query_id
int64 0
7k
| database_id
stringclasses 140
values | table_id
sequencelengths 1
5
| query
stringlengths 16
224
| answer
stringlengths 18
577
| difficulty
stringclasses 4
values |
|---|---|---|---|---|---|
800 | coffee_shop | [
"shop"
] | What are the average score and average staff number of all shops? | SELECT avg(num_of_staff) , avg(score) FROM shop | medium |
801 | coffee_shop | [
"shop"
] | Find the id and address of the shops whose score is below the average score. | SELECT shop_id , address FROM shop WHERE score < (SELECT avg(score) FROM shop) | extra |
802 | coffee_shop | [
"happy_hour",
"shop"
] | Find the address and staff number of the shops that do not have any happy hour. | SELECT address , num_of_staff FROM shop WHERE shop_id NOT IN (SELECT shop_id FROM happy_hour) | extra |
803 | coffee_shop | [
"happy_hour",
"shop"
] | What are the id and address of the shops which have a happy hour in May? | 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' | medium |
804 | coffee_shop | [
"happy_hour"
] | which shop has happy hour most frequently? List its id and number of happy hours. | SELECT shop_id , count(*) FROM happy_hour GROUP BY shop_id ORDER BY count(*) DESC LIMIT 1 | hard |
805 | coffee_shop | [
"happy_hour"
] | Which month has the most happy hours? | SELECT MONTH FROM happy_hour GROUP BY MONTH ORDER BY count(*) DESC LIMIT 1 | hard |
806 | coffee_shop | [
"happy_hour"
] | Which months have more than 2 happy hours? | SELECT MONTH FROM happy_hour GROUP BY MONTH HAVING count(*) > 2 | easy |
807 | chinook_1 | [
"ALBUM"
] | How many albums are there? | SELECT count(*) FROM ALBUM | easy |
808 | chinook_1 | [
"ALBUM"
] | Find the number of albums. | SELECT count(*) FROM ALBUM | easy |
809 | chinook_1 | [
"GENRE"
] | List the names of all music genres. | SELECT Name FROM GENRE | easy |
810 | chinook_1 | [
"GENRE"
] | What are the names of different music genres? | SELECT Name FROM GENRE | easy |
811 | chinook_1 | [
"CUSTOMER"
] | Find all the customer information in state NY. | SELECT * FROM CUSTOMER WHERE State = "NY" | easy |
812 | chinook_1 | [
"CUSTOMER"
] | What is all the customer information for customers in NY state? | SELECT * FROM CUSTOMER WHERE State = "NY" | easy |
813 | chinook_1 | [
"EMPLOYEE"
] | What are the first names and last names of the employees who live in Calgary city. | SELECT FirstName , LastName FROM EMPLOYEE WHERE City = "Calgary" | medium |
814 | chinook_1 | [
"EMPLOYEE"
] | Find the full names of employees living in the city of Calgary. | SELECT FirstName , LastName FROM EMPLOYEE WHERE City = "Calgary" | medium |
815 | chinook_1 | [
"INVOICE"
] | What are the distinct billing countries of the invoices? | SELECT distinct(BillingCountry) FROM INVOICE | easy |
816 | chinook_1 | [
"INVOICE"
] | Find the different billing countries for all invoices. | SELECT distinct(BillingCountry) FROM INVOICE | easy |
817 | chinook_1 | [
"ARTIST"
] | Find the names of all artists that have "a" in their names. | SELECT Name FROM ARTIST WHERE Name LIKE "%a%" | medium |
818 | chinook_1 | [
"ARTIST"
] | What are the names of artist who have the letter 'a' in their names? | SELECT Name FROM ARTIST WHERE Name LIKE "%a%" | medium |
819 | chinook_1 | [
"ARTIST",
"ALBUM"
] | Find the title of all the albums of the artist "AC/DC". | SELECT Title FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "AC/DC" | medium |
820 | chinook_1 | [
"ARTIST",
"ALBUM"
] | What are the titles of albums by the artist "AC/DC"? | SELECT Title FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "AC/DC" | medium |
821 | chinook_1 | [
"ARTIST",
"ALBUM"
] | Hom many albums does the artist "Metallica" have? | SELECT COUNT(*) FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "Metallica" | medium |
822 | chinook_1 | [
"ARTIST",
"ALBUM"
] | Find the number of albums by the artist "Metallica". | SELECT COUNT(*) FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "Metallica" | medium |
823 | chinook_1 | [
"ARTIST",
"ALBUM"
] | Which artist does the album "Balls to the Wall" belong to? | SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T1.Title = "Balls to the Wall" | medium |
824 | chinook_1 | [
"ARTIST",
"ALBUM"
] | Find the name of the artist who made the album "Balls to the Wall". | SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T1.Title = "Balls to the Wall" | medium |
825 | chinook_1 | [
"ARTIST",
"ALBUM"
] | Which artist has the most albums? | 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 | extra |
826 | chinook_1 | [
"ARTIST",
"ALBUM"
] | What is the name of the artist with the greatest number of albums? | 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 | extra |
827 | chinook_1 | [
"TRACK"
] | Find the names of all the tracks that contain the word "you". | SELECT Name FROM TRACK WHERE Name LIKE '%you%' | medium |
828 | chinook_1 | [
"TRACK"
] | What are the names of tracks that contain the the word you in them? | SELECT Name FROM TRACK WHERE Name LIKE '%you%' | medium |
829 | chinook_1 | [
"TRACK"
] | What is the average unit price of all the tracks? | SELECT AVG(UnitPrice) FROM TRACK | easy |
830 | chinook_1 | [
"TRACK"
] | Find the average unit price for a track. | SELECT AVG(UnitPrice) FROM TRACK | easy |
831 | chinook_1 | [
"TRACK"
] | What are the durations of the longest and the shortest tracks in milliseconds? | SELECT max(Milliseconds) , min(Milliseconds) FROM TRACK | medium |
832 | chinook_1 | [
"TRACK"
] | Find the maximum and minimum durations of tracks in milliseconds. | SELECT max(Milliseconds) , min(Milliseconds) FROM TRACK | medium |
833 | chinook_1 | [
"TRACK",
"ALBUM"
] | Show the album names, ids and the number of tracks for each album. | SELECT T1.Title , T2.AlbumID , COUNT(*) FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId GROUP BY T2.AlbumID | medium |
834 | chinook_1 | [
"TRACK",
"ALBUM"
] | What are the names and ids of the different albums, and how many tracks are on each? | SELECT T1.Title , T2.AlbumID , COUNT(*) FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId GROUP BY T2.AlbumID | medium |
835 | chinook_1 | [
"GENRE",
"TRACK"
] | What is the name of the most common genre in all tracks? | 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 | extra |
836 | chinook_1 | [
"GENRE",
"TRACK"
] | Find the name of the genre that is most frequent across all tracks. | 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 | extra |
837 | chinook_1 | [
"TRACK",
"MEDIATYPE"
] | What is the least common media type in all tracks? | 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 | extra |
838 | chinook_1 | [
"TRACK",
"MEDIATYPE"
] | What is the name of the media type that is least common across all tracks? | 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 | extra |
839 | chinook_1 | [
"TRACK",
"ALBUM"
] | Show the album names and ids for albums that contain tracks with unit price bigger than 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 | hard |
840 | chinook_1 | [
"TRACK",
"ALBUM"
] | What are the titles and ids for albums containing tracks with unit price greater than 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 | hard |
841 | chinook_1 | [
"GENRE",
"TRACK"
] | How many tracks belong to rock genre? | SELECT COUNT(*) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock" | medium |
842 | chinook_1 | [
"GENRE",
"TRACK"
] | Count the number of tracks that are part of the rock genre. | SELECT COUNT(*) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock" | medium |
843 | chinook_1 | [
"GENRE",
"TRACK"
] | What is the average unit price of tracks that belong to Jazz genre? | SELECT AVG(UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Jazz" | medium |
844 | chinook_1 | [
"GENRE",
"TRACK"
] | Find the average unit price of jazz tracks. | SELECT AVG(UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Jazz" | medium |
845 | chinook_1 | [
"CUSTOMER"
] | What is the first name and last name of the customer that has email "[email protected]"? | SELECT FirstName , LastName FROM CUSTOMER WHERE Email = "[email protected]" | medium |
846 | chinook_1 | [
"CUSTOMER"
] | Find the full name of the customer with the email "[email protected]". | SELECT FirstName , LastName FROM CUSTOMER WHERE Email = "[email protected]" | medium |
847 | chinook_1 | [
"CUSTOMER"
] | How many customers have email that contains "gmail.com"? | SELECT COUNT(*) FROM CUSTOMER WHERE Email LIKE "%gmail.com%" | medium |
848 | chinook_1 | [
"CUSTOMER"
] | Count the number of customers that have an email containing "gmail.com". | SELECT COUNT(*) FROM CUSTOMER WHERE Email LIKE "%gmail.com%" | medium |
849 | chinook_1 | [
"EMPLOYEE",
"CUSTOMER"
] | What is the first name and last name employee helps the customer with first name Leonie? | SELECT T2.FirstName , T2.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.FirstName = "Leonie" | medium |
850 | chinook_1 | [
"EMPLOYEE",
"CUSTOMER"
] | Find the full names of employees who help customers with the first name Leonie. | SELECT T2.FirstName , T2.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.FirstName = "Leonie" | medium |
851 | chinook_1 | [
"EMPLOYEE",
"CUSTOMER"
] | What city does the employee who helps the customer with postal code 70174 live in? | SELECT T2.City FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.PostalCode = "70174" | medium |
852 | chinook_1 | [
"EMPLOYEE",
"CUSTOMER"
] | Find the cities corresponding to employees who help customers with the postal code 70174. | SELECT T2.City FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.PostalCode = "70174" | medium |
853 | chinook_1 | [
"EMPLOYEE"
] | How many distinct cities does the employees live in? | SELECT COUNT(DISTINCT city) FROM EMPLOYEE | easy |
854 | chinook_1 | [
"EMPLOYEE"
] | Find the number of different cities that employees live in. | SELECT COUNT(DISTINCT city) FROM EMPLOYEE | easy |
855 | chinook_1 | [
"INVOICE",
"CUSTOMER"
] | Find all invoice dates corresponding to customers with first name Astrid and last name Gruber. | SELECT T2.InvoiceDate FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.FirstName = "Astrid" AND LastName = "Gruber" | medium |
856 | chinook_1 | [
"INVOICE",
"CUSTOMER"
] | What are the invoice dates for customers with the first name Astrid and the last name Gruber? | SELECT T2.InvoiceDate FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.FirstName = "Astrid" AND LastName = "Gruber" | medium |
857 | chinook_1 | [
"Invoice",
"CUSTOMER"
] | Find all the customer last names that do not have invoice totals larger than 20. | 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 | hard |
858 | chinook_1 | [
"Invoice",
"CUSTOMER"
] | What are the last names of customers without invoice totals exceeding 20? | 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 | hard |
859 | chinook_1 | [
"INVOICE",
"CUSTOMER"
] | Find the first names of all customers that live in Brazil and have an invoice. | SELECT DISTINCT T1.FirstName FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Brazil" | medium |
860 | chinook_1 | [
"INVOICE",
"CUSTOMER"
] | What are the different first names for customers from Brazil who have also had an invoice? | SELECT DISTINCT T1.FirstName FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Brazil" | medium |
861 | chinook_1 | [
"INVOICE",
"CUSTOMER"
] | Find the address of all customers that live in Germany and have invoice. | SELECT DISTINCT T1.Address FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Germany" | medium |
862 | chinook_1 | [
"INVOICE",
"CUSTOMER"
] | What are the addresses of customers living in Germany who have had an invoice? | SELECT DISTINCT T1.Address FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Germany" | medium |
863 | chinook_1 | [
"EMPLOYEE"
] | List the phone numbers of all employees. | SELECT Phone FROM EMPLOYEE | easy |
864 | chinook_1 | [
"EMPLOYEE"
] | What are the phone numbers for each employee? | SELECT Phone FROM EMPLOYEE | easy |
865 | chinook_1 | [
"TRACK",
"MEDIATYPE"
] | How many tracks are in the AAC audio file media type? | SELECT COUNT(*) FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId WHERE T1.Name = "AAC audio file" | medium |
866 | chinook_1 | [
"TRACK",
"MEDIATYPE"
] | Count the number of tracks that are of the media type "AAC audio file". | SELECT COUNT(*) FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId WHERE T1.Name = "AAC audio file" | medium |
867 | chinook_1 | [
"GENRE",
"TRACK"
] | What is the average duration in milliseconds of tracks that belong to Latin or Pop genre? | SELECT AVG(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Latin" OR T1.Name = "Pop" | hard |
868 | chinook_1 | [
"GENRE",
"TRACK"
] | Find the average millisecond length of Latin and Pop tracks. | SELECT AVG(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Latin" OR T1.Name = "Pop" | hard |
869 | chinook_1 | [
"EMPLOYEE",
"CUSTOMER"
] | Please show the employee first names and ids of employees who serve at least 10 customers. | 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 | medium |
870 | chinook_1 | [
"EMPLOYEE",
"CUSTOMER"
] | What are the first names and support rep ids for employees serving 10 or more customers? | 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 | medium |
871 | chinook_1 | [
"EMPLOYEE",
"CUSTOMER"
] | Please show the employee last names that serves no more than 20 customers. | SELECT T1.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) <= 20 | medium |
872 | chinook_1 | [
"EMPLOYEE",
"CUSTOMER"
] | What are the last names of employees who serve at most 20 customers? | SELECT T1.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) <= 20 | medium |
873 | chinook_1 | [
"ALBUM"
] | Please list all album titles in alphabetical order. | SELECT Title FROM ALBUM ORDER BY Title | easy |
874 | chinook_1 | [
"ALBUM"
] | What are all the album titles, in alphabetical order? | SELECT Title FROM ALBUM ORDER BY Title | easy |
875 | chinook_1 | [
"ARTIST",
"ALBUM"
] | Please list the name and id of all artists that have at least 3 albums in alphabetical order. | 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 | hard |
876 | chinook_1 | [
"ARTIST",
"ALBUM"
] | What are the names and ids of artists with 3 or more albums, listed in alphabetical order? | 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 | hard |
877 | chinook_1 | [
"ARTIST",
"ALBUM"
] | Find the names of artists that do not have any albums. | SELECT Name FROM ARTIST EXCEPT SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId | hard |
878 | chinook_1 | [
"ARTIST",
"ALBUM"
] | What are the names of artists who have not released any albums? | SELECT Name FROM ARTIST EXCEPT SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId | hard |
879 | chinook_1 | [
"GENRE",
"TRACK"
] | What is the average unit price of rock tracks? | SELECT AVG(T2.UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock" | medium |
880 | chinook_1 | [
"GENRE",
"TRACK"
] | Find the average unit price of tracks from the Rock genre. | SELECT AVG(T2.UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock" | medium |
881 | chinook_1 | [
"GENRE",
"TRACK"
] | What are the duration of the longest and shortest pop tracks in milliseconds? | SELECT max(Milliseconds) , min(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Pop" | extra |
882 | chinook_1 | [
"GENRE",
"TRACK"
] | Find the maximum and minimum millisecond lengths of pop tracks. | SELECT max(Milliseconds) , min(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Pop" | extra |
883 | chinook_1 | [
"EMPLOYEE"
] | What are the birth dates of employees living in Edmonton? | SELECT BirthDate FROM EMPLOYEE WHERE City = "Edmonton" | easy |
884 | chinook_1 | [
"EMPLOYEE"
] | Find the birth dates corresponding to employees who live in the city of Edmonton. | SELECT BirthDate FROM EMPLOYEE WHERE City = "Edmonton" | easy |
885 | chinook_1 | [
"TRACK"
] | What are the distinct unit prices of all tracks? | SELECT distinct(UnitPrice) FROM TRACK | easy |
886 | chinook_1 | [
"TRACK"
] | Find the distinct unit prices for tracks. | SELECT distinct(UnitPrice) FROM TRACK | easy |
887 | chinook_1 | [
"ARTIST",
"ALBUM"
] | How many artists do not have any album? | SELECT count(*) FROM ARTIST WHERE artistid NOT IN(SELECT artistid FROM ALBUM) | extra |
888 | chinook_1 | [
"ARTIST",
"ALBUM"
] | Cound the number of artists who have not released an album. | SELECT count(*) FROM ARTIST WHERE artistid NOT IN(SELECT artistid FROM ALBUM) | extra |
889 | chinook_1 | [
"Track",
"Album",
"Genre"
] | What are the album titles for albums containing both 'Reggae' and 'Rock' genre tracks? | 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' | extra |
890 | chinook_1 | [
"Track",
"Album",
"Genre"
] | Find the titles of albums that contain tracks of both the Reggae and Rock genres. | 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' | extra |
891 | insurance_fnol | [
"available_policies"
] | Find all the phone numbers. | SELECT customer_phone FROM available_policies | easy |
892 | insurance_fnol | [
"available_policies"
] | What are all the phone numbers? | SELECT customer_phone FROM available_policies | easy |
893 | insurance_fnol | [
"available_policies"
] | What are the customer phone numbers under the policy "Life Insurance"? | SELECT customer_phone FROM available_policies WHERE policy_type_code = "Life Insurance" | easy |
894 | insurance_fnol | [
"available_policies"
] | What are the phone numbers of customers using the policy with the code "Life Insurance"? | SELECT customer_phone FROM available_policies WHERE policy_type_code = "Life Insurance" | easy |
895 | insurance_fnol | [
"available_policies"
] | Which policy type has the most records in the database? | SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY count(*) DESC LIMIT 1 | hard |
896 | insurance_fnol | [
"available_policies"
] | Which policy type appears most frequently in the available policies? | SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY count(*) DESC LIMIT 1 | hard |
897 | insurance_fnol | [
"available_policies"
] | What are all the customer phone numbers under the most popular policy type? | 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) | hard |
898 | insurance_fnol | [
"available_policies"
] | Find the phone numbers of customers using the most common policy type among the available policies. | 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) | hard |
899 | insurance_fnol | [
"available_policies"
] | Find the policy type used by more than 4 customers. | SELECT policy_type_code FROM available_policies GROUP BY policy_type_code HAVING count(*) > 4 | easy |
Subsets and Splits