Re: help on a query
От | Thomas F.O'Connell |
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Тема | Re: help on a query |
Дата | |
Msg-id | B26D13E6-193B-11D9-936A-000D93AE0944@sitening.com обсуждение исходный текст |
Ответ на | Re: help on a query ("CHRIS HOOVER" <CHRIS.HOOVER@companiongroup.com>) |
Ответы |
Re: help on a query
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Список | 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]@sitening.com.comp, mmurrain@dbdes.com.comp > Cc: pgsql-sql@postgresql.org.comp > From: sad@bankir.ru.comp > Sender: pgsql-sql-owner+m19150@postgresql.org.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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