action传来的josn数据放到写好的table中，js怎么写

Question

代码
jQuery(function(){
          $.ajax({
           url: "/cjTest/worker_list.do",
           type:'post', 
           dataType: "json",
           success: function(data){
             $.each(data.user, function(i) {
                   $("#workerlistTbl").html(data.workerJson);
              });
           }
         }); 
    }); table：<table id="workerlistTbl" style="width:450px;height:auto"
title="Editable DataGrid" iconCls="icon-edit" singleSelect="true"
idField="name" >
<tr>
<th field="name" width="80"  >员工姓名</th>  
<th field="department" width="100" formatter="productFormatter" editor="{type:'combobox',options:{valueField:'department',textField:'name',data:departments,required:true}}">所属部门</th>
<th field="sex" width="80"  formatter="sexeach" editor="{type:'combobox',options:{valueField:'sexid',textField:'sex',data:sexsel,required:true}}" >员工性别</th>                                                                                                                                                                    
<th field="entrydate" width="60" align="center" >入职时间</th>
</tr>
</table>

Answer 1

 <table id="dg" title="实时监测数据" class="easyui-datagrid" style="width:777px;height:460px"  
            url="/SZGW/RTDServlet"  
           toolbar="#toolbar" pagination="true"  
            rownumbers="true" fitColumns="true" singleSelect="true">  
        <thead>  
            <tr>  
                <th field="rid" width="60">ID</th>  
                <th field="did" width="60">DID</th>  
                <th field="value" width="60">Value</th>  
                <th field="date" width="120">Time</th>  
                <th field="deviceid" width="60">Deviceid</th>
           
            </tr>  
        </thead>  
    </table>  
直接url对应就好了

Answer 2

谢谢你的 帮忙，我前台这样写的  
            JSONArray jsonArray = JSONArray.fromObject(workerlist);   
            JSONObject jsonObject = new JSONObject();   
            jsonObject.put("workerJson", jsonArray);   
            response.getWriter().write(jsonObject.toString());    
只写 url  不行吧 ，怎么知道获取的是workerJson？

Answer 3

js 方法中返回一个 josn 格式的data