Re: group by day
От
Rodrigo De León
Тема
Re: group by day
Дата
Msg-id
a55915760705241245q521967e1rfc46d7639ebb8bc2@mail.gmail.com
Ответ на
group by day (Edward W. Rouse)
Список
Дерево обсуждения
group by day "Edward W. Rouse" <erouse@comsquared.com>
Re: group by day "A. Kretschmer" <andreas.kretschmer@schollglas.com>
Re: group by day Richard Huxton <dev@archonet.com>
Re: group by day "Rodrigo De León" <rdeleonp@gmail.com>
On 5/24/07, Edward W. Rouse wrote: > > > I have an audit table that I am trying to get a count of the number of > distinct entries per day by the external table key field. I can do a > > select count(distinct(id)) from audit where timestamp >= '01-may-2007' > > and get a total count. What I need is a way to group on each day and get a > count per day such that the result would be something like > > date count > 01-may-2007 107 > 02-may-2007 215 > 03-may-2007 96 > 04-may-2007 0 > > > I would prefer the 0 entries be included but can live without them. Thanks. > > Oh, postgres 7.4 by the way. > > > > Edward W. Rouse > > ComSquared Systems, Inc. > > 770-734-5301 SELECT TIMESTAMP, COUNT(DISTINCT(ID)) FROM AUDIT GROUP BY TIMESTAMP ORDER BY TIMESTAMP
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