db_id
stringclasses 146
values | query
stringlengths 18
577
| question
stringlengths 3
224
| index
int64 0
7k
|
|---|---|---|---|
coffee_shop | SELECT avg(num_of_staff) , avg(score) FROM shop | What are the average score and average staff number of all shops? | 800 |
coffee_shop | SELECT shop_id , address FROM shop WHERE score < (SELECT avg(score) FROM shop) | Find the id and address of the shops whose score is below the average score. | 801 |
coffee_shop | SELECT address , num_of_staff FROM shop WHERE shop_id NOT IN (SELECT shop_id FROM happy_hour) | Find the address and staff number of the shops that do not have any happy hour. | 802 |
coffee_shop | SELECT t1.address , t1.shop_id FROM shop AS t1 JOIN happy_hour AS t2 ON t1.shop_id = t2.shop_id WHERE MONTH = 'May' | What are the id and address of the shops which have a happy hour in May? | 803 |
coffee_shop | SELECT shop_id , count(*) FROM happy_hour GROUP BY shop_id ORDER BY count(*) DESC LIMIT 1 | which shop has happy hour most frequently? List its id and number of happy hours. | 804 |
coffee_shop | SELECT MONTH FROM happy_hour GROUP BY MONTH ORDER BY count(*) DESC LIMIT 1 | Which month has the most happy hours? | 805 |
coffee_shop | SELECT MONTH FROM happy_hour GROUP BY MONTH HAVING count(*) > 2 | Which months have more than 2 happy hours? | 806 |
chinook_1 | SELECT count(*) FROM ALBUM | How many albums are there? | 807 |
chinook_1 | SELECT count(*) FROM ALBUM | Find the number of albums. | 808 |
chinook_1 | SELECT Name FROM GENRE | List the names of all music genres. | 809 |
chinook_1 | SELECT Name FROM GENRE | What are the names of different music genres? | 810 |
chinook_1 | SELECT * FROM CUSTOMER WHERE State = "NY" | Find all the customer information in state NY. | 811 |
chinook_1 | SELECT * FROM CUSTOMER WHERE State = "NY" | What is all the customer information for customers in NY state? | 812 |
chinook_1 | SELECT FirstName , LastName FROM EMPLOYEE WHERE City = "Calgary" | What are the first names and last names of the employees who live in Calgary city. | 813 |
chinook_1 | SELECT FirstName , LastName FROM EMPLOYEE WHERE City = "Calgary" | Find the full names of employees living in the city of Calgary. | 814 |
chinook_1 | SELECT distinct(BillingCountry) FROM INVOICE | What are the distinct billing countries of the invoices? | 815 |
chinook_1 | SELECT distinct(BillingCountry) FROM INVOICE | Find the different billing countries for all invoices. | 816 |
chinook_1 | SELECT Name FROM ARTIST WHERE Name LIKE "%a%" | Find the names of all artists that have "a" in their names. | 817 |
chinook_1 | SELECT Name FROM ARTIST WHERE Name LIKE "%a%" | What are the names of artist who have the letter 'a' in their names? | 818 |
chinook_1 | SELECT Title FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "AC/DC" | Find the title of all the albums of the artist "AC/DC". | 819 |
chinook_1 | SELECT Title FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "AC/DC" | What are the titles of albums by the artist "AC/DC"? | 820 |
chinook_1 | SELECT COUNT(*) FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "Metallica" | Hom many albums does the artist "Metallica" have? | 821 |
chinook_1 | SELECT COUNT(*) FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "Metallica" | Find the number of albums by the artist "Metallica". | 822 |
chinook_1 | SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T1.Title = "Balls to the Wall" | Which artist does the album "Balls to the Wall" belong to? | 823 |
chinook_1 | SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T1.Title = "Balls to the Wall" | Find the name of the artist who made the album "Balls to the Wall". | 824 |
chinook_1 | SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId GROUP BY T2.Name ORDER BY COUNT(*) DESC LIMIT 1 | Which artist has the most albums? | 825 |
chinook_1 | SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId GROUP BY T2.Name ORDER BY COUNT(*) DESC LIMIT 1 | What is the name of the artist with the greatest number of albums? | 826 |
chinook_1 | SELECT Name FROM TRACK WHERE Name LIKE '%you%' | Find the names of all the tracks that contain the word "you". | 827 |
chinook_1 | SELECT Name FROM TRACK WHERE Name LIKE '%you%' | What are the names of tracks that contain the the word you in them? | 828 |
chinook_1 | SELECT AVG(UnitPrice) FROM TRACK | What is the average unit price of all the tracks? | 829 |
chinook_1 | SELECT AVG(UnitPrice) FROM TRACK | Find the average unit price for a track. | 830 |
chinook_1 | SELECT max(Milliseconds) , min(Milliseconds) FROM TRACK | What are the durations of the longest and the shortest tracks in milliseconds? | 831 |
chinook_1 | SELECT max(Milliseconds) , min(Milliseconds) FROM TRACK | Find the maximum and minimum durations of tracks in milliseconds. | 832 |
chinook_1 | SELECT T1.Title , T2.AlbumID , COUNT(*) FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId GROUP BY T2.AlbumID | Show the album names, ids and the number of tracks for each album. | 833 |
chinook_1 | SELECT T1.Title , T2.AlbumID , COUNT(*) FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId GROUP BY T2.AlbumID | What are the names and ids of the different albums, and how many tracks are on each? | 834 |
chinook_1 | SELECT T1.Name FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId GROUP BY T2.GenreId ORDER BY COUNT(*) DESC LIMIT 1 | What is the name of the most common genre in all tracks? | 835 |
chinook_1 | SELECT T1.Name FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId GROUP BY T2.GenreId ORDER BY COUNT(*) DESC LIMIT 1 | Find the name of the genre that is most frequent across all tracks. | 836 |
chinook_1 | SELECT T1.Name FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId GROUP BY T2.MediaTypeId ORDER BY COUNT(*) ASC LIMIT 1 | What is the least common media type in all tracks? | 837 |
chinook_1 | SELECT T1.Name FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId GROUP BY T2.MediaTypeId ORDER BY COUNT(*) ASC LIMIT 1 | What is the name of the media type that is least common across all tracks? | 838 |
chinook_1 | SELECT T1.Title , T2.AlbumID FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId WHERE T2.UnitPrice > 1 GROUP BY T2.AlbumID | Show the album names and ids for albums that contain tracks with unit price bigger than 1. | 839 |
chinook_1 | SELECT T1.Title , T2.AlbumID FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId WHERE T2.UnitPrice > 1 GROUP BY T2.AlbumID | What are the titles and ids for albums containing tracks with unit price greater than 1? | 840 |
chinook_1 | SELECT COUNT(*) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock" | How many tracks belong to rock genre? | 841 |
chinook_1 | SELECT COUNT(*) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock" | Count the number of tracks that are part of the rock genre. | 842 |
chinook_1 | SELECT AVG(UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Jazz" | What is the average unit price of tracks that belong to Jazz genre? | 843 |
chinook_1 | SELECT AVG(UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Jazz" | Find the average unit price of jazz tracks. | 844 |
chinook_1 | SELECT FirstName , LastName FROM CUSTOMER WHERE Email = "[email protected]" | What is the first name and last name of the customer that has email "[email protected]"? | 845 |
chinook_1 | SELECT FirstName , LastName FROM CUSTOMER WHERE Email = "[email protected]" | Find the full name of the customer with the email "[email protected]". | 846 |
chinook_1 | SELECT COUNT(*) FROM CUSTOMER WHERE Email LIKE "%gmail.com%" | How many customers have email that contains "gmail.com"? | 847 |
chinook_1 | SELECT COUNT(*) FROM CUSTOMER WHERE Email LIKE "%gmail.com%" | Count the number of customers that have an email containing "gmail.com". | 848 |
chinook_1 | SELECT T2.FirstName , T2.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.FirstName = "Leonie" | What is the first name and last name employee helps the customer with first name Leonie? | 849 |
chinook_1 | SELECT T2.FirstName , T2.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.FirstName = "Leonie" | Find the full names of employees who help customers with the first name Leonie. | 850 |
chinook_1 | SELECT T2.City FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.PostalCode = "70174" | What city does the employee who helps the customer with postal code 70174 live in? | 851 |
chinook_1 | SELECT T2.City FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.PostalCode = "70174" | Find the cities corresponding to employees who help customers with the postal code 70174. | 852 |
chinook_1 | SELECT COUNT(DISTINCT city) FROM EMPLOYEE | How many distinct cities does the employees live in? | 853 |
chinook_1 | SELECT COUNT(DISTINCT city) FROM EMPLOYEE | Find the number of different cities that employees live in. | 854 |
chinook_1 | SELECT T2.InvoiceDate FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.FirstName = "Astrid" AND LastName = "Gruber" | Find all invoice dates corresponding to customers with first name Astrid and last name Gruber. | 855 |
chinook_1 | SELECT T2.InvoiceDate FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.FirstName = "Astrid" AND LastName = "Gruber" | What are the invoice dates for customers with the first name Astrid and the last name Gruber? | 856 |
chinook_1 | SELECT LastName FROM CUSTOMER EXCEPT SELECT T1.LastName FROM CUSTOMER AS T1 JOIN Invoice AS T2 ON T1.CustomerId = T2.CustomerId WHERE T2.total > 20 | Find all the customer last names that do not have invoice totals larger than 20. | 857 |
chinook_1 | SELECT LastName FROM CUSTOMER EXCEPT SELECT T1.LastName FROM CUSTOMER AS T1 JOIN Invoice AS T2 ON T1.CustomerId = T2.CustomerId WHERE T2.total > 20 | What are the last names of customers without invoice totals exceeding 20? | 858 |
chinook_1 | SELECT DISTINCT T1.FirstName FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Brazil" | Find the first names of all customers that live in Brazil and have an invoice. | 859 |
chinook_1 | SELECT DISTINCT T1.FirstName FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Brazil" | What are the different first names for customers from Brazil who have also had an invoice? | 860 |
chinook_1 | SELECT DISTINCT T1.Address FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Germany" | Find the address of all customers that live in Germany and have invoice. | 861 |
chinook_1 | SELECT DISTINCT T1.Address FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Germany" | What are the addresses of customers living in Germany who have had an invoice? | 862 |
chinook_1 | SELECT Phone FROM EMPLOYEE | List the phone numbers of all employees. | 863 |
chinook_1 | SELECT Phone FROM EMPLOYEE | What are the phone numbers for each employee? | 864 |
chinook_1 | SELECT COUNT(*) FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId WHERE T1.Name = "AAC audio file" | How many tracks are in the AAC audio file media type? | 865 |
chinook_1 | SELECT COUNT(*) FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId WHERE T1.Name = "AAC audio file" | Count the number of tracks that are of the media type "AAC audio file". | 866 |
chinook_1 | SELECT AVG(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Latin" OR T1.Name = "Pop" | What is the average duration in milliseconds of tracks that belong to Latin or Pop genre? | 867 |
chinook_1 | SELECT AVG(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Latin" OR T1.Name = "Pop" | Find the average millisecond length of Latin and Pop tracks. | 868 |
chinook_1 | SELECT T1.FirstName , T1.SupportRepId FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) >= 10 | Please show the employee first names and ids of employees who serve at least 10 customers. | 869 |
chinook_1 | SELECT T1.FirstName , T1.SupportRepId FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) >= 10 | What are the first names and support rep ids for employees serving 10 or more customers? | 870 |
chinook_1 | SELECT T1.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) <= 20 | Please show the employee last names that serves no more than 20 customers. | 871 |
chinook_1 | SELECT T1.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) <= 20 | What are the last names of employees who serve at most 20 customers? | 872 |
chinook_1 | SELECT Title FROM ALBUM ORDER BY Title | Please list all album titles in alphabetical order. | 873 |
chinook_1 | SELECT Title FROM ALBUM ORDER BY Title | What are all the album titles, in alphabetical order? | 874 |
chinook_1 | SELECT T2.Name , T1.ArtistId FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistID GROUP BY T1.ArtistId HAVING COUNT(*) >= 3 ORDER BY T2.Name | Please list the name and id of all artists that have at least 3 albums in alphabetical order. | 875 |
chinook_1 | SELECT T2.Name , T1.ArtistId FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistID GROUP BY T1.ArtistId HAVING COUNT(*) >= 3 ORDER BY T2.Name | What are the names and ids of artists with 3 or more albums, listed in alphabetical order? | 876 |
chinook_1 | SELECT Name FROM ARTIST EXCEPT SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId | Find the names of artists that do not have any albums. | 877 |
chinook_1 | SELECT Name FROM ARTIST EXCEPT SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId | What are the names of artists who have not released any albums? | 878 |
chinook_1 | SELECT AVG(T2.UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock" | What is the average unit price of rock tracks? | 879 |
chinook_1 | SELECT AVG(T2.UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock" | Find the average unit price of tracks from the Rock genre. | 880 |
chinook_1 | SELECT max(Milliseconds) , min(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Pop" | What are the duration of the longest and shortest pop tracks in milliseconds? | 881 |
chinook_1 | SELECT max(Milliseconds) , min(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Pop" | Find the maximum and minimum millisecond lengths of pop tracks. | 882 |
chinook_1 | SELECT BirthDate FROM EMPLOYEE WHERE City = "Edmonton" | What are the birth dates of employees living in Edmonton? | 883 |
chinook_1 | SELECT BirthDate FROM EMPLOYEE WHERE City = "Edmonton" | Find the birth dates corresponding to employees who live in the city of Edmonton. | 884 |
chinook_1 | SELECT distinct(UnitPrice) FROM TRACK | What are the distinct unit prices of all tracks? | 885 |
chinook_1 | SELECT distinct(UnitPrice) FROM TRACK | Find the distinct unit prices for tracks. | 886 |
chinook_1 | SELECT count(*) FROM ARTIST WHERE artistid NOT IN(SELECT artistid FROM ALBUM) | How many artists do not have any album? | 887 |
chinook_1 | SELECT count(*) FROM ARTIST WHERE artistid NOT IN(SELECT artistid FROM ALBUM) | Cound the number of artists who have not released an album. | 888 |
chinook_1 | SELECT T1.Title FROM Album AS T1 JOIN Track AS T2 ON T1.AlbumId = T2.AlbumId JOIN Genre AS T3 ON T2.GenreID = T3.GenreID WHERE T3.Name = 'Reggae' INTERSECT SELECT T1.Title FROM Album AS T1 JOIN Track AS T2 ON T1.AlbumId = T2.AlbumId JOIN Genre AS T3 ON T2.GenreID = T3.GenreID WHERE T3.Name = 'Rock' | What are the album titles for albums containing both 'Reggae' and 'Rock' genre tracks? | 889 |
chinook_1 | SELECT T1.Title FROM Album AS T1 JOIN Track AS T2 ON T1.AlbumId = T2.AlbumId JOIN Genre AS T3 ON T2.GenreID = T3.GenreID WHERE T3.Name = 'Reggae' INTERSECT SELECT T1.Title FROM Album AS T1 JOIN Track AS T2 ON T1.AlbumId = T2.AlbumId JOIN Genre AS T3 ON T2.GenreID = T3.GenreID WHERE T3.Name = 'Rock' | Find the titles of albums that contain tracks of both the Reggae and Rock genres. | 890 |
insurance_fnol | SELECT customer_phone FROM available_policies | Find all the phone numbers. | 891 |
insurance_fnol | SELECT customer_phone FROM available_policies | What are all the phone numbers? | 892 |
insurance_fnol | SELECT customer_phone FROM available_policies WHERE policy_type_code = "Life Insurance" | What are the customer phone numbers under the policy "Life Insurance"? | 893 |
insurance_fnol | SELECT customer_phone FROM available_policies WHERE policy_type_code = "Life Insurance" | What are the phone numbers of customers using the policy with the code "Life Insurance"? | 894 |
insurance_fnol | SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY count(*) DESC LIMIT 1 | Which policy type has the most records in the database? | 895 |
insurance_fnol | SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY count(*) DESC LIMIT 1 | Which policy type appears most frequently in the available policies? | 896 |
insurance_fnol | SELECT customer_phone FROM available_policies WHERE policy_type_code = (SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY count(*) DESC LIMIT 1) | What are all the customer phone numbers under the most popular policy type? | 897 |
insurance_fnol | SELECT customer_phone FROM available_policies WHERE policy_type_code = (SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY count(*) DESC LIMIT 1) | Find the phone numbers of customers using the most common policy type among the available policies. | 898 |
insurance_fnol | SELECT policy_type_code FROM available_policies GROUP BY policy_type_code HAVING count(*) > 4 | Find the policy type used by more than 4 customers. | 899 |