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Re: Doppelte Datensätze
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Re: Doppelte Datensätze



On Fri, 9 Jul 1999 10:06:48 +0200, Stefan Hundenborn wrote:

>select partyname, count(*) as anzahl  from partybase
>group by partyname
>having anzahl > 1

Ohje, so einfach wars!

thnx

amalesh
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