Passing variable?

[Anon]

Hi people,
Can anyone tell me why the $pay_id value that is sent with the header() gives a "0" value all the time. Why can't i get the correct id value to show?

<?
{

  $pay_id = doubleval($pay_id);
  $name = addslashes($name);
  $date_in = date("Ymd");


  $db = mysql_connect('localhost', '', '');

  if (!$db)
  {
    echo "Error: Could not connect to database. Please try again later.";
	exit;
  }

  mysql_select_db("rip");

  $query = "insert into securepay(pay_id, name, address, city, postcode, telephone, instructions, ponum, optional_info, amount, merchant_id, date_in) 
  values('$pay_id','$name','$address','$city','$postcode','$telephone','$instructions','$ponum','$optional_info','$amount','$merchant_id','$date_in')";

  $result=mysql_query($query);

  if($result)
   header ("Location: sales_cc.php?id=$pay_id");

}

echo "<br><a href=\"membership.php\">Back to Members</a>";
?>

Any thoughts or advice?
Suzie

[rrijnders]

You have an unconditioned block {curly braces} around most of the code???

You also never set the value of $pay_id to anything. Where does it originally come from?

In your header() function, you are passing the variable as "id" not "pay_id"?

-- Rich

[Anon]

Thanks for the reply.
I still don't understand how to set the $pay_id value. This value goes into the pay_id field with the database. It is set to auto_increment, primary key. So the $pay_id value is obviously made up.
Do you have any ideas about this.
I removed the {curly braces} and set the "id" to "pay_id"
Cheers,
suzie