find the number of employees whose title is IT Staff from each city? ### Answer: SELECT COUNT(*), city FROM employees WHERE title = 'IT Staff' GROUP BY city
Which employee manage most number of peoples? List employee's first and last name, and number of people report to that employee.
CREATE TABLE employees (first_name VARCHAR, last_name VARCHAR, id VARCHAR); CREATE TABLE employees (reports_to VARCHAR)
SELECT T2.first_name, T2.last_name, COUNT(T1.reports_to) FROM employees AS T1 JOIN employees AS T2 ON T1.reports_to = T2.id GROUP BY T1.reports_to ORDER BY COUNT(T1.reports_to) DESC LIMIT 1
### Context: CREATE TABLE employees (first_name VARCHAR, last_name VARCHAR, id VARCHAR); CREATE TABLE employees (reports_to VARCHAR) ### Question: Which employee manage most number of peoples? List employee's first and last name, and number of people report to that employee. ### Answer: SELECT T2.first_name, T2.last_name, COUNT(T1.reports_to) FROM employees AS T1 JOIN employees AS T2 ON T1.reports_to = T2.id GROUP BY T1.reports_to ORDER BY COUNT(T1.reports_to) DESC LIMIT 1
How many orders does Lucas Mancini has?
CREATE TABLE invoices (customer_id VARCHAR); CREATE TABLE customers (id VARCHAR, first_name VARCHAR, last_name VARCHAR)
SELECT COUNT(*) FROM customers AS T1 JOIN invoices AS T2 ON T1.id = T2.customer_id WHERE T1.first_name = "Lucas" AND T1.last_name = "Mancini"
### Context: CREATE TABLE invoices (customer_id VARCHAR); CREATE TABLE customers (id VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: How many orders does Lucas Mancini has? ### Answer: SELECT COUNT(*) FROM customers AS T1 JOIN invoices AS T2 ON T1.id = T2.customer_id WHERE T1.first_name = "Lucas" AND T1.last_name = "Mancini"
What is the total amount of money spent by Lucas Mancini?
CREATE TABLE invoices (total INTEGER, customer_id VARCHAR); CREATE TABLE customers (id VARCHAR, first_name VARCHAR, last_name VARCHAR)
SELECT SUM(T2.total) FROM customers AS T1 JOIN invoices AS T2 ON T1.id = T2.customer_id WHERE T1.first_name = "Lucas" AND T1.last_name = "Mancini"
### Context: CREATE TABLE invoices (total INTEGER, customer_id VARCHAR); CREATE TABLE customers (id VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: What is the total amount of money spent by Lucas Mancini? ### Answer: SELECT SUM(T2.total) FROM customers AS T1 JOIN invoices AS T2 ON T1.id = T2.customer_id WHERE T1.first_name = "Lucas" AND T1.last_name = "Mancini"
List all media types.
CREATE TABLE media_types (name VARCHAR)
SELECT name FROM media_types
### Context: CREATE TABLE media_types (name VARCHAR) ### Question: List all media types. ### Answer: SELECT name FROM media_types
List all different genre types.
CREATE TABLE genres (name VARCHAR)
SELECT DISTINCT name FROM genres
### Context: CREATE TABLE genres (name VARCHAR) ### Question: List all different genre types. ### Answer: SELECT DISTINCT name FROM genres
List the name of all playlist.
CREATE TABLE playlists (name VARCHAR)
SELECT name FROM playlists
### Context: CREATE TABLE playlists (name VARCHAR) ### Question: List the name of all playlist. ### Answer: SELECT name FROM playlists
Who is the composer of track Fast As a Shark?
CREATE TABLE tracks (composer VARCHAR, name VARCHAR)
SELECT composer FROM tracks WHERE name = "Fast As a Shark"
### Context: CREATE TABLE tracks (composer VARCHAR, name VARCHAR) ### Question: Who is the composer of track Fast As a Shark? ### Answer: SELECT composer FROM tracks WHERE name = "Fast As a Shark"
How long does track Fast As a Shark has?
CREATE TABLE tracks (milliseconds VARCHAR, name VARCHAR)
SELECT milliseconds FROM tracks WHERE name = "Fast As a Shark"
### Context: CREATE TABLE tracks (milliseconds VARCHAR, name VARCHAR) ### Question: How long does track Fast As a Shark has? ### Answer: SELECT milliseconds FROM tracks WHERE name = "Fast As a Shark"
What is the name of tracks whose genre is Rock?
CREATE TABLE genres (id VARCHAR, name VARCHAR); CREATE TABLE tracks (name VARCHAR, genre_id VARCHAR)
SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T1.name = "Rock"
### Context: CREATE TABLE genres (id VARCHAR, name VARCHAR); CREATE TABLE tracks (name VARCHAR, genre_id VARCHAR) ### Question: What is the name of tracks whose genre is Rock? ### Answer: SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T1.name = "Rock"
What is title of album which track Balls to the Wall belongs to?
CREATE TABLE tracks (genre_id VARCHAR, name VARCHAR); CREATE TABLE albums (title VARCHAR, id VARCHAR)
SELECT T1.title FROM albums AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T2.name = "Balls to the Wall"
### Context: CREATE TABLE tracks (genre_id VARCHAR, name VARCHAR); CREATE TABLE albums (title VARCHAR, id VARCHAR) ### Question: What is title of album which track Balls to the Wall belongs to? ### Answer: SELECT T1.title FROM albums AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T2.name = "Balls to the Wall"
List name of all tracks in Balls to the Wall.
CREATE TABLE tracks (name VARCHAR, genre_id VARCHAR); CREATE TABLE albums (id VARCHAR, title VARCHAR)
SELECT T2.name FROM albums AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T1.title = "Balls to the Wall"
### Context: CREATE TABLE tracks (name VARCHAR, genre_id VARCHAR); CREATE TABLE albums (id VARCHAR, title VARCHAR) ### Question: List name of all tracks in Balls to the Wall. ### Answer: SELECT T2.name FROM albums AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T1.title = "Balls to the Wall"
List title of albums have the number of tracks greater than 10.
CREATE TABLE tracks (album_id VARCHAR); CREATE TABLE albums (title VARCHAR, id VARCHAR)
SELECT T1.title FROM albums AS T1 JOIN tracks AS T2 ON T1.id = T2.album_id GROUP BY T1.id HAVING COUNT(T1.id) > 10
### Context: CREATE TABLE tracks (album_id VARCHAR); CREATE TABLE albums (title VARCHAR, id VARCHAR) ### Question: List title of albums have the number of tracks greater than 10. ### Answer: SELECT T1.title FROM albums AS T1 JOIN tracks AS T2 ON T1.id = T2.album_id GROUP BY T1.id HAVING COUNT(T1.id) > 10
List the name of tracks belongs to genre Rock and whose media type is MPEG audio file.
CREATE TABLE genres (id VARCHAR, name VARCHAR); CREATE TABLE tracks (name VARCHAR, genre_id VARCHAR, media_type_id VARCHAR); CREATE TABLE media_types (id VARCHAR, name VARCHAR)
SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id JOIN media_types AS T3 ON T3.id = T2.media_type_id WHERE T1.name = "Rock" AND T3.name = "MPEG audio file"
### Context: CREATE TABLE genres (id VARCHAR, name VARCHAR); CREATE TABLE tracks (name VARCHAR, genre_id VARCHAR, media_type_id VARCHAR); CREATE TABLE media_types (id VARCHAR, name VARCHAR) ### Question: List the name of tracks belongs to genre Rock and whose media type is MPEG audio file. ### Answer: SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id JOIN media_types AS T3 ON T3.id = T2.media_type_id WHERE T1.name = "Rock" AND T3.name = "MPEG audio file"
List the name of tracks belongs to genre Rock or media type is MPEG audio file.
CREATE TABLE genres (id VARCHAR, name VARCHAR); CREATE TABLE tracks (name VARCHAR, genre_id VARCHAR, media_type_id VARCHAR); CREATE TABLE media_types (id VARCHAR, name VARCHAR)
SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id JOIN media_types AS T3 ON T3.id = T2.media_type_id WHERE T1.name = "Rock" OR T3.name = "MPEG audio file"
### Context: CREATE TABLE genres (id VARCHAR, name VARCHAR); CREATE TABLE tracks (name VARCHAR, genre_id VARCHAR, media_type_id VARCHAR); CREATE TABLE media_types (id VARCHAR, name VARCHAR) ### Question: List the name of tracks belongs to genre Rock or media type is MPEG audio file. ### Answer: SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id JOIN media_types AS T3 ON T3.id = T2.media_type_id WHERE T1.name = "Rock" OR T3.name = "MPEG audio file"
List the name of tracks belongs to genre Rock or genre Jazz.
CREATE TABLE genres (id VARCHAR, name VARCHAR); CREATE TABLE tracks (name VARCHAR, genre_id VARCHAR)
SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T1.name = "Rock" OR T1.name = "Jazz"
### Context: CREATE TABLE genres (id VARCHAR, name VARCHAR); CREATE TABLE tracks (name VARCHAR, genre_id VARCHAR) ### Question: List the name of tracks belongs to genre Rock or genre Jazz. ### Answer: SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T1.name = "Rock" OR T1.name = "Jazz"
List the name of all tracks in the playlists of Movies.
CREATE TABLE playlists (id VARCHAR, name VARCHAR); CREATE TABLE playlist_tracks (track_id VARCHAR, playlist_id VARCHAR); CREATE TABLE tracks (name VARCHAR, id VARCHAR)
SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T3.id = T2.playlist_id WHERE T3.name = "Movies"
### Context: CREATE TABLE playlists (id VARCHAR, name VARCHAR); CREATE TABLE playlist_tracks (track_id VARCHAR, playlist_id VARCHAR); CREATE TABLE tracks (name VARCHAR, id VARCHAR) ### Question: List the name of all tracks in the playlists of Movies. ### Answer: SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T3.id = T2.playlist_id WHERE T3.name = "Movies"
List the name of playlist which has number of tracks greater than 100.
CREATE TABLE playlist_tracks (playlist_id VARCHAR, track_id VARCHAR); CREATE TABLE playlists (name VARCHAR, id VARCHAR)
SELECT T2.name FROM playlist_tracks AS T1 JOIN playlists AS T2 ON T2.id = T1.playlist_id GROUP BY T1.playlist_id HAVING COUNT(T1.track_id) > 100
### Context: CREATE TABLE playlist_tracks (playlist_id VARCHAR, track_id VARCHAR); CREATE TABLE playlists (name VARCHAR, id VARCHAR) ### Question: List the name of playlist which has number of tracks greater than 100. ### Answer: SELECT T2.name FROM playlist_tracks AS T1 JOIN playlists AS T2 ON T2.id = T1.playlist_id GROUP BY T1.playlist_id HAVING COUNT(T1.track_id) > 100
List all tracks bought by customer Daan Peeters.
CREATE TABLE invoices (id VARCHAR, customer_id VARCHAR); CREATE TABLE invoice_lines (track_id VARCHAR, invoice_id VARCHAR); CREATE TABLE tracks (name VARCHAR, id VARCHAR); CREATE TABLE customers (id VARCHAR, first_name VARCHAR, last_name VARCHAR)
SELECT T1.name FROM tracks AS T1 JOIN invoice_lines AS T2 ON T1.id = T2.track_id JOIN invoices AS T3 ON T3.id = T2.invoice_id JOIN customers AS T4 ON T4.id = T3.customer_id WHERE T4.first_name = "Daan" AND T4.last_name = "Peeters"
### Context: CREATE TABLE invoices (id VARCHAR, customer_id VARCHAR); CREATE TABLE invoice_lines (track_id VARCHAR, invoice_id VARCHAR); CREATE TABLE tracks (name VARCHAR, id VARCHAR); CREATE TABLE customers (id VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: List all tracks bought by customer Daan Peeters. ### Answer: SELECT T1.name FROM tracks AS T1 JOIN invoice_lines AS T2 ON T1.id = T2.track_id JOIN invoices AS T3 ON T3.id = T2.invoice_id JOIN customers AS T4 ON T4.id = T3.customer_id WHERE T4.first_name = "Daan" AND T4.last_name = "Peeters"
How much is the track Fast As a Shark?
CREATE TABLE tracks (unit_price VARCHAR, name VARCHAR)
SELECT unit_price FROM tracks WHERE name = "Fast As a Shark"
### Context: CREATE TABLE tracks (unit_price VARCHAR, name VARCHAR) ### Question: How much is the track Fast As a Shark? ### Answer: SELECT unit_price FROM tracks WHERE name = "Fast As a Shark"
Find the name of tracks which are in Movies playlist but not in music playlist.
CREATE TABLE playlists (id VARCHAR, name VARCHAR); CREATE TABLE playlist_tracks (track_id VARCHAR, playlist_id VARCHAR); CREATE TABLE tracks (name VARCHAR, id VARCHAR)
SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Movies' EXCEPT SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Music'
### Context: CREATE TABLE playlists (id VARCHAR, name VARCHAR); CREATE TABLE playlist_tracks (track_id VARCHAR, playlist_id VARCHAR); CREATE TABLE tracks (name VARCHAR, id VARCHAR) ### Question: Find the name of tracks which are in Movies playlist but not in music playlist. ### Answer: SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Movies' EXCEPT SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Music'
Find the name of tracks which are in both Movies and music playlists.
CREATE TABLE playlists (id VARCHAR, name VARCHAR); CREATE TABLE playlist_tracks (track_id VARCHAR, playlist_id VARCHAR); CREATE TABLE tracks (name VARCHAR, id VARCHAR)
SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Movies' INTERSECT SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Music'
### Context: CREATE TABLE playlists (id VARCHAR, name VARCHAR); CREATE TABLE playlist_tracks (track_id VARCHAR, playlist_id VARCHAR); CREATE TABLE tracks (name VARCHAR, id VARCHAR) ### Question: Find the name of tracks which are in both Movies and music playlists. ### Answer: SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Movies' INTERSECT SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Music'
Find number of tracks in each genre?
CREATE TABLE tracks (genre_id VARCHAR); CREATE TABLE genres (name VARCHAR, id VARCHAR)
SELECT COUNT(*), T1.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id GROUP BY T1.name
### Context: CREATE TABLE tracks (genre_id VARCHAR); CREATE TABLE genres (name VARCHAR, id VARCHAR) ### Question: Find number of tracks in each genre? ### Answer: SELECT COUNT(*), T1.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id GROUP BY T1.name
How many editors are there?
CREATE TABLE editor (Id VARCHAR)
SELECT COUNT(*) FROM editor
### Context: CREATE TABLE editor (Id VARCHAR) ### Question: How many editors are there? ### Answer: SELECT COUNT(*) FROM editor
List the names of editors in ascending order of age.
CREATE TABLE editor (Name VARCHAR, Age VARCHAR)
SELECT Name FROM editor ORDER BY Age
### Context: CREATE TABLE editor (Name VARCHAR, Age VARCHAR) ### Question: List the names of editors in ascending order of age. ### Answer: SELECT Name FROM editor ORDER BY Age
What are the names and ages of editors?
CREATE TABLE editor (Name VARCHAR, Age VARCHAR)
SELECT Name, Age FROM editor
### Context: CREATE TABLE editor (Name VARCHAR, Age VARCHAR) ### Question: What are the names and ages of editors? ### Answer: SELECT Name, Age FROM editor
List the names of editors who are older than 25.
CREATE TABLE editor (Name VARCHAR, Age INTEGER)
SELECT Name FROM editor WHERE Age > 25
### Context: CREATE TABLE editor (Name VARCHAR, Age INTEGER) ### Question: List the names of editors who are older than 25. ### Answer: SELECT Name FROM editor WHERE Age > 25
Show the names of editors of age either 24 or 25.
CREATE TABLE editor (Name VARCHAR, Age VARCHAR)
SELECT Name FROM editor WHERE Age = 24 OR Age = 25
### Context: CREATE TABLE editor (Name VARCHAR, Age VARCHAR) ### Question: Show the names of editors of age either 24 or 25. ### Answer: SELECT Name FROM editor WHERE Age = 24 OR Age = 25
What is the name of the youngest editor?
CREATE TABLE editor (Name VARCHAR, Age VARCHAR)
SELECT Name FROM editor ORDER BY Age LIMIT 1
### Context: CREATE TABLE editor (Name VARCHAR, Age VARCHAR) ### Question: What is the name of the youngest editor? ### Answer: SELECT Name FROM editor ORDER BY Age LIMIT 1
What are the different ages of editors? Show each age along with the number of editors of that age.
CREATE TABLE editor (Age VARCHAR)
SELECT Age, COUNT(*) FROM editor GROUP BY Age
### Context: CREATE TABLE editor (Age VARCHAR) ### Question: What are the different ages of editors? Show each age along with the number of editors of that age. ### Answer: SELECT Age, COUNT(*) FROM editor GROUP BY Age
Please show the most common age of editors.
CREATE TABLE editor (Age VARCHAR)
SELECT Age FROM editor GROUP BY Age ORDER BY COUNT(*) DESC LIMIT 1
### Context: CREATE TABLE editor (Age VARCHAR) ### Question: Please show the most common age of editors. ### Answer: SELECT Age FROM editor GROUP BY Age ORDER BY COUNT(*) DESC LIMIT 1
Show the distinct themes of journals.
CREATE TABLE journal (Theme VARCHAR)
SELECT DISTINCT Theme FROM journal
### Context: CREATE TABLE journal (Theme VARCHAR) ### Question: Show the distinct themes of journals. ### Answer: SELECT DISTINCT Theme FROM journal
Show the names of editors and the theme of journals for which they serve on committees.
CREATE TABLE journal_committee (Editor_ID VARCHAR, Journal_ID VARCHAR); CREATE TABLE editor (Name VARCHAR, Editor_ID VARCHAR); CREATE TABLE journal (Theme VARCHAR, Journal_ID VARCHAR)
SELECT T2.Name, T3.Theme FROM journal_committee AS T1 JOIN editor AS T2 ON T1.Editor_ID = T2.Editor_ID JOIN journal AS T3 ON T1.Journal_ID = T3.Journal_ID
### Context: CREATE TABLE journal_committee (Editor_ID VARCHAR, Journal_ID VARCHAR); CREATE TABLE editor (Name VARCHAR, Editor_ID VARCHAR); CREATE TABLE journal (Theme VARCHAR, Journal_ID VARCHAR) ### Question: Show the names of editors and the theme of journals for which they serve on committees. ### Answer: SELECT T2.Name, T3.Theme FROM journal_committee AS T1 JOIN editor AS T2 ON T1.Editor_ID = T2.Editor_ID JOIN journal AS T3 ON T1.Journal_ID = T3.Journal_ID
Show the names and ages of editors and the theme of journals for which they serve on committees, in ascending alphabetical order of theme.
CREATE TABLE journal_committee (Editor_ID VARCHAR, Journal_ID VARCHAR); CREATE TABLE editor (Name VARCHAR, age VARCHAR, Editor_ID VARCHAR); CREATE TABLE journal (Theme VARCHAR, Journal_ID VARCHAR)
SELECT T2.Name, T2.age, T3.Theme FROM journal_committee AS T1 JOIN editor AS T2 ON T1.Editor_ID = T2.Editor_ID JOIN journal AS T3 ON T1.Journal_ID = T3.Journal_ID ORDER BY T3.Theme
### Context: CREATE TABLE journal_committee (Editor_ID VARCHAR, Journal_ID VARCHAR); CREATE TABLE editor (Name VARCHAR, age VARCHAR, Editor_ID VARCHAR); CREATE TABLE journal (Theme VARCHAR, Journal_ID VARCHAR) ### Question: Show the names and ages of editors and the theme of journals for which they serve on committees, in ascending alphabetical order of theme. ### Answer: SELECT T2.Name, T2.age, T3.Theme FROM journal_committee AS T1 JOIN editor AS T2 ON T1.Editor_ID = T2.Editor_ID JOIN journal AS T3 ON T1.Journal_ID = T3.Journal_ID ORDER BY T3.Theme
Show the names of editors that are on the committee of journals with sales bigger than 3000.
CREATE TABLE journal_committee (Editor_ID VARCHAR, Journal_ID VARCHAR); CREATE TABLE editor (Name VARCHAR, Editor_ID VARCHAR); CREATE TABLE journal (Journal_ID VARCHAR, Sales INTEGER)
SELECT T2.Name FROM journal_committee AS T1 JOIN editor AS T2 ON T1.Editor_ID = T2.Editor_ID JOIN journal AS T3 ON T1.Journal_ID = T3.Journal_ID WHERE T3.Sales > 3000
### Context: CREATE TABLE journal_committee (Editor_ID VARCHAR, Journal_ID VARCHAR); CREATE TABLE editor (Name VARCHAR, Editor_ID VARCHAR); CREATE TABLE journal (Journal_ID VARCHAR, Sales INTEGER) ### Question: Show the names of editors that are on the committee of journals with sales bigger than 3000. ### Answer: SELECT T2.Name FROM journal_committee AS T1 JOIN editor AS T2 ON T1.Editor_ID = T2.Editor_ID JOIN journal AS T3 ON T1.Journal_ID = T3.Journal_ID WHERE T3.Sales > 3000
Show the id, name of each editor and the number of journal committees they are on.
CREATE TABLE editor (editor_id VARCHAR, Name VARCHAR, Editor_ID VARCHAR); CREATE TABLE journal_committee (Editor_ID VARCHAR)
SELECT T1.editor_id, T1.Name, COUNT(*) FROM editor AS T1 JOIN journal_committee AS T2 ON T1.Editor_ID = T2.Editor_ID GROUP BY T1.editor_id
### Context: CREATE TABLE editor (editor_id VARCHAR, Name VARCHAR, Editor_ID VARCHAR); CREATE TABLE journal_committee (Editor_ID VARCHAR) ### Question: Show the id, name of each editor and the number of journal committees they are on. ### Answer: SELECT T1.editor_id, T1.Name, COUNT(*) FROM editor AS T1 JOIN journal_committee AS T2 ON T1.Editor_ID = T2.Editor_ID GROUP BY T1.editor_id
Show the names of editors that are on at least two journal committees.
CREATE TABLE editor (Name VARCHAR, Editor_ID VARCHAR); CREATE TABLE journal_committee (Editor_ID VARCHAR)
SELECT T1.Name FROM editor AS T1 JOIN journal_committee AS T2 ON T1.Editor_ID = T2.Editor_ID GROUP BY T1.Name HAVING COUNT(*) >= 2
### Context: CREATE TABLE editor (Name VARCHAR, Editor_ID VARCHAR); CREATE TABLE journal_committee (Editor_ID VARCHAR) ### Question: Show the names of editors that are on at least two journal committees. ### Answer: SELECT T1.Name FROM editor AS T1 JOIN journal_committee AS T2 ON T1.Editor_ID = T2.Editor_ID GROUP BY T1.Name HAVING COUNT(*) >= 2
List the names of editors that are not on any journal committee.
CREATE TABLE editor (Name VARCHAR, editor_id VARCHAR); CREATE TABLE journal_committee (Name VARCHAR, editor_id VARCHAR)
SELECT Name FROM editor WHERE NOT editor_id IN (SELECT editor_id FROM journal_committee)
### Context: CREATE TABLE editor (Name VARCHAR, editor_id VARCHAR); CREATE TABLE journal_committee (Name VARCHAR, editor_id VARCHAR) ### Question: List the names of editors that are not on any journal committee. ### Answer: SELECT Name FROM editor WHERE NOT editor_id IN (SELECT editor_id FROM journal_committee)
List the date, theme and sales of the journal which did not have any of the listed editors serving on committee.
CREATE TABLE journal_committee (journal_ID VARCHAR); CREATE TABLE journal (date VARCHAR, theme VARCHAR, sales VARCHAR); CREATE TABLE journal (date VARCHAR, theme VARCHAR, sales VARCHAR, journal_ID VARCHAR)
SELECT date, theme, sales FROM journal EXCEPT SELECT T1.date, T1.theme, T1.sales FROM journal AS T1 JOIN journal_committee AS T2 ON T1.journal_ID = T2.journal_ID
### Context: CREATE TABLE journal_committee (journal_ID VARCHAR); CREATE TABLE journal (date VARCHAR, theme VARCHAR, sales VARCHAR); CREATE TABLE journal (date VARCHAR, theme VARCHAR, sales VARCHAR, journal_ID VARCHAR) ### Question: List the date, theme and sales of the journal which did not have any of the listed editors serving on committee. ### Answer: SELECT date, theme, sales FROM journal EXCEPT SELECT T1.date, T1.theme, T1.sales FROM journal AS T1 JOIN journal_committee AS T2 ON T1.journal_ID = T2.journal_ID
What is the average sales of the journals that have an editor whose work type is 'Photo'?
CREATE TABLE journal_committee (journal_ID VARCHAR, work_type VARCHAR); CREATE TABLE journal (sales INTEGER, journal_ID VARCHAR)
SELECT AVG(T1.sales) FROM journal AS T1 JOIN journal_committee AS T2 ON T1.journal_ID = T2.journal_ID WHERE T2.work_type = 'Photo'
### Context: CREATE TABLE journal_committee (journal_ID VARCHAR, work_type VARCHAR); CREATE TABLE journal (sales INTEGER, journal_ID VARCHAR) ### Question: What is the average sales of the journals that have an editor whose work type is 'Photo'? ### Answer: SELECT AVG(T1.sales) FROM journal AS T1 JOIN journal_committee AS T2 ON T1.journal_ID = T2.journal_ID WHERE T2.work_type = 'Photo'
How many accounts do we have?
CREATE TABLE Accounts (Id VARCHAR)
SELECT COUNT(*) FROM Accounts
### Context: CREATE TABLE Accounts (Id VARCHAR) ### Question: How many accounts do we have? ### Answer: SELECT COUNT(*) FROM Accounts
Show ids, customer ids, names for all accounts.
CREATE TABLE Accounts (account_id VARCHAR, customer_id VARCHAR, account_name VARCHAR)
SELECT account_id, customer_id, account_name FROM Accounts
### Context: CREATE TABLE Accounts (account_id VARCHAR, customer_id VARCHAR, account_name VARCHAR) ### Question: Show ids, customer ids, names for all accounts. ### Answer: SELECT account_id, customer_id, account_name FROM Accounts
Show other account details for account with name 338.
CREATE TABLE Accounts (other_account_details VARCHAR, account_name VARCHAR)
SELECT other_account_details FROM Accounts WHERE account_name = "338"
### Context: CREATE TABLE Accounts (other_account_details VARCHAR, account_name VARCHAR) ### Question: Show other account details for account with name 338. ### Answer: SELECT other_account_details FROM Accounts WHERE account_name = "338"
What is the first name, last name, and phone of the customer with account name 162?
CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR); CREATE TABLE Accounts (customer_id VARCHAR, account_name VARCHAR)
SELECT T2.customer_first_name, T2.customer_last_name, T2.customer_phone FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.account_name = "162"
### Context: CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR); CREATE TABLE Accounts (customer_id VARCHAR, account_name VARCHAR) ### Question: What is the first name, last name, and phone of the customer with account name 162? ### Answer: SELECT T2.customer_first_name, T2.customer_last_name, T2.customer_phone FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.account_name = "162"
How many accounts does the customer with first name Art and last name Turcotte have?
CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR)
SELECT COUNT(*) FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = "Art" AND T2.customer_last_name = "Turcotte"
### Context: CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR) ### Question: How many accounts does the customer with first name Art and last name Turcotte have? ### Answer: SELECT COUNT(*) FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = "Art" AND T2.customer_last_name = "Turcotte"
Show all customer ids and the number of accounts for each customer.
CREATE TABLE Accounts (customer_id VARCHAR)
SELECT customer_id, COUNT(*) FROM Accounts GROUP BY customer_id
### Context: CREATE TABLE Accounts (customer_id VARCHAR) ### Question: Show all customer ids and the number of accounts for each customer. ### Answer: SELECT customer_id, COUNT(*) FROM Accounts GROUP BY customer_id
Show the customer id and number of accounts with most accounts.
CREATE TABLE Accounts (customer_id VARCHAR)
SELECT customer_id, COUNT(*) FROM Accounts GROUP BY customer_id ORDER BY COUNT(*) DESC LIMIT 1
### Context: CREATE TABLE Accounts (customer_id VARCHAR) ### Question: Show the customer id and number of accounts with most accounts. ### Answer: SELECT customer_id, COUNT(*) FROM Accounts GROUP BY customer_id ORDER BY COUNT(*) DESC LIMIT 1
What is the customer first, last name and id with least number of accounts.
CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR)
SELECT T2.customer_first_name, T2.customer_last_name, T1.customer_id FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) LIMIT 1
### Context: CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR) ### Question: What is the customer first, last name and id with least number of accounts. ### Answer: SELECT T2.customer_first_name, T2.customer_last_name, T1.customer_id FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) LIMIT 1
Show the number of all customers without an account.
CREATE TABLE Customers (customer_id VARCHAR); CREATE TABLE Accounts (customer_id VARCHAR)
SELECT COUNT(*) FROM Customers WHERE NOT customer_id IN (SELECT customer_id FROM Accounts)
### Context: CREATE TABLE Customers (customer_id VARCHAR); CREATE TABLE Accounts (customer_id VARCHAR) ### Question: Show the number of all customers without an account. ### Answer: SELECT COUNT(*) FROM Customers WHERE NOT customer_id IN (SELECT customer_id FROM Accounts)
Show the first names and last names of customers without any account.
CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR)
SELECT customer_first_name, customer_last_name FROM Customers EXCEPT SELECT T1.customer_first_name, T1.customer_last_name FROM Customers AS T1 JOIN Accounts AS T2 ON T1.customer_id = T2.customer_id
### Context: CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR) ### Question: Show the first names and last names of customers without any account. ### Answer: SELECT customer_first_name, customer_last_name FROM Customers EXCEPT SELECT T1.customer_first_name, T1.customer_last_name FROM Customers AS T1 JOIN Accounts AS T2 ON T1.customer_id = T2.customer_id
Show distinct first and last names for all customers with an account.
CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR)
SELECT DISTINCT T1.customer_first_name, T1.customer_last_name FROM Customers AS T1 JOIN Accounts AS T2 ON T1.customer_id = T2.customer_id
### Context: CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR) ### Question: Show distinct first and last names for all customers with an account. ### Answer: SELECT DISTINCT T1.customer_first_name, T1.customer_last_name FROM Customers AS T1 JOIN Accounts AS T2 ON T1.customer_id = T2.customer_id
How many customers have an account?
CREATE TABLE Accounts (customer_id VARCHAR)
SELECT COUNT(DISTINCT customer_id) FROM Accounts
### Context: CREATE TABLE Accounts (customer_id VARCHAR) ### Question: How many customers have an account? ### Answer: SELECT COUNT(DISTINCT customer_id) FROM Accounts
How many customers do we have?
CREATE TABLE Customers (Id VARCHAR)
SELECT COUNT(*) FROM Customers
### Context: CREATE TABLE Customers (Id VARCHAR) ### Question: How many customers do we have? ### Answer: SELECT COUNT(*) FROM Customers
Show ids, first names, last names, and phones for all customers.
CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR, customer_phone VARCHAR)
SELECT customer_id, customer_first_name, customer_last_name, customer_phone FROM Customers
### Context: CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR, customer_phone VARCHAR) ### Question: Show ids, first names, last names, and phones for all customers. ### Answer: SELECT customer_id, customer_first_name, customer_last_name, customer_phone FROM Customers
What is the phone and email for customer with first name Aniyah and last name Feest?
CREATE TABLE Customers (customer_phone VARCHAR, customer_email VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR)
SELECT customer_phone, customer_email FROM Customers WHERE customer_first_name = "Aniyah" AND customer_last_name = "Feest"
### Context: CREATE TABLE Customers (customer_phone VARCHAR, customer_email VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR) ### Question: What is the phone and email for customer with first name Aniyah and last name Feest? ### Answer: SELECT customer_phone, customer_email FROM Customers WHERE customer_first_name = "Aniyah" AND customer_last_name = "Feest"
Show the number of customer cards.
CREATE TABLE Customers_cards (Id VARCHAR)
SELECT COUNT(*) FROM Customers_cards
### Context: CREATE TABLE Customers_cards (Id VARCHAR) ### Question: Show the number of customer cards. ### Answer: SELECT COUNT(*) FROM Customers_cards
Show ids, customer ids, card type codes, card numbers for all cards.
CREATE TABLE Customers_cards (card_id VARCHAR, customer_id VARCHAR, card_type_code VARCHAR, card_number VARCHAR)
SELECT card_id, customer_id, card_type_code, card_number FROM Customers_cards
### Context: CREATE TABLE Customers_cards (card_id VARCHAR, customer_id VARCHAR, card_type_code VARCHAR, card_number VARCHAR) ### Question: Show ids, customer ids, card type codes, card numbers for all cards. ### Answer: SELECT card_id, customer_id, card_type_code, card_number FROM Customers_cards
Show the date valid from and the date valid to for the card with card number '4560596484842'.
CREATE TABLE Customers_cards (date_valid_from VARCHAR, date_valid_to VARCHAR, card_number VARCHAR)
SELECT date_valid_from, date_valid_to FROM Customers_cards WHERE card_number = "4560596484842"
### Context: CREATE TABLE Customers_cards (date_valid_from VARCHAR, date_valid_to VARCHAR, card_number VARCHAR) ### Question: Show the date valid from and the date valid to for the card with card number '4560596484842'. ### Answer: SELECT date_valid_from, date_valid_to FROM Customers_cards WHERE card_number = "4560596484842"
What is the first name, last name, and phone of the customer with card 4560596484842.
CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR); CREATE TABLE Customers_cards (customer_id VARCHAR, card_number VARCHAR)
SELECT T2.customer_first_name, T2.customer_last_name, T2.customer_phone FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.card_number = "4560596484842"
### Context: CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR); CREATE TABLE Customers_cards (customer_id VARCHAR, card_number VARCHAR) ### Question: What is the first name, last name, and phone of the customer with card 4560596484842. ### Answer: SELECT T2.customer_first_name, T2.customer_last_name, T2.customer_phone FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.card_number = "4560596484842"
How many cards does customer Art Turcotte have?
CREATE TABLE Customers_cards (customer_id VARCHAR); CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR)
SELECT COUNT(*) FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = "Art" AND T2.customer_last_name = "Turcotte"
### Context: CREATE TABLE Customers_cards (customer_id VARCHAR); CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR) ### Question: How many cards does customer Art Turcotte have? ### Answer: SELECT COUNT(*) FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = "Art" AND T2.customer_last_name = "Turcotte"
How many debit cards do we have?
CREATE TABLE Customers_cards (card_type_code VARCHAR)
SELECT COUNT(*) FROM Customers_cards WHERE card_type_code = "Debit"
### Context: CREATE TABLE Customers_cards (card_type_code VARCHAR) ### Question: How many debit cards do we have? ### Answer: SELECT COUNT(*) FROM Customers_cards WHERE card_type_code = "Debit"
How many credit cards does customer Blanche Huels have?
CREATE TABLE Customers_cards (customer_id VARCHAR, card_type_code VARCHAR); CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR)
SELECT COUNT(*) FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = "Blanche" AND T2.customer_last_name = "Huels" AND T1.card_type_code = "Credit"
### Context: CREATE TABLE Customers_cards (customer_id VARCHAR, card_type_code VARCHAR); CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR) ### Question: How many credit cards does customer Blanche Huels have? ### Answer: SELECT COUNT(*) FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = "Blanche" AND T2.customer_last_name = "Huels" AND T1.card_type_code = "Credit"
Show all customer ids and the number of cards owned by each customer.
CREATE TABLE Customers_cards (customer_id VARCHAR)
SELECT customer_id, COUNT(*) FROM Customers_cards GROUP BY customer_id
### Context: CREATE TABLE Customers_cards (customer_id VARCHAR) ### Question: Show all customer ids and the number of cards owned by each customer. ### Answer: SELECT customer_id, COUNT(*) FROM Customers_cards GROUP BY customer_id
What is the customer id with most number of cards, and how many does he have?
CREATE TABLE Customers_cards (customer_id VARCHAR)
SELECT customer_id, COUNT(*) FROM Customers_cards GROUP BY customer_id ORDER BY COUNT(*) DESC LIMIT 1
### Context: CREATE TABLE Customers_cards (customer_id VARCHAR) ### Question: What is the customer id with most number of cards, and how many does he have? ### Answer: SELECT customer_id, COUNT(*) FROM Customers_cards GROUP BY customer_id ORDER BY COUNT(*) DESC LIMIT 1
Show id, first and last names for all customers with at least two cards.
CREATE TABLE Customers_cards (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR)
SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id HAVING COUNT(*) >= 2
### Context: CREATE TABLE Customers_cards (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR) ### Question: Show id, first and last names for all customers with at least two cards. ### Answer: SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id HAVING COUNT(*) >= 2
What is the customer id, first and last name with least number of accounts.
CREATE TABLE Customers_cards (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR)
SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) LIMIT 1
### Context: CREATE TABLE Customers_cards (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR) ### Question: What is the customer id, first and last name with least number of accounts. ### Answer: SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) LIMIT 1
Show all card type codes and the number of cards in each type.
CREATE TABLE Customers_cards (card_type_code VARCHAR)
SELECT card_type_code, COUNT(*) FROM Customers_cards GROUP BY card_type_code
### Context: CREATE TABLE Customers_cards (card_type_code VARCHAR) ### Question: Show all card type codes and the number of cards in each type. ### Answer: SELECT card_type_code, COUNT(*) FROM Customers_cards GROUP BY card_type_code