[error_22]

Yes I'm aware of that. Now I'm looking for a solution.....

I tried the following:

PHP Code:

echo "<form action='page.php?action=order' method='POST'>"; 
$sql = "SELECT * FROM `page` ORDER BY `id` DESC"; 
    $result = mysql_query($sql) or die(mysql_error()); 
    while ($row = mysql_fetch_object($result)) 
{ 
echo "<input type='text' name='order[$row->id]' value='$row->order'>"; 
} 
echo "<input type='submit' value='save'></form>"; 


In page.php?action=order:

PHP Code:

    $sql = "SELECT * FROM `page` ORDER BY `id` DESC";
    $result = mysql_query($sql) or die(mysql_error());
    while ($row = mysql_fetch_object($result))
    {
        $insert = $_POST["order[$row->id]"];
    
        $sql = "UPDATE `page` SET `order` = '$insert' ORDER BY `id` DESC";
        mysql_query($sql) or die(mysql_error());
    } 


Sadly, it didnt work. Didnt think it would but it was worth a try atleast. For some reason, the mysql query can only store one value, and tha value is stored in every row.