as 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
>
; 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
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 (c) as temp
-Yayati
Dave Torr wrote:
Probably simple but I can't figure i
pgrade
> 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
>
Just a note...
> 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!
This isn't a subquery -- this is a Derived Table.
With regards,
Martijn Tonies
Database Workbench - developer tool for
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 fi
NT 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
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.
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.
Wha
