I've found this drag and drop script and it's brilliant but I can't seem to get it to update the results in the database.
I've got this on the page:
the alert box does give the correct values, but when I call the ajaxtest.php page nothing happens.Code:<script language="Javascript"> $(document).ready(function() { // Initialise the table $('#images').tableDnD({ onDragClass: "drag", onDrop: function(table, row) { alert($.tableDnD.serialize()); $.ajax({ type: "POST", url: "http:/www.domain.com/new_site/admin/ajaxtest.php", data: "" + $.tableDnD.serialize(), success: function(html){ } }); } }); }); </script>
The ajaxtest.php page has this:
I'm pretty sure the problem is with the $images[] = $_POST['images']; but of code, but I'm not sure exactly where and would be extremely grateful for any help.Code:<?php include '../includes/db.php'; $images[] = $_POST['images']; $i = 0; if(!empty($images[0])){ foreach($images as $value) { foreach($value as $row){ $i++; if($row != 'tHeader'){ $this->Image->set('id', $row); $this->Image->set('order', $i); $this->Image->save(); } } } } $con = mysql_connect($host,$username,$password); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db($database, $con); $sql= sprintf("UPDATE `images` SET `order`='$i' WHERE `id`='$row'"); if (!mysql_query($sql,$con)) mysql_close($con) ?>



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