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jquery ajaxSubmit 异步提交的简单实现

时间:2021-07-01 10:21:17 帮助过:39人阅读

前台js
代码如下:
$("#nickForm").ajaxSubmit({
     type: "post",
     url: "http://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();

}

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