老师帮我看看该怎样修改
来源:2-8 自由编程
Wonwayshon
2020-10-12 16:03:23
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Ajax</title>
<style type="text/css">
input{
width:200px;
height:25px;
}
</style>
</head>
<body>
<div style="text-align:center;">
<input type="button" value="员工列表" class="btn" name="0">
<input type="button" value="职位列表" class="btn" name="1">
<input type="button" value="部门列表" class="btn" name="2">
<div id="div1"></div>
</div>
<script type="text/javascript">
var btn=document.getElementByClassName("btn");
for(var i=0;i<btn.length;i++){
btn[i].onclick=function(){
var xmlhttp,
name=this.name;
if(window.XMLHttpRequest){
xmlhttp=new XMLHttpRequest();
}else{
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.open("get","/Ajax/Ajax?name="+name,true);
xmlhttp.send();
xmlhttp.onreadystatechange=function(){
if(xmlhttp.readyState==4 && xmlhttp.status==200){
var json=JSON.parse(xmlhttp.responseText);
str="";
for(int i=0;i<json.length;i++){
str+=json[i]+"</br>";
}
document.getElementById("div1").innerHTML=str;
}
}
}
}
</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 Ajax
*/
@WebServlet("/Ajax")
public class Ajax extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public Ajax() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
List el=new ArrayList();
List pl=new ArrayList();
List dl=new ArrayList();
el.add("小红");
el.add("小明");
el.add("小白");
pl.add("职员");
pl.add("经理");
dl.add("人事部");
dl.add("技术部");
dl.add("无线事业部");
String els=JSON.toJSONString(el);
String pls=JSON.toJSONString(pl);
String dls=JSON.toJSONString(dl);
String name=(String)(request.getParameter("name"));
String responseString=null;
if(name.equals("0")) {
responseString=els;
}else if(name.equals("1")) {
responseString=pls;
}else if(name.equals("2")) {
responseString=dls;
}
response.setContentType("text/html;charset=utf-8");
response.getWriter().println(responseString);
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
doGet(request, response);
}
}1回答
同学你好,1. getElementsByClassName少写了一个s,建议同学添加上。如下所示:

2. 在JavaScript中定义变量使用var即可,并不需要使用数据类型。如下所示:

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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