Re: sub query
От | Martin Kuria |
---|---|
Тема | Re: sub query |
Дата | |
Msg-id | Sea2-F63mFZjvEGP0Jk0000b45e@hotmail.com обсуждение исходный текст |
Ответ на | sub query ("Martin Kuria" <martinkuria@hotmail.com>) |
Список | pgsql-sql |
Thanks Haller, the second one worked thanks a million be blessed Regards +-----------------------------------------------------+ | Martin W. Kuria (Mr.) martin.kuria@unon.org +----------------------------------------------------+ >From: Christoph Haller <ch@rodos.fzk.de> >To: pgsql-sql@postgresql.org >CC: martinkuria@hotmail.com >Subject: Re: [SQL] sub query >Date: Wed, 17 Sep 2003 10:54:49 +0200 > > > > > > Hi I have this problem, when I try to run this query: > > > > > > SELECT MAX(d), host_position FROM (SELECT host_position, > > > COUNT(host_position) as d FROM sss_host GROUP BY host_position) as >e; > > > > > > am getting and ERROR: Attribute e.host_position must be GROUPed or > > used in > > > an aggregate function. > > > > > > Please to advice what could be the problem and how can I rewrite it >to > > work > > > thanks in advance. > > > > > As the error message says: e.host_position must be GROUPed > > > > so (supposing you want a one row result showing the maximum count) > > > > SELECT MAX(e.d), e.host_position FROM (SELECT host_position, > > COUNT(host_position) as d FROM sss_host GROUP BY host_position) as e > > GROUP BY e.host_position ORDER BY 1 LIMIT 1; > > > > should match your intentions. > > >Just thought about another (less complex) way: > >SELECT COUNT(host_position), host_position FROM >sss_host GROUP BY host_position ORDER BY 1 DESC LIMIT 1; > >Regards, Christoph > > > >---------------------------(end of broadcast)--------------------------- >TIP 3: if posting/reading through Usenet, please send an appropriate > subscribe-nomail command to majordomo@postgresql.org so that your > message can get through to the mailing list cleanly _________________________________________________________________ Protect your PC - get McAfee.com VirusScan Online http://clinic.mcafee.com/clinic/ibuy/campaign.asp?cid=3963
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