jquery ajaxSubmit 异步提交的简单实现

网络编程 2024/11/15 佚名

3 2 1

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, 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();

}

jquery,ajaxSubmit,异步提交