Well, in MSSQL 6.5 and 7 there are system tables you can query to get the
number of records in an index. Works if you have a primary key. Not sure how
wise it is to query system tables, but I prefered it to having to query
large tables.

-steve

-----Original Message-----
From: Bob Silverberg [mailto:[EMAIL PROTECTED]]
Sent: Monday, February 25, 2002 1:36 AM
To: [EMAIL PROTECTED]
Subject: RE: [CFTALKTor] RecordCount


Bill,

Although you cannot get a count of the number of rows in a table without
performing some kind of query, the following is a bit better:

<CFQUERY name="name" datasource="datasource">
SELECT  Count(*) AS RecordCnt
FROM    tblFoo
</CFQUERY>

You can then refer to name.RecordCnt for the number of records.  This is
better in that the database only has to return one record to CF, whereas
using the other method causes the database to return many records.

I have seen many debates about whether you should use Count(*) or Count(ID).
This supposedly differs between databases, but I'd guess that any db worth
its salt will be able to properly optimize a Count(*) query.

Bob

P.S.  Sorry I haven't posted that promised "Caching with Structures" example
code yet.  I just returned from a week's vacation, so I hope to get it done
soon.


> -----Original Message-----
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On
> Behalf Of Bill Sourour
> Sent: Monday, 25 February 2002 6:56 p.m.
> To: [EMAIL PROTECTED]
> Subject: [CFTALKTor] RecordCount
>
>
> Before my question, I am obliged to say:
>
> "GO CANADA! GO! ... WE ARE THE BEST HOCKEY PLAYERS ON EARTH BABY!!!"
>
> ok, now that that's out of the way.
>
> Is there a way to return a record count for a given table whithout
> performing a query? To clarify:
>
> here's what I do now:
>
> <CFQUERY name="name" datasource="datasource">
> SELECT  ID
> FROM    tblFoo
>
> </CFQUERY>
>
> Then I can use #name.recordcount#. But I never use the ID for
> anything, the
> sole purpose of that query is to get a record count. This strikes me as
> inefficient. Any suggestions?
>
> Thanks in advance.
>
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