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Thread: group by day

group by day

From
"Edward W. Rouse"
Date:
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

 

 

Re: group by day

From
"A. Kretschmer"
Date:
am  Thu, dem 24.05.2007, um 14:49:47 -0400 mailte Edward W. Rouse folgendes:
> 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.

You are searching for GROUP BY.

A simple example:

test=*# select * from foo;        ts          | val
---------------------+-----2007-05-01 08:00:00 |  102007-05-01 08:00:00 |  202007-05-02 10:00:00 |  202007-05-02
11:00:00|  30
 
(4 rows)

Time: 1.079 ms
test=*# select ts::date, sum(val) from foo group by 1;    ts     | sum
------------+-----2007-05-02 |  502007-05-01 |  30
(2 rows)


Andreas
-- 
Andreas Kretschmer
Kontakt:  Heynitz: 035242/47150,   D1: 0160/7141639 (mehr: -> Header)
GnuPG-ID:   0x3FFF606C, privat 0x7F4584DA   http://wwwkeys.de.pgp.net

Re: group by day

From
"Rodrigo De León"
Date:
On 5/24/07, Edward W. Rouse <erouse@comsquared.com> 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

Re: group by day

From
Richard Huxton
Date:
Edward W. Rouse wrote:
>  
> Oh, postgres 7.4 by the way.

The latest release there is 7.4.17 - make sure you're running that 
revision. See the website for what bugs have been fixed.

--   Richard Huxton  Archonet Ltd