Re: help on a query
От | CHRIS HOOVER |
---|---|
Тема | Re: help on a query |
Дата | |
Msg-id | NYe01121-06b74e7b@companiongroup.com обсуждение исходный текст |
Ответ на | help on a query (Michelle Murrain <tech@murrain.net>) |
Ответы |
Re: help on a query
Re: help on a query |
Список | pgsql-sql |
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_idFROM registrations AS r LEFT OUTER JOIN receipts AS recON rec.registration_id = r.registration_id WHERE rec.registration_id IS NULL; ---------------------------(end of broadcast)--------------------------- TIP 6: Have you searched our list archives? http://archives.postgresql.org
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