Cannot get Ajax update form to work with unserialize - php

Question

I'm trying to put together a working demo (including mysql) of a form that uses jquery to serialize form data and php to retrieve the data so that it can be passed to a mysql query.
The form seems to be posting data correctly but I'm not sure how to set up the processing script which sees $_POST to unserialize the data so that I can pass it to mysql. You can see the demo at http://www.dottedi.biz/demo/code/ajax/serialize .
I have tried using:
$data=unserialize($_POST['data']);
to unserialize the data but it comes back empty. A simple print_r ($_POST); returns the array data from the form. You can see that if you test the demo. Suggestions please?
Added info - the contents of the script.js file:
$(document).ready(function() {
$('form').submit(function(evt) {
evt.preventDefault();
$.each(this, function() {
var input = $(this);
var value = input.val();
var column = input.attr('name');
var form = input.parents('form');
var linked_col = form.find('#associated').attr('name');
var linked_val = form.find('#associated').val();
// var serializedData = $(this).serialize();
$("#Status").html( "" );
$.ajax({
url: "update.php",
data: {
val: value,
col: column,
id: linked_col,
id_val: linked_val
},
type: "POST",
success: function(html) {
$("#Status").html( html );
}
});
});
});
});
9/22 - shortened script.js
$(document).ready(function() {
$('form').submit(function(evt) {
evt.preventDefault();
$.each(this, function() {
$("#Result").html( "" );
$.ajax({
url: "update.php",
data: $('form').serialize(), // works to post data
type: "POST",
success: function(html) {
$("#Result").html( html );
}
});
});
});
});
Comment - I tested and it seems that the same data is posted using serialize as above vs creating a variable like var serializedData = $(this).serialize() and posting the variable, but this is shorter.

Answer 1

Perhaps you should
$('form').submit(function(evt) {
// Serialize the data in the form
var serializedData = $(this).serialize();
//send off serializedData in your ajax
}
THEN
php script will have
$data=json_decode($_POST['data']);
NEW WAY
$.ajax({
url: "update.php",
data: {'data': $('form').serialize()}, // works to post data
type: "POST",
success: function(html) {
$("#Result").html( html );
}