jquery,在Ajax中获取数据(完成),然后将其显示为star(错误)

我使用的是星级,100%有效,除非我用javascript插入值:

例子: http://www.gamer-certified.fr/test.html

    <script type="text/javascript">
jQuery(document).ready(function() { 

jQuery.ajax({
      type: "get",
      dataType: "jsonp",
      url: "http://www.gamer-certified.fr/export/widget.php",
      data: {demandeur: "esl" },
      cache: true, 
      success: function(data, textStatus, XMLHttpRequest){

        var obj = null, length = data.length;
        for (var i = 0; i < length; i++) {

            widget = "<span class='stars'>"+(data[i].qualite / 2)+"</span>" // automatic not working
            widget = "<span class='stars'>4</span>" // manually not working
            jQuery('#gamer-certified'+i).html(widget);
            }    
    } 
});
});
</script>

唯一有效的方法是将数据直接放入小部件外部的跨度值中。+=

<span class="stars">4</span> // is working when directly put on the HTML side

例子: http://www.gamer-certified.fr/test.html

谢谢!