ajax得到django的回复，但我无法添加到模板中

Question

我有这样的代码.. JS

    $(".id_sort").bind('keypress',function(e){
    if (e.keyCode == '13'){
        var all_name = [];
        var all_pass = [];
        var index_fix = [];
        var index_ori = [];
        for (i=0;i < $("tbody tr").length; i++){
            index_ori.push(i);
        }
        $(".id_sort").each(function(){
            index_fix.push(Number($(this).val())-1); 
        });
        if (JSON.stringify(index_fix) !== JSON.stringify(index_ori)){
            data = { key : 'sort', index : JSON.stringify(index_fix)};
            $.ajax({
                url : "/ajax/",
                data : data,
                dataType : "json",
                type : "POST",
                cache : false,
                success : function (resp){
                    $(".data").html(resp);
                } 
            });     
        }
        else {
            alert("data sama coy");
        }
    }
});

和django查看

    if request.POST.get('key') == 'sort':
        index = json.loads(request.POST.get('index'))
        urut_fix = [int(i) for i in index]
        files = open('log.txt','r')
        read = files.readlines()
        data = []
        try :
            for i in read:
                data.append(i)
        except ValueError:
            pass
        urut_prev = range(len(data))
        for i in urut_prev :
            if i != (urut_fix[i]):
                get_data2 = data[i]
                data.pop(i)
                data.insert(urut_fix[i],get_data2)
        files.close()
        reads = open('log.txt', 'w')
        for y in data:
            reads.write(y)
        foo = []
        for i in data:
            foo.append(json.loads(i))
        return render_to_response("sort.html",locals())

code sort.html

    {% for i in foo%}
    <tr>
      {% if i.name and i.pwd%}
      <td class="name">{{ i.name}}</td>
      <td class="pass">{{i.pwd}}</td>
      <td>
        <input class="id_sort" type="text" name="sort" value="{{forloop.counter}}"/>
      </td>
      <td class="up" value="{{forloop.counter}}"onclick="up(this);">up</td>
      <td class="down" value="{{forloop.counter}}" onclick="down(this);">down</td>
      <td class="del" value="{{forloop.counter}}" onclick="del(this);">del</td>
    {% endif%}
    </tr>
  {% endfor%}

和我这样的模板

    <div class="hasil">
<table>
  <thead>
    <tr>
      <th>Name</th>
      <th>Password</th>
      <th>Sort</th>
      <th>Up</th>
      <th>Down</th>
      <th>Delete</th>
    </tr>
  </thead>
  <tbody class="data">
  {% for i in foo%}
    <tr>
      {% if i.name and i.pwd%}
      <td class="name">{{ i.name}}</td>
      <td class="pass">{{i.pwd}}</td>
      <td>
        <input class="id_sort" type="text" name="sort" value="{{forloop.counter}}"/>
      </td>
      <td class="up" value="{{forloop.counter}}"onclick="up(this);">up</td>
      <td class="down" value="{{forloop.counter}}" onclick="down(this);">down</td>
      <td class="del" value="{{forloop.counter}}" onclick="del(this);">del</td>
    {% endif%}
    </tr>
  {% endfor%}
  </tbody>
</table>

我希望添加对类数据的响应，.i获取响应但不添加到类，我的代码中有什么错误？我非常困惑我在django的新手..谢谢提前：D