[RESOLVED] DELETE return value

i have the following that i can delete my records with no problem but if you enter a records that doesnt exist then i need to be able to get a return value that tells the user you cant delete that cause it doesn't exist. with my SELECT statements or UPDATE i do $num=mysql_numrows($result); and test the vaule of $num, but i cant with the following

Code:

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$query = "DELETE FROM WorkOrders WHERE id='$TicketNumber' AND siteid='$siteid'";

$result=mysql_query($query);

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i want to be able to via the $result value see if the record even exist then i can tell the user that it either doesn't or it was deleted.

use mysql_affected_rows() for INSERT, UPDATE and DELETE. mysql_num_rows() only returns the number of rows found (meaning only for SELECT).

Kows,

thanks once again. that did it. for those interested here is my entire delete with a redirect if not found.

Code:

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$query = "DELETE FROM WorkOrders WHERE id='$TicketNumber' AND siteid='$siteid'";

$result=mysql_query($query);

$result = mysql_affected_rows();

mysql_close();



if($result == "0"){

        echo "<br><center><font size=\"2\"><b>That Ticket number doesn't exist in your datebase. It may have already been deleted. You will now be redirected back to the main page.</b></font></center>";
       
        //below will redirect the uese to another page when completed
$url = 'http://www.yoursite.com/main.php';
 
//directory.

$content = 4;
exit("
      <meta
      http-equiv=\"refresh\"
      content=\"{$content};url={$url}\"
      >
      <script>
      function doRefresh() {
          window.location = '{$url}';
      }
      doRefreshTimeout = setTimeout(\"doRefresh()\" , ".
($content*200000).");
      </script>
  ");
//done with redirecting
}

else{       


echo "<br><center><font size=\"2\"><b>Ticket # $TicketNumber Has Been Deleted. </b></font></center>";

}

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it's better to use headers than JavaScript if you want to redirect; you can always show the error message on the next page.

PHP Code:

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// Redirect
header("Location: " . $url); 


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