jquery ajaxSubmit 异步提交的简单实现

jquery ajaxsubmit 异步提交的简单实现代码如下:

$("#nickform").ajaxsubmit({
     type: "post",
     url: "https://localhost:8080/test/myspace.do?method=updatenick¶m=1",
     datatype: "json",
     success: function(result){

 

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

后台封装：

. 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();