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jqueryajaxSubmit异步提交的简单实现_jquery

时间:2021-07-01 10:21:17 帮助过:31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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