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Find the first name of the band mate that has performed in most songs. | CREATE TABLE Songs (SongId VARCHAR); CREATE TABLE Band (firstname VARCHAR, id VARCHAR); CREATE TABLE Performance (bandmate VARCHAR) | SELECT t2.firstname FROM Performance AS t1 JOIN Band AS t2 ON t1.bandmate = t2.id JOIN Songs AS T3 ON T3.SongId = T1.SongId GROUP BY firstname ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE Songs (SongId VARCHAR); CREATE TABLE Band (firstname VARCHAR, id VARCHAR); CREATE TABLE Performance (bandmate VARCHAR) ### Question: Find the first name of the band mate that has performed in most songs. ### Answer: SELECT t2.firstname FROM Performance AS t1 JOIN Band AS t2 ON t1.bandmate = t2.id JOIN Songs AS T3 ON T3.SongId = T1.SongId GROUP BY firstname ORDER BY COUNT(*) DESC LIMIT 1 |
Which vocal type has the band mate with first name "Marianne" played the most? | CREATE TABLE vocals (bandmate VARCHAR); CREATE TABLE band (id VARCHAR) | SELECT TYPE FROM vocals AS T1 JOIN band AS T2 ON T1.bandmate = T2.id WHERE firstname = "Marianne" GROUP BY TYPE ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE vocals (bandmate VARCHAR); CREATE TABLE band (id VARCHAR) ### Question: Which vocal type has the band mate with first name "Marianne" played the most? ### Answer: SELECT TYPE FROM vocals AS T1 JOIN band AS T2 ON T1.bandmate = T2.id WHERE firstname = "Marianne" GROUP BY TYPE ORDER BY COUNT(*) DESC LIMIT 1 |
Who is performing in the back stage position for the song "Der Kapitan"? Show the first name and last name. | CREATE TABLE Band (firstname VARCHAR, lastname VARCHAR, id VARCHAR); CREATE TABLE Performance (bandmate VARCHAR, SongId VARCHAR, StagePosition VARCHAR); CREATE TABLE Songs (SongId VARCHAR, Title VARCHAR) | SELECT T2.firstname, T2.lastname FROM Performance AS T1 JOIN Band AS T2 ON T1.bandmate = T2.id JOIN Songs AS T3 ON T3.SongId = T1.SongId WHERE T3.Title = "Der Kapitan" AND T1.StagePosition = "back" | ### Context: CREATE TABLE Band (firstname VARCHAR, lastname VARCHAR, id VARCHAR); CREATE TABLE Performance (bandmate VARCHAR, SongId VARCHAR, StagePosition VARCHAR); CREATE TABLE Songs (SongId VARCHAR, Title VARCHAR) ### Question: Who is performing in the back stage position for the song "Der Kapitan"? Show the first name and last name. ### Answer: SELECT T2.firstname, T2.lastname FROM Performance AS T1 JOIN Band AS T2 ON T1.bandmate = T2.id JOIN Songs AS T3 ON T3.SongId = T1.SongId WHERE T3.Title = "Der Kapitan" AND T1.StagePosition = "back" |
What are the songs in album "A Kiss Before You Go: Live in Hamburg"? | CREATE TABLE songs (title VARCHAR, songid VARCHAR); CREATE TABLE albums (aid VARCHAR, title VARCHAR); CREATE TABLE tracklists (albumid VARCHAR, songid VARCHAR) | SELECT T3.title FROM albums AS T1 JOIN tracklists AS T2 ON T1.aid = T2.albumid JOIN songs AS T3 ON T2.songid = T3.songid WHERE T1.title = "A Kiss Before You Go: Live in Hamburg" | ### Context: CREATE TABLE songs (title VARCHAR, songid VARCHAR); CREATE TABLE albums (aid VARCHAR, title VARCHAR); CREATE TABLE tracklists (albumid VARCHAR, songid VARCHAR) ### Question: What are the songs in album "A Kiss Before You Go: Live in Hamburg"? ### Answer: SELECT T3.title FROM albums AS T1 JOIN tracklists AS T2 ON T1.aid = T2.albumid JOIN songs AS T3 ON T2.songid = T3.songid WHERE T1.title = "A Kiss Before You Go: Live in Hamburg" |
What are all the songs in albums under label "Universal Music Group"? | CREATE TABLE songs (title VARCHAR, songid VARCHAR); CREATE TABLE albums (aid VARCHAR); CREATE TABLE tracklists (albumid VARCHAR, songid VARCHAR) | SELECT T3.title FROM albums AS T1 JOIN tracklists AS T2 ON T1.aid = T2.albumid JOIN songs AS T3 ON T2.songid = T3.songid WHERE t1.label = "Universal Music Group" | ### Context: CREATE TABLE songs (title VARCHAR, songid VARCHAR); CREATE TABLE albums (aid VARCHAR); CREATE TABLE tracklists (albumid VARCHAR, songid VARCHAR) ### Question: What are all the songs in albums under label "Universal Music Group"? ### Answer: SELECT T3.title FROM albums AS T1 JOIN tracklists AS T2 ON T1.aid = T2.albumid JOIN songs AS T3 ON T2.songid = T3.songid WHERE t1.label = "Universal Music Group" |
Find the number of songs in all the studio albums. | CREATE TABLE songs (title VARCHAR, songid VARCHAR); CREATE TABLE albums (aid VARCHAR); CREATE TABLE tracklists (albumid VARCHAR, songid VARCHAR) | SELECT COUNT(DISTINCT T3.title) FROM albums AS T1 JOIN tracklists AS T2 ON T1.aid = T2.albumid JOIN songs AS T3 ON T2.songid = T3.songid WHERE t1.type = "Studio" | ### Context: CREATE TABLE songs (title VARCHAR, songid VARCHAR); CREATE TABLE albums (aid VARCHAR); CREATE TABLE tracklists (albumid VARCHAR, songid VARCHAR) ### Question: Find the number of songs in all the studio albums. ### Answer: SELECT COUNT(DISTINCT T3.title) FROM albums AS T1 JOIN tracklists AS T2 ON T1.aid = T2.albumid JOIN songs AS T3 ON T2.songid = T3.songid WHERE t1.type = "Studio" |
Who is the founder of Sony? | CREATE TABLE manufacturers (founder VARCHAR, name VARCHAR) | SELECT founder FROM manufacturers WHERE name = 'Sony' | ### Context: CREATE TABLE manufacturers (founder VARCHAR, name VARCHAR) ### Question: Who is the founder of Sony? ### Answer: SELECT founder FROM manufacturers WHERE name = 'Sony' |
Where is the headquarter of the company founded by James? | CREATE TABLE manufacturers (headquarter VARCHAR, founder VARCHAR) | SELECT headquarter FROM manufacturers WHERE founder = 'James' | ### Context: CREATE TABLE manufacturers (headquarter VARCHAR, founder VARCHAR) ### Question: Where is the headquarter of the company founded by James? ### Answer: SELECT headquarter FROM manufacturers WHERE founder = 'James' |
Find all manufacturers' names and their headquarters, sorted by the ones with highest revenue first. | CREATE TABLE manufacturers (name VARCHAR, headquarter VARCHAR, revenue VARCHAR) | SELECT name, headquarter FROM manufacturers ORDER BY revenue DESC | ### Context: CREATE TABLE manufacturers (name VARCHAR, headquarter VARCHAR, revenue VARCHAR) ### Question: Find all manufacturers' names and their headquarters, sorted by the ones with highest revenue first. ### Answer: SELECT name, headquarter FROM manufacturers ORDER BY revenue DESC |
What are the average, maximum and total revenues of all companies? | CREATE TABLE manufacturers (revenue INTEGER) | SELECT AVG(revenue), MAX(revenue), SUM(revenue) FROM manufacturers | ### Context: CREATE TABLE manufacturers (revenue INTEGER) ### Question: What are the average, maximum and total revenues of all companies? ### Answer: SELECT AVG(revenue), MAX(revenue), SUM(revenue) FROM manufacturers |
How many companies were created by Andy? | CREATE TABLE manufacturers (founder VARCHAR) | SELECT COUNT(*) FROM manufacturers WHERE founder = 'Andy' | ### Context: CREATE TABLE manufacturers (founder VARCHAR) ### Question: How many companies were created by Andy? ### Answer: SELECT COUNT(*) FROM manufacturers WHERE founder = 'Andy' |
Find the total revenue created by the companies whose headquarter is located at Austin. | CREATE TABLE manufacturers (revenue INTEGER, headquarter VARCHAR) | SELECT SUM(revenue) FROM manufacturers WHERE headquarter = 'Austin' | ### Context: CREATE TABLE manufacturers (revenue INTEGER, headquarter VARCHAR) ### Question: Find the total revenue created by the companies whose headquarter is located at Austin. ### Answer: SELECT SUM(revenue) FROM manufacturers WHERE headquarter = 'Austin' |
What are the different cities listed? | CREATE TABLE manufacturers (headquarter VARCHAR) | SELECT DISTINCT headquarter FROM manufacturers | ### Context: CREATE TABLE manufacturers (headquarter VARCHAR) ### Question: What are the different cities listed? ### Answer: SELECT DISTINCT headquarter FROM manufacturers |
Find the number of manufactures that are based in Tokyo or Beijing. | CREATE TABLE manufacturers (headquarter VARCHAR) | SELECT COUNT(*) FROM manufacturers WHERE headquarter = 'Tokyo' OR headquarter = 'Beijing' | ### Context: CREATE TABLE manufacturers (headquarter VARCHAR) ### Question: Find the number of manufactures that are based in Tokyo or Beijing. ### Answer: SELECT COUNT(*) FROM manufacturers WHERE headquarter = 'Tokyo' OR headquarter = 'Beijing' |
Find the founder of the company whose name begins with the letter 'S'. | CREATE TABLE manufacturers (founder VARCHAR, name VARCHAR) | SELECT founder FROM manufacturers WHERE name LIKE 'S%' | ### Context: CREATE TABLE manufacturers (founder VARCHAR, name VARCHAR) ### Question: Find the founder of the company whose name begins with the letter 'S'. ### Answer: SELECT founder FROM manufacturers WHERE name LIKE 'S%' |
Find the name of companies whose revenue is between 100 and 150. | CREATE TABLE manufacturers (name VARCHAR, revenue INTEGER) | SELECT name FROM manufacturers WHERE revenue BETWEEN 100 AND 150 | ### Context: CREATE TABLE manufacturers (name VARCHAR, revenue INTEGER) ### Question: Find the name of companies whose revenue is between 100 and 150. ### Answer: SELECT name FROM manufacturers WHERE revenue BETWEEN 100 AND 150 |
What is the total revenue of all companies whose main office is at Tokyo or Taiwan? | CREATE TABLE manufacturers (revenue INTEGER, Headquarter VARCHAR) | SELECT SUM(revenue) FROM manufacturers WHERE Headquarter = 'Tokyo' OR Headquarter = 'Taiwan' | ### Context: CREATE TABLE manufacturers (revenue INTEGER, Headquarter VARCHAR) ### Question: What is the total revenue of all companies whose main office is at Tokyo or Taiwan? ### Answer: SELECT SUM(revenue) FROM manufacturers WHERE Headquarter = 'Tokyo' OR Headquarter = 'Taiwan' |
Find the name of product that is produced by both companies Creative Labs and Sony. | CREATE TABLE manufacturers (code VARCHAR, name VARCHAR); CREATE TABLE products (name VARCHAR, Manufacturer VARCHAR) | SELECT T1.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T2.name = 'Creative Labs' INTERSECT SELECT T1.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T2.name = 'Sony' | ### Context: CREATE TABLE manufacturers (code VARCHAR, name VARCHAR); CREATE TABLE products (name VARCHAR, Manufacturer VARCHAR) ### Question: Find the name of product that is produced by both companies Creative Labs and Sony. ### Answer: SELECT T1.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T2.name = 'Creative Labs' INTERSECT SELECT T1.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T2.name = 'Sony' |
Find the name, headquarter and founder of the manufacturer that has the highest revenue. | CREATE TABLE manufacturers (name VARCHAR, headquarter VARCHAR, founder VARCHAR, revenue VARCHAR) | SELECT name, headquarter, founder FROM manufacturers ORDER BY revenue DESC LIMIT 1 | ### Context: CREATE TABLE manufacturers (name VARCHAR, headquarter VARCHAR, founder VARCHAR, revenue VARCHAR) ### Question: Find the name, headquarter and founder of the manufacturer that has the highest revenue. ### Answer: SELECT name, headquarter, founder FROM manufacturers ORDER BY revenue DESC LIMIT 1 |
Find the name, headquarter and revenue of all manufacturers sorted by their revenue in the descending order. | CREATE TABLE manufacturers (name VARCHAR, headquarter VARCHAR, revenue VARCHAR) | SELECT name, headquarter, revenue FROM manufacturers ORDER BY revenue DESC | ### Context: CREATE TABLE manufacturers (name VARCHAR, headquarter VARCHAR, revenue VARCHAR) ### Question: Find the name, headquarter and revenue of all manufacturers sorted by their revenue in the descending order. ### Answer: SELECT name, headquarter, revenue FROM manufacturers ORDER BY revenue DESC |
Find the name of companies whose revenue is greater than the average revenue of all companies. | CREATE TABLE manufacturers (name VARCHAR, revenue INTEGER) | SELECT name FROM manufacturers WHERE revenue > (SELECT AVG(revenue) FROM manufacturers) | ### Context: CREATE TABLE manufacturers (name VARCHAR, revenue INTEGER) ### Question: Find the name of companies whose revenue is greater than the average revenue of all companies. ### Answer: SELECT name FROM manufacturers WHERE revenue > (SELECT AVG(revenue) FROM manufacturers) |
Find the name of companies whose revenue is smaller than the revenue of all companies based in Austin. | CREATE TABLE manufacturers (name VARCHAR, revenue INTEGER, headquarter VARCHAR) | SELECT name FROM manufacturers WHERE revenue < (SELECT MIN(revenue) FROM manufacturers WHERE headquarter = 'Austin') | ### Context: CREATE TABLE manufacturers (name VARCHAR, revenue INTEGER, headquarter VARCHAR) ### Question: Find the name of companies whose revenue is smaller than the revenue of all companies based in Austin. ### Answer: SELECT name FROM manufacturers WHERE revenue < (SELECT MIN(revenue) FROM manufacturers WHERE headquarter = 'Austin') |
Find the total revenue of companies whose revenue is larger than the revenue of some companies based in Austin. | CREATE TABLE manufacturers (revenue INTEGER, headquarter VARCHAR) | SELECT SUM(revenue) FROM manufacturers WHERE revenue > (SELECT MIN(revenue) FROM manufacturers WHERE headquarter = 'Austin') | ### Context: CREATE TABLE manufacturers (revenue INTEGER, headquarter VARCHAR) ### Question: Find the total revenue of companies whose revenue is larger than the revenue of some companies based in Austin. ### Answer: SELECT SUM(revenue) FROM manufacturers WHERE revenue > (SELECT MIN(revenue) FROM manufacturers WHERE headquarter = 'Austin') |
Find the total revenue of companies of each founder. | CREATE TABLE manufacturers (founder VARCHAR, revenue INTEGER) | SELECT SUM(revenue), founder FROM manufacturers GROUP BY founder | ### Context: CREATE TABLE manufacturers (founder VARCHAR, revenue INTEGER) ### Question: Find the total revenue of companies of each founder. ### Answer: SELECT SUM(revenue), founder FROM manufacturers GROUP BY founder |
Find the name and revenue of the company that earns the highest revenue in each city. | CREATE TABLE manufacturers (name VARCHAR, Headquarter VARCHAR, revenue INTEGER) | SELECT name, MAX(revenue), Headquarter FROM manufacturers GROUP BY Headquarter | ### Context: CREATE TABLE manufacturers (name VARCHAR, Headquarter VARCHAR, revenue INTEGER) ### Question: Find the name and revenue of the company that earns the highest revenue in each city. ### Answer: SELECT name, MAX(revenue), Headquarter FROM manufacturers GROUP BY Headquarter |
Find the total revenue for each manufacturer. | CREATE TABLE manufacturers (name VARCHAR, revenue INTEGER) | SELECT SUM(revenue), name FROM manufacturers GROUP BY name | ### Context: CREATE TABLE manufacturers (name VARCHAR, revenue INTEGER) ### Question: Find the total revenue for each manufacturer. ### Answer: SELECT SUM(revenue), name FROM manufacturers GROUP BY name |
Find the average prices of all products from each manufacture, and list each company's name. | CREATE TABLE products (price INTEGER, Manufacturer VARCHAR); CREATE TABLE manufacturers (name VARCHAR, code VARCHAR) | SELECT AVG(T1.price), T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.name | ### Context: CREATE TABLE products (price INTEGER, Manufacturer VARCHAR); CREATE TABLE manufacturers (name VARCHAR, code VARCHAR) ### Question: Find the average prices of all products from each manufacture, and list each company's name. ### Answer: SELECT AVG(T1.price), T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.name |
Find the number of different products that are produced by companies at different headquarter cities. | CREATE TABLE manufacturers (Headquarter VARCHAR, code VARCHAR); CREATE TABLE products (name VARCHAR, Manufacturer VARCHAR) | SELECT COUNT(DISTINCT T1.name), T2.Headquarter FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.Headquarter | ### Context: CREATE TABLE manufacturers (Headquarter VARCHAR, code VARCHAR); CREATE TABLE products (name VARCHAR, Manufacturer VARCHAR) ### Question: Find the number of different products that are produced by companies at different headquarter cities. ### Answer: SELECT COUNT(DISTINCT T1.name), T2.Headquarter FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.Headquarter |
Find number of products which Sony does not make. | CREATE TABLE manufacturers (code VARCHAR, name VARCHAR); CREATE TABLE products (name VARCHAR); CREATE TABLE products (name VARCHAR, Manufacturer VARCHAR) | SELECT COUNT(DISTINCT name) FROM products WHERE NOT name IN (SELECT T1.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T2.name = 'Sony') | ### Context: CREATE TABLE manufacturers (code VARCHAR, name VARCHAR); CREATE TABLE products (name VARCHAR); CREATE TABLE products (name VARCHAR, Manufacturer VARCHAR) ### Question: Find number of products which Sony does not make. ### Answer: SELECT COUNT(DISTINCT name) FROM products WHERE NOT name IN (SELECT T1.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T2.name = 'Sony') |
Find the name of companies that do not make DVD drive. | CREATE TABLE manufacturers (name VARCHAR, code VARCHAR); CREATE TABLE products (Manufacturer VARCHAR, name VARCHAR); CREATE TABLE manufacturers (name VARCHAR) | SELECT name FROM manufacturers EXCEPT SELECT T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T1.name = 'DVD drive' | ### Context: CREATE TABLE manufacturers (name VARCHAR, code VARCHAR); CREATE TABLE products (Manufacturer VARCHAR, name VARCHAR); CREATE TABLE manufacturers (name VARCHAR) ### Question: Find the name of companies that do not make DVD drive. ### Answer: SELECT name FROM manufacturers EXCEPT SELECT T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T1.name = 'DVD drive' |
Find the number of products for each manufacturer, showing the name of each company. | CREATE TABLE products (Manufacturer VARCHAR); CREATE TABLE manufacturers (name VARCHAR, code VARCHAR) | SELECT COUNT(*), T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.name | ### Context: CREATE TABLE products (Manufacturer VARCHAR); CREATE TABLE manufacturers (name VARCHAR, code VARCHAR) ### Question: Find the number of products for each manufacturer, showing the name of each company. ### Answer: SELECT COUNT(*), T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.name |
Select the names of all the products in the store. | CREATE TABLE Products (Name VARCHAR) | SELECT Name FROM Products | ### Context: CREATE TABLE Products (Name VARCHAR) ### Question: Select the names of all the products in the store. ### Answer: SELECT Name FROM Products |
Select the names and the prices of all the products in the store. | CREATE TABLE products (name VARCHAR, price VARCHAR) | SELECT name, price FROM products | ### Context: CREATE TABLE products (name VARCHAR, price VARCHAR) ### Question: Select the names and the prices of all the products in the store. ### Answer: SELECT name, price FROM products |
Select the name of the products with a price less than or equal to $200. | CREATE TABLE products (name VARCHAR, price VARCHAR) | SELECT name FROM products WHERE price <= 200 | ### Context: CREATE TABLE products (name VARCHAR, price VARCHAR) ### Question: Select the name of the products with a price less than or equal to $200. ### Answer: SELECT name FROM products WHERE price <= 200 |
Find all information of all the products with a price between $60 and $120. | CREATE TABLE products (price INTEGER) | SELECT * FROM products WHERE price BETWEEN 60 AND 120 | ### Context: CREATE TABLE products (price INTEGER) ### Question: Find all information of all the products with a price between $60 and $120. ### Answer: SELECT * FROM products WHERE price BETWEEN 60 AND 120 |
Compute the average price of all the products. | CREATE TABLE products (price INTEGER) | SELECT AVG(price) FROM products | ### Context: CREATE TABLE products (price INTEGER) ### Question: Compute the average price of all the products. ### Answer: SELECT AVG(price) FROM products |
Compute the average price of all products with manufacturer code equal to 2. | CREATE TABLE products (price INTEGER, Manufacturer VARCHAR) | SELECT AVG(price) FROM products WHERE Manufacturer = 2 | ### Context: CREATE TABLE products (price INTEGER, Manufacturer VARCHAR) ### Question: Compute the average price of all products with manufacturer code equal to 2. ### Answer: SELECT AVG(price) FROM products WHERE Manufacturer = 2 |
Compute the number of products with a price larger than or equal to $180. | CREATE TABLE products (price VARCHAR) | SELECT COUNT(*) FROM products WHERE price >= 180 | ### Context: CREATE TABLE products (price VARCHAR) ### Question: Compute the number of products with a price larger than or equal to $180. ### Answer: SELECT COUNT(*) FROM products WHERE price >= 180 |
Select the name and price of all products with a price larger than or equal to $180, and sort first by price (in descending order), and then by name (in ascending order). | CREATE TABLE products (name VARCHAR, price VARCHAR) | SELECT name, price FROM products WHERE price >= 180 ORDER BY price DESC, name | ### Context: CREATE TABLE products (name VARCHAR, price VARCHAR) ### Question: Select the name and price of all products with a price larger than or equal to $180, and sort first by price (in descending order), and then by name (in ascending order). ### Answer: SELECT name, price FROM products WHERE price >= 180 ORDER BY price DESC, name |
Select all the data from the products and each product's manufacturer. | CREATE TABLE products (manufacturer VARCHAR); CREATE TABLE Manufacturers (code VARCHAR) | SELECT * FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code | ### Context: CREATE TABLE products (manufacturer VARCHAR); CREATE TABLE Manufacturers (code VARCHAR) ### Question: Select all the data from the products and each product's manufacturer. ### Answer: SELECT * FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code |
Select the average price of each manufacturer's products, showing only the manufacturer's code. | CREATE TABLE Products (Manufacturer VARCHAR, Price INTEGER) | SELECT AVG(Price), Manufacturer FROM Products GROUP BY Manufacturer | ### Context: CREATE TABLE Products (Manufacturer VARCHAR, Price INTEGER) ### Question: Select the average price of each manufacturer's products, showing only the manufacturer's code. ### Answer: SELECT AVG(Price), Manufacturer FROM Products GROUP BY Manufacturer |
Select the average price of each manufacturer's products, showing the manufacturer's name. | CREATE TABLE Manufacturers (name VARCHAR, code VARCHAR); CREATE TABLE products (Price INTEGER, manufacturer VARCHAR) | SELECT AVG(T1.Price), T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name | ### Context: CREATE TABLE Manufacturers (name VARCHAR, code VARCHAR); CREATE TABLE products (Price INTEGER, manufacturer VARCHAR) ### Question: Select the average price of each manufacturer's products, showing the manufacturer's name. ### Answer: SELECT AVG(T1.Price), T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name |
Select the names of manufacturer whose products have an average price higher than or equal to $150. | CREATE TABLE Manufacturers (name VARCHAR, code VARCHAR); CREATE TABLE products (Price INTEGER, manufacturer VARCHAR, price INTEGER) | SELECT AVG(T1.Price), T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name HAVING AVG(T1.price) >= 150 | ### Context: CREATE TABLE Manufacturers (name VARCHAR, code VARCHAR); CREATE TABLE products (Price INTEGER, manufacturer VARCHAR, price INTEGER) ### Question: Select the names of manufacturer whose products have an average price higher than or equal to $150. ### Answer: SELECT AVG(T1.Price), T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name HAVING AVG(T1.price) >= 150 |
Select the name and price of the cheapest product. | CREATE TABLE Products (name VARCHAR, price VARCHAR) | SELECT name, price FROM Products ORDER BY price LIMIT 1 | ### Context: CREATE TABLE Products (name VARCHAR, price VARCHAR) ### Question: Select the name and price of the cheapest product. ### Answer: SELECT name, price FROM Products ORDER BY price LIMIT 1 |
Select the name of each manufacturer along with the name and price of its most expensive product. | CREATE TABLE Manufacturers (name VARCHAR, code VARCHAR); CREATE TABLE products (Name VARCHAR, Price INTEGER, manufacturer VARCHAR) | SELECT T1.Name, MAX(T1.Price), T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name | ### Context: CREATE TABLE Manufacturers (name VARCHAR, code VARCHAR); CREATE TABLE products (Name VARCHAR, Price INTEGER, manufacturer VARCHAR) ### Question: Select the name of each manufacturer along with the name and price of its most expensive product. ### Answer: SELECT T1.Name, MAX(T1.Price), T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name |
Select the code of the product that is cheapest in each product category. | CREATE TABLE products (code VARCHAR, name VARCHAR, price INTEGER) | SELECT code, name, MIN(price) FROM products GROUP BY name | ### Context: CREATE TABLE products (code VARCHAR, name VARCHAR, price INTEGER) ### Question: Select the code of the product that is cheapest in each product category. ### Answer: SELECT code, name, MIN(price) FROM products GROUP BY name |
What is the id of the problem log that is created most recently? | CREATE TABLE problem_log (problem_log_id VARCHAR, log_entry_date VARCHAR) | SELECT problem_log_id FROM problem_log ORDER BY log_entry_date DESC LIMIT 1 | ### Context: CREATE TABLE problem_log (problem_log_id VARCHAR, log_entry_date VARCHAR) ### Question: What is the id of the problem log that is created most recently? ### Answer: SELECT problem_log_id FROM problem_log ORDER BY log_entry_date DESC LIMIT 1 |
What is the oldest log id and its corresponding problem id? | CREATE TABLE problem_log (problem_log_id VARCHAR, problem_id VARCHAR, log_entry_date VARCHAR) | SELECT problem_log_id, problem_id FROM problem_log ORDER BY log_entry_date LIMIT 1 | ### Context: CREATE TABLE problem_log (problem_log_id VARCHAR, problem_id VARCHAR, log_entry_date VARCHAR) ### Question: What is the oldest log id and its corresponding problem id? ### Answer: SELECT problem_log_id, problem_id FROM problem_log ORDER BY log_entry_date LIMIT 1 |
Find all the ids and dates of the logs for the problem whose id is 10. | CREATE TABLE problem_log (problem_log_id VARCHAR, log_entry_date VARCHAR, problem_id VARCHAR) | SELECT problem_log_id, log_entry_date FROM problem_log WHERE problem_id = 10 | ### Context: CREATE TABLE problem_log (problem_log_id VARCHAR, log_entry_date VARCHAR, problem_id VARCHAR) ### Question: Find all the ids and dates of the logs for the problem whose id is 10. ### Answer: SELECT problem_log_id, log_entry_date FROM problem_log WHERE problem_id = 10 |
List all the log ids and their descriptions from the problem logs. | CREATE TABLE problem_log (problem_log_id VARCHAR, log_entry_description VARCHAR) | SELECT problem_log_id, log_entry_description FROM problem_log | ### Context: CREATE TABLE problem_log (problem_log_id VARCHAR, log_entry_description VARCHAR) ### Question: List all the log ids and their descriptions from the problem logs. ### Answer: SELECT problem_log_id, log_entry_description FROM problem_log |
List the first and last names of all distinct staff members who are assigned to the problem whose id is 1. | CREATE TABLE problem_log (assigned_to_staff_id VARCHAR, problem_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR) | SELECT DISTINCT staff_first_name, staff_last_name FROM staff AS T1 JOIN problem_log AS T2 ON T1.staff_id = T2.assigned_to_staff_id WHERE T2.problem_id = 1 | ### Context: CREATE TABLE problem_log (assigned_to_staff_id VARCHAR, problem_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR) ### Question: List the first and last names of all distinct staff members who are assigned to the problem whose id is 1. ### Answer: SELECT DISTINCT staff_first_name, staff_last_name FROM staff AS T1 JOIN problem_log AS T2 ON T1.staff_id = T2.assigned_to_staff_id WHERE T2.problem_id = 1 |
List the problem id and log id which are assigned to the staff named Rylan Homenick. | CREATE TABLE problem_log (problem_id VARCHAR, problem_log_id VARCHAR, assigned_to_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR) | SELECT DISTINCT T2.problem_id, T2.problem_log_id FROM staff AS T1 JOIN problem_log AS T2 ON T1.staff_id = T2.assigned_to_staff_id WHERE T1.staff_first_name = "Rylan" AND T1.staff_last_name = "Homenick" | ### Context: CREATE TABLE problem_log (problem_id VARCHAR, problem_log_id VARCHAR, assigned_to_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR) ### Question: List the problem id and log id which are assigned to the staff named Rylan Homenick. ### Answer: SELECT DISTINCT T2.problem_id, T2.problem_log_id FROM staff AS T1 JOIN problem_log AS T2 ON T1.staff_id = T2.assigned_to_staff_id WHERE T1.staff_first_name = "Rylan" AND T1.staff_last_name = "Homenick" |
How many problems are there for product voluptatem? | CREATE TABLE problems (product_id VARCHAR); CREATE TABLE product (product_id VARCHAR, product_name VARCHAR) | SELECT COUNT(*) FROM product AS T1 JOIN problems AS T2 ON T1.product_id = T2.product_id WHERE T1.product_name = "voluptatem" | ### Context: CREATE TABLE problems (product_id VARCHAR); CREATE TABLE product (product_id VARCHAR, product_name VARCHAR) ### Question: How many problems are there for product voluptatem? ### Answer: SELECT COUNT(*) FROM product AS T1 JOIN problems AS T2 ON T1.product_id = T2.product_id WHERE T1.product_name = "voluptatem" |
How many problems does the product with the most problems have? List the number of the problems and product name. | CREATE TABLE problems (product_id VARCHAR); CREATE TABLE product (product_name VARCHAR, product_id VARCHAR) | SELECT COUNT(*), T1.product_name FROM product AS T1 JOIN problems AS T2 ON T1.product_id = T2.product_id GROUP BY T1.product_name ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE problems (product_id VARCHAR); CREATE TABLE product (product_name VARCHAR, product_id VARCHAR) ### Question: How many problems does the product with the most problems have? List the number of the problems and product name. ### Answer: SELECT COUNT(*), T1.product_name FROM product AS T1 JOIN problems AS T2 ON T1.product_id = T2.product_id GROUP BY T1.product_name ORDER BY COUNT(*) DESC LIMIT 1 |
Give me a list of descriptions of the problems that are reported by the staff whose first name is Christop. | CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR); CREATE TABLE problems (problem_description VARCHAR, reported_by_staff_id VARCHAR) | SELECT T1.problem_description FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Christop" | ### Context: CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR); CREATE TABLE problems (problem_description VARCHAR, reported_by_staff_id VARCHAR) ### Question: Give me a list of descriptions of the problems that are reported by the staff whose first name is Christop. ### Answer: SELECT T1.problem_description FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Christop" |
Find the ids of the problems that are reported by the staff whose last name is Bosco. | CREATE TABLE problems (problem_id VARCHAR, reported_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_last_name VARCHAR) | SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_last_name = "Bosco" | ### Context: CREATE TABLE problems (problem_id VARCHAR, reported_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_last_name VARCHAR) ### Question: Find the ids of the problems that are reported by the staff whose last name is Bosco. ### Answer: SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_last_name = "Bosco" |
What are the ids of the problems which are reported after 1978-06-26? | CREATE TABLE problems (problem_id VARCHAR, date_problem_reported INTEGER) | SELECT problem_id FROM problems WHERE date_problem_reported > "1978-06-26" | ### Context: CREATE TABLE problems (problem_id VARCHAR, date_problem_reported INTEGER) ### Question: What are the ids of the problems which are reported after 1978-06-26? ### Answer: SELECT problem_id FROM problems WHERE date_problem_reported > "1978-06-26" |
What are the ids of the problems which are reported before 1978-06-26? | CREATE TABLE problems (problem_id VARCHAR, date_problem_reported INTEGER) | SELECT problem_id FROM problems WHERE date_problem_reported < "1978-06-26" | ### Context: CREATE TABLE problems (problem_id VARCHAR, date_problem_reported INTEGER) ### Question: What are the ids of the problems which are reported before 1978-06-26? ### Answer: SELECT problem_id FROM problems WHERE date_problem_reported < "1978-06-26" |
For each product which has problems, what are the number of problems and the product id? | CREATE TABLE problems (product_id VARCHAR); CREATE TABLE product (product_id VARCHAR) | SELECT COUNT(*), T2.product_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id GROUP BY T2.product_id | ### Context: CREATE TABLE problems (product_id VARCHAR); CREATE TABLE product (product_id VARCHAR) ### Question: For each product which has problems, what are the number of problems and the product id? ### Answer: SELECT COUNT(*), T2.product_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id GROUP BY T2.product_id |
For each product that has problems, find the number of problems reported after 1986-11-13 and the product id? | CREATE TABLE product (product_id VARCHAR); CREATE TABLE problems (product_id VARCHAR, date_problem_reported INTEGER) | SELECT COUNT(*), T2.product_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id WHERE T1.date_problem_reported > "1986-11-13" GROUP BY T2.product_id | ### Context: CREATE TABLE product (product_id VARCHAR); CREATE TABLE problems (product_id VARCHAR, date_problem_reported INTEGER) ### Question: For each product that has problems, find the number of problems reported after 1986-11-13 and the product id? ### Answer: SELECT COUNT(*), T2.product_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id WHERE T1.date_problem_reported > "1986-11-13" GROUP BY T2.product_id |
List the names of all the distinct product names in alphabetical order? | CREATE TABLE product (product_name VARCHAR) | SELECT DISTINCT product_name FROM product ORDER BY product_name | ### Context: CREATE TABLE product (product_name VARCHAR) ### Question: List the names of all the distinct product names in alphabetical order? ### Answer: SELECT DISTINCT product_name FROM product ORDER BY product_name |
List all the distinct product names ordered by product id? | CREATE TABLE product (product_name VARCHAR, product_id VARCHAR) | SELECT DISTINCT product_name FROM product ORDER BY product_id | ### Context: CREATE TABLE product (product_name VARCHAR, product_id VARCHAR) ### Question: List all the distinct product names ordered by product id? ### Answer: SELECT DISTINCT product_name FROM product ORDER BY product_id |
What are the id of problems reported by the staff named Dameon Frami or Jolie Weber? | CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR); CREATE TABLE problems (reported_by_staff_id VARCHAR) | SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Dameon" AND T2.staff_last_name = "Frami" UNION SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Jolie" AND T2.staff_last_name = "Weber" | ### Context: CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR); CREATE TABLE problems (reported_by_staff_id VARCHAR) ### Question: What are the id of problems reported by the staff named Dameon Frami or Jolie Weber? ### Answer: SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Dameon" AND T2.staff_last_name = "Frami" UNION SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Jolie" AND T2.staff_last_name = "Weber" |
What are the product ids for the problems reported by Christop Berge with closure authorised by Ashley Medhurst? | CREATE TABLE problems (reported_by_staff_id VARCHAR, closure_authorised_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR) | SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Christop" AND T2.staff_last_name = "Berge" INTERSECT SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.closure_authorised_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Ashley" AND T2.staff_last_name = "Medhurst" | ### Context: CREATE TABLE problems (reported_by_staff_id VARCHAR, closure_authorised_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR) ### Question: What are the product ids for the problems reported by Christop Berge with closure authorised by Ashley Medhurst? ### Answer: SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Christop" AND T2.staff_last_name = "Berge" INTERSECT SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.closure_authorised_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Ashley" AND T2.staff_last_name = "Medhurst" |
What are the ids of the problems reported before the date of any problem reported by Lysanne Turcotte? | CREATE TABLE problems (problem_id VARCHAR, reported_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR); CREATE TABLE problems (reported_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR) | SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE date_problem_reported < (SELECT MIN(date_problem_reported) FROM problems AS T3 JOIN staff AS T4 ON T3.reported_by_staff_id = T4.staff_id WHERE T4.staff_first_name = "Lysanne" AND T4.staff_last_name = "Turcotte") | ### Context: CREATE TABLE problems (problem_id VARCHAR, reported_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR); CREATE TABLE problems (reported_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR) ### Question: What are the ids of the problems reported before the date of any problem reported by Lysanne Turcotte? ### Answer: SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE date_problem_reported < (SELECT MIN(date_problem_reported) FROM problems AS T3 JOIN staff AS T4 ON T3.reported_by_staff_id = T4.staff_id WHERE T4.staff_first_name = "Lysanne" AND T4.staff_last_name = "Turcotte") |
What are the ids of the problems reported after the date of any problems reported by Rylan Homenick? | CREATE TABLE problems (problem_id VARCHAR, reported_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR); CREATE TABLE problems (reported_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR) | SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE date_problem_reported > (SELECT MAX(date_problem_reported) FROM problems AS T3 JOIN staff AS T4 ON T3.reported_by_staff_id = T4.staff_id WHERE T4.staff_first_name = "Rylan" AND T4.staff_last_name = "Homenick") | ### Context: CREATE TABLE problems (problem_id VARCHAR, reported_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR); CREATE TABLE problems (reported_by_staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR) ### Question: What are the ids of the problems reported after the date of any problems reported by Rylan Homenick? ### Answer: SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE date_problem_reported > (SELECT MAX(date_problem_reported) FROM problems AS T3 JOIN staff AS T4 ON T3.reported_by_staff_id = T4.staff_id WHERE T4.staff_first_name = "Rylan" AND T4.staff_last_name = "Homenick") |
Find the top 3 products which have the largest number of problems? | CREATE TABLE problems (product_id VARCHAR); CREATE TABLE product (product_name VARCHAR, product_id VARCHAR) | SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id GROUP BY T2.product_name ORDER BY COUNT(*) DESC LIMIT 3 | ### Context: CREATE TABLE problems (product_id VARCHAR); CREATE TABLE product (product_name VARCHAR, product_id VARCHAR) ### Question: Find the top 3 products which have the largest number of problems? ### Answer: SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id GROUP BY T2.product_name ORDER BY COUNT(*) DESC LIMIT 3 |
List the ids of the problems from the product "voluptatem" that are reported after 1995? | CREATE TABLE problems (problem_id VARCHAR, product_id VARCHAR, date_problem_reported VARCHAR); CREATE TABLE product (product_id VARCHAR, product_name VARCHAR) | SELECT T1.problem_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id WHERE T2.product_name = "voluptatem" AND T1.date_problem_reported > "1995" | ### Context: CREATE TABLE problems (problem_id VARCHAR, product_id VARCHAR, date_problem_reported VARCHAR); CREATE TABLE product (product_id VARCHAR, product_name VARCHAR) ### Question: List the ids of the problems from the product "voluptatem" that are reported after 1995? ### Answer: SELECT T1.problem_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id WHERE T2.product_name = "voluptatem" AND T1.date_problem_reported > "1995" |
Find the first and last name of the staff members who reported problems from the product "rem" but not "aut"? | CREATE TABLE product (product_name VARCHAR, product_id VARCHAR); CREATE TABLE staff (staff_first_name VARCHAR, staff_last_name VARCHAR, staff_id VARCHAR); CREATE TABLE problems (product_id VARCHAR, reported_by_staff_id VARCHAR) | SELECT T3.staff_first_name, T3.staff_last_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T2.product_name = "rem" EXCEPT SELECT T3.staff_first_name, T3.staff_last_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T2.product_name = "aut" | ### Context: CREATE TABLE product (product_name VARCHAR, product_id VARCHAR); CREATE TABLE staff (staff_first_name VARCHAR, staff_last_name VARCHAR, staff_id VARCHAR); CREATE TABLE problems (product_id VARCHAR, reported_by_staff_id VARCHAR) ### Question: Find the first and last name of the staff members who reported problems from the product "rem" but not "aut"? ### Answer: SELECT T3.staff_first_name, T3.staff_last_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T2.product_name = "rem" EXCEPT SELECT T3.staff_first_name, T3.staff_last_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T2.product_name = "aut" |
Find the products which have problems reported by both Lacey Bosco and Kenton Champlin? | CREATE TABLE product (product_name VARCHAR, product_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR); CREATE TABLE problems (product_id VARCHAR, reported_by_staff_id VARCHAR) | SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T3.staff_first_name = "Lacey" AND T3.staff_last_name = "Bosco" INTERSECT SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T3.staff_first_name = "Kenton" AND T3.staff_last_name = "Champlin" | ### Context: CREATE TABLE product (product_name VARCHAR, product_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, staff_first_name VARCHAR, staff_last_name VARCHAR); CREATE TABLE problems (product_id VARCHAR, reported_by_staff_id VARCHAR) ### Question: Find the products which have problems reported by both Lacey Bosco and Kenton Champlin? ### Answer: SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T3.staff_first_name = "Lacey" AND T3.staff_last_name = "Bosco" INTERSECT SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T3.staff_first_name = "Kenton" AND T3.staff_last_name = "Champlin" |
How many branches where have more than average number of memberships are there? | CREATE TABLE branch (membership_amount INTEGER) | SELECT COUNT(*) FROM branch WHERE membership_amount > (SELECT AVG(membership_amount) FROM branch) | ### Context: CREATE TABLE branch (membership_amount INTEGER) ### Question: How many branches where have more than average number of memberships are there? ### Answer: SELECT COUNT(*) FROM branch WHERE membership_amount > (SELECT AVG(membership_amount) FROM branch) |
Show name, address road, and city for all branches sorted by open year. | CREATE TABLE branch (name VARCHAR, address_road VARCHAR, city VARCHAR, open_year VARCHAR) | SELECT name, address_road, city FROM branch ORDER BY open_year | ### Context: CREATE TABLE branch (name VARCHAR, address_road VARCHAR, city VARCHAR, open_year VARCHAR) ### Question: Show name, address road, and city for all branches sorted by open year. ### Answer: SELECT name, address_road, city FROM branch ORDER BY open_year |
What are names for top three branches with most number of membership? | CREATE TABLE branch (name VARCHAR, membership_amount VARCHAR) | SELECT name FROM branch ORDER BY membership_amount DESC LIMIT 3 | ### Context: CREATE TABLE branch (name VARCHAR, membership_amount VARCHAR) ### Question: What are names for top three branches with most number of membership? ### Answer: SELECT name FROM branch ORDER BY membership_amount DESC LIMIT 3 |
Show all distinct city where branches with at least 100 memberships are located. | CREATE TABLE branch (city VARCHAR, membership_amount VARCHAR) | SELECT DISTINCT city FROM branch WHERE membership_amount >= 100 | ### Context: CREATE TABLE branch (city VARCHAR, membership_amount VARCHAR) ### Question: Show all distinct city where branches with at least 100 memberships are located. ### Answer: SELECT DISTINCT city FROM branch WHERE membership_amount >= 100 |
List all open years when at least two shops are opened. | CREATE TABLE branch (open_year VARCHAR) | SELECT open_year FROM branch GROUP BY open_year HAVING COUNT(*) >= 2 | ### Context: CREATE TABLE branch (open_year VARCHAR) ### Question: List all open years when at least two shops are opened. ### Answer: SELECT open_year FROM branch GROUP BY open_year HAVING COUNT(*) >= 2 |
Show minimum and maximum amount of memberships for all branches opened in 2011 or located at city London. | CREATE TABLE branch (membership_amount INTEGER, open_year VARCHAR, city VARCHAR) | SELECT MIN(membership_amount), MAX(membership_amount) FROM branch WHERE open_year = 2011 OR city = 'London' | ### Context: CREATE TABLE branch (membership_amount INTEGER, open_year VARCHAR, city VARCHAR) ### Question: Show minimum and maximum amount of memberships for all branches opened in 2011 or located at city London. ### Answer: SELECT MIN(membership_amount), MAX(membership_amount) FROM branch WHERE open_year = 2011 OR city = 'London' |
Show the city and the number of branches opened before 2010 for each city. | CREATE TABLE branch (city VARCHAR, open_year INTEGER) | SELECT city, COUNT(*) FROM branch WHERE open_year < 2010 GROUP BY city | ### Context: CREATE TABLE branch (city VARCHAR, open_year INTEGER) ### Question: Show the city and the number of branches opened before 2010 for each city. ### Answer: SELECT city, COUNT(*) FROM branch WHERE open_year < 2010 GROUP BY city |
How many different levels do members have? | CREATE TABLE member (LEVEL VARCHAR) | SELECT COUNT(DISTINCT LEVEL) FROM member | ### Context: CREATE TABLE member (LEVEL VARCHAR) ### Question: How many different levels do members have? ### Answer: SELECT COUNT(DISTINCT LEVEL) FROM member |
Show card number, name, and hometown for all members in a descending order of level. | CREATE TABLE member (card_number VARCHAR, name VARCHAR, hometown VARCHAR, LEVEL VARCHAR) | SELECT card_number, name, hometown FROM member ORDER BY LEVEL DESC | ### Context: CREATE TABLE member (card_number VARCHAR, name VARCHAR, hometown VARCHAR, LEVEL VARCHAR) ### Question: Show card number, name, and hometown for all members in a descending order of level. ### Answer: SELECT card_number, name, hometown FROM member ORDER BY LEVEL DESC |
Show the membership level with most number of members. | CREATE TABLE member (LEVEL VARCHAR) | SELECT LEVEL FROM member GROUP BY LEVEL ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE member (LEVEL VARCHAR) ### Question: Show the membership level with most number of members. ### Answer: SELECT LEVEL FROM member GROUP BY LEVEL ORDER BY COUNT(*) DESC LIMIT 1 |
Show all member names and registered branch names sorted by register year. | CREATE TABLE member (name VARCHAR, member_id VARCHAR); CREATE TABLE membership_register_branch (branch_id VARCHAR, member_id VARCHAR, register_year VARCHAR); CREATE TABLE branch (name VARCHAR, branch_id VARCHAR) | SELECT T3.name, T2.name FROM membership_register_branch AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id JOIN member AS T3 ON T1.member_id = T3.member_id ORDER BY T1.register_year | ### Context: CREATE TABLE member (name VARCHAR, member_id VARCHAR); CREATE TABLE membership_register_branch (branch_id VARCHAR, member_id VARCHAR, register_year VARCHAR); CREATE TABLE branch (name VARCHAR, branch_id VARCHAR) ### Question: Show all member names and registered branch names sorted by register year. ### Answer: SELECT T3.name, T2.name FROM membership_register_branch AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id JOIN member AS T3 ON T1.member_id = T3.member_id ORDER BY T1.register_year |
Show all branch names with the number of members in each branch registered after 2015. | CREATE TABLE branch (name VARCHAR, branch_id VARCHAR); CREATE TABLE membership_register_branch (branch_id VARCHAR, register_year INTEGER) | SELECT T2.name, COUNT(*) FROM membership_register_branch AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id WHERE T1.register_year > 2015 GROUP BY T2.branch_id | ### Context: CREATE TABLE branch (name VARCHAR, branch_id VARCHAR); CREATE TABLE membership_register_branch (branch_id VARCHAR, register_year INTEGER) ### Question: Show all branch names with the number of members in each branch registered after 2015. ### Answer: SELECT T2.name, COUNT(*) FROM membership_register_branch AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id WHERE T1.register_year > 2015 GROUP BY T2.branch_id |
Show member names without any registered branch. | CREATE TABLE member (name VARCHAR, member_id VARCHAR); CREATE TABLE membership_register_branch (name VARCHAR, member_id VARCHAR) | SELECT name FROM member WHERE NOT member_id IN (SELECT member_id FROM membership_register_branch) | ### Context: CREATE TABLE member (name VARCHAR, member_id VARCHAR); CREATE TABLE membership_register_branch (name VARCHAR, member_id VARCHAR) ### Question: Show member names without any registered branch. ### Answer: SELECT name FROM member WHERE NOT member_id IN (SELECT member_id FROM membership_register_branch) |
List the branch name and city without any registered members. | CREATE TABLE membership_register_branch (name VARCHAR, city VARCHAR, branch_id VARCHAR); CREATE TABLE branch (name VARCHAR, city VARCHAR, branch_id VARCHAR) | SELECT name, city FROM branch WHERE NOT branch_id IN (SELECT branch_id FROM membership_register_branch) | ### Context: CREATE TABLE membership_register_branch (name VARCHAR, city VARCHAR, branch_id VARCHAR); CREATE TABLE branch (name VARCHAR, city VARCHAR, branch_id VARCHAR) ### Question: List the branch name and city without any registered members. ### Answer: SELECT name, city FROM branch WHERE NOT branch_id IN (SELECT branch_id FROM membership_register_branch) |
What is the name and open year for the branch with most number of memberships registered in 2016? | CREATE TABLE membership_register_branch (branch_id VARCHAR, register_year VARCHAR); CREATE TABLE branch (name VARCHAR, open_year VARCHAR, branch_id VARCHAR) | SELECT T2.name, T2.open_year FROM membership_register_branch AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id WHERE T1.register_year = 2016 GROUP BY T2.branch_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE membership_register_branch (branch_id VARCHAR, register_year VARCHAR); CREATE TABLE branch (name VARCHAR, open_year VARCHAR, branch_id VARCHAR) ### Question: What is the name and open year for the branch with most number of memberships registered in 2016? ### Answer: SELECT T2.name, T2.open_year FROM membership_register_branch AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id WHERE T1.register_year = 2016 GROUP BY T2.branch_id ORDER BY COUNT(*) DESC LIMIT 1 |
Show the member name and hometown who registered a branch in 2016. | CREATE TABLE member (name VARCHAR, hometown VARCHAR, member_id VARCHAR); CREATE TABLE membership_register_branch (member_id VARCHAR, register_year VARCHAR) | SELECT T2.name, T2.hometown FROM membership_register_branch AS T1 JOIN member AS T2 ON T1.member_id = T2.member_id WHERE T1.register_year = 2016 | ### Context: CREATE TABLE member (name VARCHAR, hometown VARCHAR, member_id VARCHAR); CREATE TABLE membership_register_branch (member_id VARCHAR, register_year VARCHAR) ### Question: Show the member name and hometown who registered a branch in 2016. ### Answer: SELECT T2.name, T2.hometown FROM membership_register_branch AS T1 JOIN member AS T2 ON T1.member_id = T2.member_id WHERE T1.register_year = 2016 |
Show all city with a branch opened in 2001 and a branch with more than 100 membership. | CREATE TABLE branch (city VARCHAR, open_year VARCHAR, membership_amount VARCHAR) | SELECT city FROM branch WHERE open_year = 2001 AND membership_amount > 100 | ### Context: CREATE TABLE branch (city VARCHAR, open_year VARCHAR, membership_amount VARCHAR) ### Question: Show all city with a branch opened in 2001 and a branch with more than 100 membership. ### Answer: SELECT city FROM branch WHERE open_year = 2001 AND membership_amount > 100 |
Show all cities without a branch having more than 100 memberships. | CREATE TABLE branch (city VARCHAR, membership_amount INTEGER) | SELECT city FROM branch EXCEPT SELECT city FROM branch WHERE membership_amount > 100 | ### Context: CREATE TABLE branch (city VARCHAR, membership_amount INTEGER) ### Question: Show all cities without a branch having more than 100 memberships. ### Answer: SELECT city FROM branch EXCEPT SELECT city FROM branch WHERE membership_amount > 100 |
What is the sum of total pounds of purchase in year 2018 for all branches in London? | CREATE TABLE purchase (branch_id VARCHAR, year VARCHAR); CREATE TABLE branch (branch_id VARCHAR, city VARCHAR) | SELECT SUM(total_pounds) FROM purchase AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id WHERE T2.city = 'London' AND T1.year = 2018 | ### Context: CREATE TABLE purchase (branch_id VARCHAR, year VARCHAR); CREATE TABLE branch (branch_id VARCHAR, city VARCHAR) ### Question: What is the sum of total pounds of purchase in year 2018 for all branches in London? ### Answer: SELECT SUM(total_pounds) FROM purchase AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id WHERE T2.city = 'London' AND T1.year = 2018 |
What is the total number of purchases for members with level 6? | CREATE TABLE member (member_id VARCHAR, level VARCHAR); CREATE TABLE purchase (member_id VARCHAR) | SELECT COUNT(*) FROM purchase AS T1 JOIN member AS T2 ON T1.member_id = T2.member_id WHERE T2.level = 6 | ### Context: CREATE TABLE member (member_id VARCHAR, level VARCHAR); CREATE TABLE purchase (member_id VARCHAR) ### Question: What is the total number of purchases for members with level 6? ### Answer: SELECT COUNT(*) FROM purchase AS T1 JOIN member AS T2 ON T1.member_id = T2.member_id WHERE T2.level = 6 |
Find the name of branches where have some members whose hometown is in Louisville, Kentucky and some in Hiram, Georgia. | CREATE TABLE member (member_id VARCHAR, Hometown VARCHAR); CREATE TABLE branch (name VARCHAR, branch_id VARCHAR); CREATE TABLE membership_register_branch (branch_id VARCHAR, member_id VARCHAR) | SELECT T2.name FROM membership_register_branch AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id JOIN member AS T3 ON T1.member_id = T3.member_id WHERE T3.Hometown = 'Louisville , Kentucky' INTERSECT SELECT T2.name FROM membership_register_branch AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id JOIN member AS T3 ON T1.member_id = T3.member_id WHERE T3.Hometown = 'Hiram , Georgia' | ### Context: CREATE TABLE member (member_id VARCHAR, Hometown VARCHAR); CREATE TABLE branch (name VARCHAR, branch_id VARCHAR); CREATE TABLE membership_register_branch (branch_id VARCHAR, member_id VARCHAR) ### Question: Find the name of branches where have some members whose hometown is in Louisville, Kentucky and some in Hiram, Georgia. ### Answer: SELECT T2.name FROM membership_register_branch AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id JOIN member AS T3 ON T1.member_id = T3.member_id WHERE T3.Hometown = 'Louisville , Kentucky' INTERSECT SELECT T2.name FROM membership_register_branch AS T1 JOIN branch AS T2 ON T1.branch_id = T2.branch_id JOIN member AS T3 ON T1.member_id = T3.member_id WHERE T3.Hometown = 'Hiram , Georgia' |
list the card number of all members whose hometown address includes word "Kentucky". | CREATE TABLE member (card_number VARCHAR, Hometown VARCHAR) | SELECT card_number FROM member WHERE Hometown LIKE "%Kentucky%" | ### Context: CREATE TABLE member (card_number VARCHAR, Hometown VARCHAR) ### Question: list the card number of all members whose hometown address includes word "Kentucky". ### Answer: SELECT card_number FROM member WHERE Hometown LIKE "%Kentucky%" |
Find the number of students in total. | CREATE TABLE STUDENT (Id VARCHAR) | SELECT COUNT(*) FROM STUDENT | ### Context: CREATE TABLE STUDENT (Id VARCHAR) ### Question: Find the number of students in total. ### Answer: SELECT COUNT(*) FROM STUDENT |
Find the number of voting records in total. | CREATE TABLE VOTING_RECORD (Id VARCHAR) | SELECT COUNT(*) FROM VOTING_RECORD | ### Context: CREATE TABLE VOTING_RECORD (Id VARCHAR) ### Question: Find the number of voting records in total. ### Answer: SELECT COUNT(*) FROM VOTING_RECORD |
Find the distinct number of president votes. | CREATE TABLE VOTING_RECORD (President_Vote VARCHAR) | SELECT COUNT(DISTINCT President_Vote) FROM VOTING_RECORD | ### Context: CREATE TABLE VOTING_RECORD (President_Vote VARCHAR) ### Question: Find the distinct number of president votes. ### Answer: SELECT COUNT(DISTINCT President_Vote) FROM VOTING_RECORD |
Find the maximum age of all the students. | CREATE TABLE STUDENT (Age INTEGER) | SELECT MAX(Age) FROM STUDENT | ### Context: CREATE TABLE STUDENT (Age INTEGER) ### Question: Find the maximum age of all the students. ### Answer: SELECT MAX(Age) FROM STUDENT |
Find the last names of students with major 50. | CREATE TABLE STUDENT (LName VARCHAR, Major VARCHAR) | SELECT LName FROM STUDENT WHERE Major = 50 | ### Context: CREATE TABLE STUDENT (LName VARCHAR, Major VARCHAR) ### Question: Find the last names of students with major 50. ### Answer: SELECT LName FROM STUDENT WHERE Major = 50 |
Find the first names of students with age above 22. | CREATE TABLE STUDENT (Fname VARCHAR, Age INTEGER) | SELECT Fname FROM STUDENT WHERE Age > 22 | ### Context: CREATE TABLE STUDENT (Fname VARCHAR, Age INTEGER) ### Question: Find the first names of students with age above 22. ### Answer: SELECT Fname FROM STUDENT WHERE Age > 22 |
What are the majors of male (sex is M) students? | CREATE TABLE STUDENT (Major VARCHAR, Sex VARCHAR) | SELECT Major FROM STUDENT WHERE Sex = "M" | ### Context: CREATE TABLE STUDENT (Major VARCHAR, Sex VARCHAR) ### Question: What are the majors of male (sex is M) students? ### Answer: SELECT Major FROM STUDENT WHERE Sex = "M" |
What is the average age of female (sex is F) students? | CREATE TABLE STUDENT (Age INTEGER, Sex VARCHAR) | SELECT AVG(Age) FROM STUDENT WHERE Sex = "F" | ### Context: CREATE TABLE STUDENT (Age INTEGER, Sex VARCHAR) ### Question: What is the average age of female (sex is F) students? ### Answer: SELECT AVG(Age) FROM STUDENT WHERE Sex = "F" |
Subsets and Splits