jquery ajaxSubmit 异步提交的简单实现 - Web前端

前台JS
复制代码 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);
     }
 });

后台封装：
复制代码 代码如下:
public ActionForward toUpDateNickName(ActionmapPing mAPPing, ActionformForm,
   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();

}

标签：简单WebajaxSubmitjquery