jquery ajaxSubmit 异步提交的简单实现

前台js

代码如下：

$("#nickForm").ajaxSubmit({
     type: "post",
     url: "http://localhost:8080/test/myspace.do?method=updateNick&param=1",
     dataType: "json",
     success: function(result){

           //返回提示信息       
           alert(result.nickMsg);
     }
 });

代码如下：

public ActionForward toUpdateNickName(ActionMapping mapping, ActionForm form,
   HttpServletRequest request, HttpServletResponse response){

       PrintWriter pw = response.getWriter();
      JSONObject obj = new JSONObject();


      obj.put("nickMsg", "昵称修改成功！");


      pw.print(obj);
      pw.close();


}