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Posted by gosha bine on 08/09/07 11:37
On 09.08.2007 13:31 upendrajpr@gmail.com wrote:
> Dear friends,
>
> I am running the code
> $check_1 = mysql_query("SELECT * FROM user WHERE u_name = '".
> $_POST['username']."'") or die(mysql_error());
>
> $check_2 = mysql_num_rows($check);
>
> but the warning comes
> " Warning: mysql_num_rows(): supplied argument is not a valid MySQL
> result resource in C:\Program Files\Apache Software Foundation
> \Apache2.2\htdocs\rajput_new\loginphp.php on line 44.."
>
> $_POST['username'] has name of form on same code the user enter
> username..
>
> What's the problem...
>
>
> Thanks in advance...
> situ
>
Find the "php.ini" file on your computer and edit "error_reporting" line
to be "error_reporting=E_ALL". Restart Apache and run your script again.
You'll see where the problem is.
--
gosha bine
makrell ~ http://www.tagarga.com/blok/makrell
php done right ;) http://code.google.com/p/pihipi
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