Re: Retrieving result of COUNT(*) with PHP

Hi,

actually, I don't think you need pg_fetch_all.
I use pg_fetch_row for the purpose. What you get with the SELECT
clause is one row.

E.g. like this:
$res = pg_fetch_row(pg_query("SELECT COUNT(*) AS result_count FROM
etc. WHERE etc.")); //Not a very good style :-), one should check if
the query runs
$count = $res[0];

Another way to calculate number of rows in the result is to use
pg_num_rows function.
$res = pg_query("SELECT COUNT(*) AS result_count FROM etc. WHERE etc.");
$count = pg_num_rows($res);

Enjoy!

Mihail

Quoting Lynna Landstreet <lynna(at)spidersilk(dot)net>:

> Hi there,
>
> I'm trying to use a SELECT COUNT(*) to count how many results would be
> retrieved from a particular query (as part of the process of paginating
> search results).
>
> But I'm having trouble figuring out how to retrieve the result of the count
> in PHP. The result on its own is a resource rather than a specific value,
> but when I try to retrieve the result via pg_fetch_result, some kind of
> weird math error happens and I get a huge number that bears no resemblance
> to the number of results the query actually gets when it runs (1,714,608 for
> a query that in actuality produces three results).
>
> I tried using an AS clause (SELECT COUNT(*) AS result_count FROM etc. WHERE
> etc.) to give the result a name, but that didn't help, and when I tried
> using pg_fetch_all on the result to see exactly what it was retrieving, I
> got this:
>
> Array
> (
> [0] => Array
> (
> [result_count] => 1714608
> )
>
> )
>
> Again with the weird number. And yet, if I run the exact same query in the
> SQL window of phpPgAdmin, I get the proper result count (3 in this
> instance).
>
> Does anyone know what's going on here? Can I just not use SELECT COUNT(*)
> with PHP at all?
>
> I originally had the script running the actual query and then counting the
> results, and then running it again with LIMIT and OFFSET to get one page's
> worth of results, but it seemed wasteful to do it that way, so I was trying
> to do it more efficiently... :-/
>
> Thanks,
>
> Lynna
>
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