Thanks - that is basically what I used to do (it works fine now on 4.1.2)
but it was very slow as there are a LOT of rows and this method returned all
of them.
From: Garth Webb <[EMAIL PROTECTED]>
To: Dave Torr <[EMAIL PROTECTED]>
CC: [EMAIL PROTECTED]
Subject: Re: How to COUNT rows when they have a COUNT in them
Date: Tue, 15 Jun 2004 09:54:19 -0700
You could also try:
SELECT owner, COUNT(*) FROM pet GROUP BY owner;
SELECT FOUND_ROWS();
On Mon, 2004-06-14 at 20:41, Dave Torr wrote:
> Thanks - this did not work for me as I am on 4.0.17 - presumably this
works
> on 4.1 (seems to need the SubQuery feature)? If so I will upgrade
> immediately!
>
>
> >From: Yayati Kasralikar <[EMAIL PROTECTED]>
> >To: Dave Torr <[EMAIL PROTECTED]>
> >CC: [EMAIL PROTECTED]
> >Subject: Re: How to COUNT rows when they have a COUNT in them
> >Date: Mon, 14 Jun 2004 23:37:15 -0400
> >
> >Following query does what you want:
> >
> >SELECT COUNT(*) from (SELECT COUNT(*) as c FROM pet GROUP BY owner
HAVING
> >c>1) as temp
> >
> >-Yayati
> >
> >Dave Torr wrote:
> >
> >>Probably simple but I can't figure it out!
> >>
> >>THe manual section 3.3.4.8 has the example
> >>
> >>SELECT owner, COUNT(*) FROM pet GROUP BY owner
> >>
> >>which is fine. Now what I want to do is count the number of rows this
> >>returns. Actually of course this is trivial - I can just count how
many
> >>owners there are.
> >>
> >>What I actually have is something similar to
> >>
> >>SELECT owner, COUNT(*) as c FROM pet GROUP BY owner HAVING c>1
> >>
> >>(ie I want to see the owners who have more than one pet). And I just
want
> >>to know how many there are - at the moment I am having to retreive the
> >>full data set (which is large in my case).
> >>
> >>What I want is something like
> >>
> >>SELECT COUNT(SELECT owner, COUNT(*) FROM pet GROUP BY owner HAVING
c>1)
> >>
> >>but that doesn't work....
> >>
> >>
> >>
> >
> >
>
>
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