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From: terry@ashtonwoodshomes.com
To: 'Matt Chatterley' <zen31329@zen.co.uk>
Cc: pgsql-sql@postgresql.org
Subject: Re: Trying to make efficient "all vendors who can provide all items"
Date: Tue, 9 Mar 2004 08:37:48 -0500
Message-ID: <042301c405db$b63c5340$2766f30a@development.greatgulfhomes.com> (raw)
In-Reply-To: <Zen-1B0RXT-0003jc-3j@heisenberg.zen.co.uk>

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
>




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