老师,请问一下我的代码哪里错了,为什么老是404错误啊,那个data类型到底该怎么填啊?
来源:1-19 自由编程
qq_慕容1283338
2019-09-21 19:33:13
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title>Ajax使用JQuery相关代码实现</title> </head> <style type="text/css"> body{ position:relative; } input{ width:250px; height:30px; } span{ width:800px; position:absolute; left:50%; margin-left:-400px; } #container{ width:250px; height:100px; margin-top:40px; text-align:center; position:absolute; left:50%; margin-left:-145px; } </style> <body> <span> <input type="button" value="流行歌曲" name="popular"> <input type="button" value="经典歌曲" name="classic"> <input type="button" value="摇滚歌曲" name="rock"> </span> <div id = "container"></div> <script type="text/javascript" src="js/jquery-3.4.1.js"></script> <script type="text/javascript"> $(function(){ $("input").click(function(){ var Name = $(this).attr("name"); console.log(bName); $.ajax({ "url":"/Jquery/song", "type":"get", "data":{"name":Name}, "dataType":"json", "sucess":function(json){ for (var i = 0 ; i < json.length ; i++){ $("#container").append("<h3>" + json[i] + "</h3>"); console.log(json[i]); } }, "error":function(xmlhttp,errorText){ if(xmlhttp.status=="404"){ alert("未找到url资源"); }else if(xmlhttp.status == "405"){ alert("无效的请求方式"); }else if(xmlhttp.status =="500"){ alert("服务器错误,请联系管理员"); }else{ alert("请联系管理员"); } } }) }) }) </script> </body> </html>
package com.imooc.ajax; import java.io.IOException; import java.util.ArrayList; import java.util.List; import javax.servlet.ServletException; import javax.servlet.annotation.WebServlet; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; import com.alibaba.fastjson.JSON; /** * Servlet implementation class songList */ @WebServlet("/song") public class songList extends HttpServlet { private static final long serialVersionUID = 1L; /** * @see HttpServlet#HttpServlet() */ public songList() { super(); // TODO Auto-generated constructor stub } /** * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response) */ protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { String name = request.getParameter("name"); List<String> list = new ArrayList<String>(); if (name.equals("popular")) { list.add("稻香"); list.add("晴天"); list.add("告白气球"); }else if (name.equals("classic")) { list.add("千千阙歌"); list.add("傻女"); list.add("七友"); }else if (name.equals("rock")) { list.add("一块红布"); list.add("假行僧"); list.add("新长征路上的摇滚"); } String json = JSON.toJSONString(list); System.out.println(json); response.setContentType("text/html;charset=UTF-8"); response.getWriter().println(json); } }
2回答
好帮手慕小班
2019-09-23
同学你好,1、复制运行贴出代码,并没有出现404错误呐,建议同学将自己项目目录、报错信息截图贴出呐。
2、贴出代码中,输出内容没有出现是因为success单词书写错误,例如:
如上修改,就可以将json内容展示到页面中。
如果我的回答解决了你的疑惑,请采纳。祝:学习愉快~
慕容6195653
2019-09-21
html中47行,"data":{"name":Name}, 不对吧,改成 "data" : "name=rock" 试试呢
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