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?