弹不出json的值
来源:3-6 自由编程
我还差得多
2020-08-19 19:24:38
package com.imooc.ajax.text;
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 jQuery_ajax
*/
@WebServlet("/jQuery/ajax")
public class jQuery_ajax extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public jQuery_ajax() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
String butt=request.getParameter("idx");
System.out.println(butt);
List list= new ArrayList();
if(butt!=null&&butt.equals("001"))
list.add("稻香<br>晴天<br>告别气球<br>");
else if(butt!=null&&butt.equals("002"))
list.add("千千厥歌<br>傻女<br>七友<br>");
else if(butt!=null&&butt.equals("003"))
list.add("一块红布<br>假行僧<br>新长征路上的摇滚<br>");
String json = JSON.toJSONString(list);
response.setContentType("text/html;charset=utf-8");
System.out.println(json);
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
doGet(request, response);
}
}<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Insert title here</title>
<style type="text/css">
.button {
width: 600px;
text-align: center;
margin: 0 auto;
}
button {
width: 150px;
}
button:hover {
border: 1px solid blue;
}
</style>
</head>
<body>
<div id="button" class="button">
<button id="001">流行歌曲</button>
<button id="002">经典歌曲</button>
<button id="003">摇滚歌曲</button>
</div>
<div id="view"></div>
<script type="text/javascript" src="../js/jquery-3.5.1.js"></script>
<script type="text/javascript">
var idx;
$(function (){
$("button").click(function (){
idx=$(this).attr("id");
$.ajax({
"url":"/ajax/jQuery/ajax",
"type":"GET",
"data":{"idx":idx},
"dataType": "json",
"success": function (json){
alert(json);
}
})
})
})
</script>
</body>
</html>不清楚哪里有错,怎么也弹不出json的值
1回答
同学你好,没有输出json数据,所以获得不到数据,无法弹出json的值。
参考代码如下:

如果我的回答解决了你的疑惑,请采纳!祝学习愉快!
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