Question

I am stumped!! I can’t figure this out.

I created an HTML form that inserts a record into mySQL. It works and I can see the new records I add/insert. BUT, I get the wrong confirmation page: I get a the FAIL PAGE instead of the SUCCESS page. I see the new record but I always get taken to the fail page. Why?

Is there something wrong with the script or a setting inside mySQL?

Here is my form post script:

<? 
$host="XXXXXXXXXXXX";
$username="XXXXXXXX";
$password="XXXXXXXX";
$db_name="XXXXXXXXX";
$tbl_name="cartons_current";

mysql_connect("$host", "$username", "$password") or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

$order = "INSERT INTO cartons_current (type, part_no, description, count, 
size, min, max, qty)

VALUES
('$_POST[type]', '$_POST[part_no]', '$_POST[description]', '$_POST[count]',
'$_POST[size]', '$_POST[min]', '$_POST[max]', '$_POST[qty]')";

$result = mysql_query($order);

$result = mysql_query($order);  //order executes
if ($result) {

$part_no = $_REQUEST['part_no'] ;

header("location: inv_fc_result_new_success.php?part_no=" . urlencode($part_no));
}
else {
header("location: inv_fc_result_new_fail.php");
}
?>

Answer 1

Your code looks OK, except for the possibility that mysql_query() gets called twice. If that is actual code, then I suspect the first call loads the record you are seeing, and the subsequent call returns the error message.

$result = mysql_query($order);  
$result = mysql_query($order);  //order executes