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Re: help on a query

From: Thomas F(dot)O'Connell <tfo(at)sitening(dot)com>
To: "CHRIS HOOVER" <CHRIS(dot)HOOVER(at)companiongroup(dot)com>
Cc: sad(at)bankir(dot)ru, pgsql-sql(at)postgresql(dot)org, mmurrain(at)dbdes(dot)com
Subject: Re: help on a query
Date: 2004-10-08 15:06:57
Message-ID: B26D13E6-193B-11D9-936A-000D93AE0944@sitening.com (view raw or flat)
Thread:
Lists: pgsql-sql
I think the OUTER JOIN version is probably more efficient, but EXPLAIN 
would tell you.

-tfo

On Oct 8, 2004, at 8:02 AM, CHRIS HOOVER wrote:

> Just curious, what is wrong with the first way of coding the solution?
> ------------------( Forwarded letter 1 follows )---------------------
> Date: Fri, 8 Oct 2004 08:44:23 +0400
> To: Thomas.F.O'Connell[tfo](at)sitening(dot)com(dot)comp, mmurrain(at)dbdes(dot)com(dot)comp
> Cc: pgsql-sql(at)postgresql(dot)org(dot)comp
> From: sad(at)bankir(dot)ru(dot)comp
> Sender: pgsql-sql-owner+m19150(at)postgresql(dot)org(dot)comp
> Subject: Re: [SQL] help on a query
>
> On Friday 08 October 2004 07:10, Thomas F.O'Connell wrote:
>> A query that should get the job done is:
>>
>> SELECT registration_id
>> FROM registrations r
>> WHERE NOT EXISTS (
>> 	SELECT 1
>> 	FROM receipts
>> 	WHERE registration_id = r.registration_id
>> );
>
> Don't, PLEASE, don't !!!
>
> drive this way :
>
> SELECT r.registration_id
>  FROM registrations AS r
> LEFT OUTER JOIN receipts AS rec
>  ON rec.registration_id = r.registration_id
> WHERE rec.registration_id IS NULL;


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