human
Schema:
Album (AlbumId, Title, ArtistId)
Artist (ArtistId, Name)
Customer (CustomerId, FirstName, LastName, Company, Address, City, State, Country, PostalCode, Phone, Fax, Email, SupportRepId)
Employee (EmployeeId, LastName, FirstName, Title, ReportsTo, BirthDate, HireDate, Address, City, State, Country, PostalCode, Phone, Fax, Email)
Genre (GenreId, Name)
Invoice (InvoiceId, CustomerId, InvoiceDate, BillingAddress, BillingCity, BillingState, BillingCountry, BillingPostalCode, Total)
InvoiceLine (InvoiceLineId, InvoiceId, TrackId, UnitPrice, Quantity)
MediaType (MediaTypeId, Name)
Playlist (PlaylistId, Name)
PlaylistTrack (PlaylistId, TrackId)
Track (TrackId, Name, AlbumId, MediaTypeId, GenreId, Composer, Milliseconds, Bytes, UnitPrice)
Please answer with SQL.
assistant
Sure, I understand.
human
What is the least common media type in all tracks
assistant
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;
human
How many different kinds of media types are there?
assistant
SELECT COUNT( * ) FROM ( SELECT T2.MediaTypeId FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId GROUP BY T2.MediaTypeId );
human
What is the most common media type?
assistant
SELECT T1.Name FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId GROUP BY T2.MediaTypeId ORDER BY COUNT( * ) DESC LIMIT 1;
human
How many tracks are there?
assistant
SELECT COUNT( * ) FROM Track ;