Re: Trying to make efficient "all vendors who can provide all items"
От | |
---|---|
Тема | Re: Trying to make efficient "all vendors who can provide all items" |
Дата | |
Msg-id | 042301c405db$b63c5340$2766f30a@development.greatgulfhomes.com обсуждение исходный текст |
Ответ на | Re: Trying to make efficient "all vendors who can provide all items" ("Matt Chatterley" <zen31329@zen.co.uk>) |
Ответы |
Re: Trying to make efficient "all vendors who can provide all items"
|
Список | pgsql-sql |
Of all the proposed solutions, this appears to run the fastest, and not require the creation of an additional table. Thanks! Terry Fielder Manager Software Development and Deployment Great Gulf Homes / Ashton Woods Homes terry@greatgulfhomes.com Fax: (416) 441-9085 > -----Original Message----- > From: pgsql-sql-owner@postgresql.org > [mailto:pgsql-sql-owner@postgresql.org]On Behalf Of Matt Chatterley > Sent: Monday, March 08, 2004 3:41 PM > To: terry@ashtonwoodshomes.com > Cc: pgsql-sql@postgresql.org > Subject: Re: [SQL] Trying to make efficient "all vendors who > can provide > all items" > > > Hmm. My PGSQL knowledge is rusty, so this may be slightly > microsoftified.. > > How about just: > > SELECT V.VendorID, V.VendorName, COUNT(IV.ItemID) > FROM Vendor V > INNER JOIN Item_Vendor IV ON IV.VendorID = V.VendorID AND > IV.ItemID IN (1, > 2, 3, 4, 5) > GROUP BY V.VendorID, V.VendorName > HAVING COUNT(IV.ItemID) = 5 >
В списке pgsql-sql по дате отправления: