Posted by upendrajpr on 08/09/07 11:31

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

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