Re: help on a query
| От | CHRIS HOOVER |
|---|---|
| Тема | Re: help on a query |
| Дата | |
| Msg-id | NYe01121-06b74e7b@companiongroup.com обсуждение |
| 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;
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