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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