Hi
it means that the query did not produce a result - usually this is because
of a problem with the query.
$sql_query = mysql_query("SELECT * FROM cm_customer WHERE emial='$user'");
I am guessing you misspelt email and this should be
$sql_query = mysql_query("SELECT * FROM cm_customer WHERE email='$user'");
if you add an error trap it makes this knd of thing easier to spot
$sql_query = mysql_query("SELECT * FROM cm_customer WHERE email='$user'") or
die(mysql_error());
HTH
Peter
> -----Original Message-----
> From: Earl Clare [mailto:eclare@xxxxxxxxxxxxx]
> Sent: 18 January 2005 06:59
> To: php-db@xxxxxxxxxxxxx
> Subject: i am lost (php warning)
> Importance: High
>
>
> 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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