I am working on a drop down box that loads its options dynamically from an xml file. The XML file has two nodes for each option: the title node is what gets displayed in the drop down and the 'xml' node which contains the data that actually gets submitted. When a user submits the form, the action is processed by the php in the top of the page. Essentially, submitted data is turned in to a session variable at the top of the page. I need to figure out how to display the user's last choice in the dropdown menu. Since the data submitted is not the same as the data displayed, I can't figure out a way to show this... Any suggestions?

Code:
<?php
session_start(); // start up your PHP session! 
$_SESSION['views'] = $_POST['selectBox'];
?>

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Sample</title>

<script type="text/javascript" src="js/jquery-1.3.2.min.js"></script>

<script type="text/javascript">
     	$(document).ready(function(){
			$.ajax({
				type: "GET",
				url: "galleries.xml",
				dataType: "xml",
				success: function(xml) {
					var select = $('#mySelect');
					$(xml).find('category').each(function(){
					var title = $(this).find('title').text();
					var val = $(this).find('xml').text();
					select.append("<option value='"+val+"'>"+title+"</option>");
					});
select.children(":first").text("please make a selection").attr("selected",true);
				}
			});
		});
</script>

</head>
<body>
		<form name="myform" action="index.php" method="post" enctype="multipart/form-data">
           <select name="selectBox" id="mySelect" onChange="document.myform.submit()">
              <option>loading</option>
           </select>
       </form>
      
</body>
</html>