On Monday 14 April 2003 09:46, JeRRy wrote:
> Hi,
>
> I tried you 'as' example and got this:
>
> Warning: mysql_fetch_array(): supplied argument is not
> a valid MySQL result resource in message.php
>
> My code is this:
>
> $qr = mysql_query("SELECT IF( emailerror = ('y' ),
> 'error yes', 'do nothing' ) AS answer FROM tipping
> WHERE sid =
> '$sid'");
> $result = mysql_query($qr);
> $row = mysql_fetch_array($result);
> echo $row['answer'];
Why are you using mysql_query() twice?
$qr = "SELECT IF( emailerror = ('y' ),
'error yes', 'do nothing' ) AS answer FROM tipping
WHERE sid = '$sid'";
$result = mysql_query($qr) or die(mysql_error());
$row = mysql_fetch_array($result) or die(mysql_error());
echo $row['answer'];
--
Jason Wong -> Gremlins Associates -> www.gremlins.biz
Open Source Software Systems Integrators
* Web Design & Hosting * Internet & Intranet Applications Development *
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