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 [EMAIL PROTECTED] Fax: (416) 441-9085 > -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] Behalf Of Matt Chatterley > Sent: Monday, March 08, 2004 3:41 PM > To: [EMAIL PROTECTED] > Cc: [EMAIL PROTECTED] > 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 > ---------------------------(end of broadcast)--------------------------- TIP 6: Have you searched our list archives? http://archives.postgresql.org
