Article View: borland.public.interbase.sql
Article #11339Re: Multiple counts
From: Arno Brinkman
Date: Fri, 22 Nov 2002 12:44
Date: Fri, 22 Nov 2002 12:44
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1388 bytes
Hi, > Does anyone know how i would count the number of records via a distinct > selection. > > i.e. Select distinct Agent_Name from Table1 brings me all the agents but I > would like to do a record count for each without having to do a select > count(*) for each one manually? If i understand you right then use GROUP BY. SELECT AGENT_NAME, COUNT(*) FROM Table1 GROUP BY AGENT_NAME Regards, Arno Brinkman
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X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: 114a0e,edf94dfb88a4c551 X-Google-Attributes: gid114a0e,public X-Google-ArrivalTime: 2002-11-22 04:45:51 PST From: "Arno Brinkman" <_firebird@abvisie.nl_> Newsgroups: borland.public.interbase.sql References: <3dde12b2$1@newsgroups.borland.com> Subject: Re: Multiple counts Date: Fri, 22 Nov 2002 13:44:44 +0100 Lines: 25 X-Priority: 3 X-MSMail-Priority: Normal X-Newsreader: Microsoft Outlook Express 5.00.2919.6700 X-MimeOLE: Produced By Microsoft MimeOLE V5.00.2919.6700 NNTP-Posting-Host: 213.116.162.149 Message-ID: <3dde26ed@newsgroups.borland.com> X-Trace: newsgroups.borland.com 1037969133 213.116.162.149 (22 Nov 2002 04:45:33 -0800) Path: arc1-ca.pugleaf.net!archiver1.google.com!news1.google.com!sn-xit-02!sn-xit-05!sn-xit-06!sn-xit-08!supernews.com!newshosting.com!news-xfer1.atl.newshosting.com!nntp-relay.ihug.net!newsfeeds.ihug.co.nz!ihug.co.nz!newsgroups.borland.com!not-for-mail