On 1/17/08, mike <mike503@xxxxxxxxx> wrote:
> On 1/17/08, Nathan Nobbe <quickshiftin@xxxxxxxxx> wrote:
>
> > > $name = $_POST['name'];
> > > if ($name) {
> > > foreach ($name as $t) {
> > >
> > > echo "$t";
> > > }
>
> > > $order = $_POST['order'];
> > > if ($order) {
> > > foreach ($order as $i) {
>
> >
> > there are a few different issues here; first of all; are you sure
> > $_POST['name']
> > and $_POST['order'] are even arrays?
>
> hint:
>
> if(isset($_POST['name']) && is_array($_POST['name']))
Steve,
// Do you have several html form elements such as <input type="text"
name="name[]"> in your html?
// Mike's suggestion...
if( isset( $_POST['name'] ) && is_array( $_POST['name'] ) ) {
// you'll never get in here if you don't...
$name = $_POST['name'];
// foreach expects an array, as Nathan states. Even if $name is an
array, $t
// will hold only the last value in the array when the foreach loop
is exited
// because $t is being overwritten with each iteration.
foreach ($name as $t) {
echo "$t";
} // end foreach ($name)
$order = $_POST['order'];
if ($order) {
// see above about arrays and foreach
foreach ($order as $i) {
//Update the table in MySQL
$i = mysql_real_escape_string($i, $cnx); // One of
Eric's suggestions
$update_data = "UPDATE sections SET `order` = '$i'
WHERE name = '$t'";
$response = mysql_query( $update_data, $cnx );
if(mysql_error()) die ('database error<br>'.
mysql_error());
echo "$i";
} //end foreach ($order)
}
}
Assuming both $_POST['name'] and $_POST['order'] are arrays, the way your
code is now structured, the table `sections` will have the record(s) where
name equals the last value in the $names array updated multiple times, once
for each value in the $order array, but all you will see is that
the record(s) will have the last value in the $order array.
See if this makes any sense and then ask more questions.
David
