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
Show me first names for employees serving 20 customers.
assistant
SELECT T1.FirstName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT( * ) = 20;
human
Show me their last names.
assistant
SELECT T1.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT( * ) = 20;
human
Show me the same for employees serving no more than 20 customers.
assistant
SELECT T1.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT( * ) <= 20 ;