RE: [PHP-DB] i am lost (php warning)

Hey thanks guys, the email was the problem I had emial instead of email Thanks for the assistance. -----Original Message----- From: Peter Lovatt [mailto:pjl@sunmaia.net] Sent: Tuesday, January 18, 2005 1:45 AM To: Earl Clare; php-db@lists.php.net Subject: RE: [PHP-DB] i am lost (php warning) 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@claremedia.bz] > Sent: 18 January 2005 06:59 > To: php-db@lists.php.net > Subject: [PHP-DB] 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"; > > } > > > >