RE: i am lost (php warning)

From: Date: Tue, 18 Jan 2005 20:16:06 +0000
Subject: RE: i am lost (php warning)
References: 1  Groups: php.db 
Request: Send a blank email to php-db+get-38215@lists.php.net to get a copy of this message
<?php if($_POST['submit'] == '>Log In'){ $user=$_POST["email"]; $pass=$_POST["pass"]; if(!$sql_query = mysql_query("SELECT * FROM cm_customer WHERE emial='".mysql_real_escape_string($user)."'")){ //This will let you know if there is an error in your query. I do not recomend leaving this here when when it is live. die(mysql_error()); } $numRows = mysql_num_rows($sql_query); if (($numRows) >0){ $valid = $user; session_start(); session_register("valid"); } } if (session_is_registered("reg")){ echo "ok"; }else{ echo "sorry"; } ?> Jeremy Shovan -----Original Message----- From: Earl Clare [mailto:eclare@claremedia.bz] Sent: Monday, January 17, 2005 10:59 PM To: php-db@lists.php.net Subject: i am lost (php warning) Hi ya'll, I am lost as to why I am getting this error in my script. Can anyone kindly explain why this is so. Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/cm/public_html/cell/login.php ------------------------------------------------------------------------ ---- --- My script ------------------------------------------------------------------------ ---- --- if($_POST['submit'] == '>Log In') { $user=$_POST["email"]; $pass=$_POST["pass"]; $sql_query = mysql_query("SELECT * FROM cm_customer WHERE emial='$user'"); $sql_query = mysql_num_rows($sql_query); if (($sql_query) >0) { $valid = $user; session_start(); session_register("valid"); } } if (session_is_registered("reg")) { echo "ok"; } else { echo "sorry"; }

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