麻烦老师指点一下 为什么点击员工列表按钮就报错
来源：2-8 自由编程
慕函数4495222
2019-05-30 00:04:28
Servlet类代码
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 Practise
*/
@WebServlet("/practice")
public class Practise extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public Practise() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
List list1 = new ArrayList<>();
list1.add("小红");
list1.add("小明");
list1.add("小白");
List list2 = new ArrayList<>();
list2.add("职员");
list2.add("经理");
List list3 = new ArrayList<>();
list3.add("人事部");
list3.add("技术部");
list3.add("无线事业部");
String json1 = JSON.toJSONString(list1);
String json2 = JSON.toJSONString(list2);
String json3 = JSON.toJSONString(list3);
response.setContentType("text/html;charset=UTF-8");
if(request.getParameter("id").equals("btn1") ) {
System.out.println(json1);
response.getWriter().println(json1);
}else if(request.getParameter("id").equals("btn2") ) {
System.out.println(json2);
response.getWriter().println(json2);
}else if(request.getParameter("id").equals("btn3")) {
System.out.println(json3);
response.getWriter().println(json3);
}
}
}
html代码
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Insert title here</title>
</head>
<body>
<style>
.container{
width:100%;
margin:0 auto;
text-align: center;
}
.btn{
width:33%;
height:50px;
}
</style>
<div class="container">
<input type="submit" value="员工列表" class="btn" id="btn1"/>
<input type="submit" value="职位列表" class="btn" id="btn2"/>
<input type="submit" value="部门列表" class="btn" id="btn3"/>
<script type="text/javascript" src="js/jquery-3.3.1.js"></script>
<script type="text/javascript">
document.getElementById("btn1").onclick=function(){
var xmlhttp;
if(window.XMLHttpRequest){
xmlhttp=new XMLHttpRequest();
}else{
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
console.log(xmlhttp);
xmlhttp.open("GET","/jquery/practice",true);
xmlhttp.send();
}
</script>
</div>
</body>
</html>
1回答
-
好帮手慕柯南
2019-05-30
同学你好！同学的eclipse的控制台是否出现了这样的异常呢
原因是因为不能通过id来回去按钮中的value呢
可以在ajax请求的地址中拼接id
如果我的回答解决了你的疑惑，请采纳，朱学习愉快~
0
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