<?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 = $user;
session_start();
session_register("valid");
}
}
if (session_is_registered("reg")){
echo "ok";
}else{
echo "sorry";
}
?>
Jeremy Shovan
-----Original Message-----
From: Earl Clare [mailto:eclare@xxxxxxxxxxxxx]
Sent: Monday, January 17, 2005 10:59 PM
To: php-db@xxxxxxxxxxxxx
Subject: i am lost (php warning)
Hi ya'll,
I am lost as to why I am getting this error in my script. Can anyone
kindly
explain why this is so.
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
resource in /home/cm/public_html/cell/login.php
------------------------------------------------------------------------
----
---
My script
------------------------------------------------------------------------
----
---
if($_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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