Re: Retrieving result of COUNT(*) with PHP

It will return the result as the column name "count" unless you specify it
as something else:

SELECT count(*) AS numrows FROM mytable;

Let me caution you that SELECT count(*) is slow on larger tables...

But as for your example:

  $result = pg_Query($conn, 'SELECT count(*) FROM table;');
  $data = pg_Fetch_Object($result, 0);
  echo 'My Count is: ' . $data->count . "<br />\n";

Should do the trick.

Hope this helps,

Gavin


On 3/28/07, Lynna Landstreet <[email protected]> wrote:
>
> 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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