sto cercando di adattare questo script
http://www.zurb.com/playground/ajax_upload
per il mio sito
il problema è che il previev non si vede...

ecco il codice
Codice PHP:

<!DOCTYPE html>

<html xmlns="http://www.w3.org/1999/xhtml">
    <head>
    
        
<script src="../programmi/imagezoom/jquery-latest.js" type="text/javascript"></script><script src="ajaxupload.js" type="text/javascript"></script>


    <script>
$(document).ready(function(){

    var thumb = $('img#thumb');    

    new AjaxUpload('imageUpload', {
        action: $('form#newHotnessForm').attr('action'),
        name: 'image',
        onSubmit: function(file, extension) {
            $('div.preview').addClass('loading');
        },
        onComplete: function(file, response) {
            thumb.load(function(){
                $('div.preview').removeClass('loading');
                thumb.unbind();
            });
            thumb.attr('src', response);
        }
    });
});
        
        function initZURB(){}
    </script>

                
    </head>
    
    <body>



        <h2>New Hotness</h2>
        <div class="column-row">
            <div class="seven columns">
                <div class="preview">
                    [img]http://www.zurb.com/images/icons/128px/zurb.png[/img]
                </div>

                <span class="wrap hotness">
                    <form id="newHotnessForm" action="/playground/ajax_upload">
                        <label>Upload a Picture of Yourself</label>
                        <input type="file" id="imageUpload" size="20" />
                        <button type="submit" class="button">Save</button>
                    </form>
                </span>
            </div> 

qualcuno puo' aiutarmi?