麻烦看下 ，我哪里错了，没显示出来？
来源：2-3 发送请求及处理响应
兮兮666
2019-09-12 22:28:34
package com.imooc.ajax;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
/**
* Servlet implementation class ContentServlet
*/
@WebServlet("/content")
public class ContentServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public ContentServlet() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.getWriter().print("<b>I'm server content</b>");
}
}
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Insert title here</title>
</head>
<body>
<input id="btnLoad" type="button" value="加载">
<div id="divContent"></div>
<script type="text/javascript">
document.getElementById("btnLoad").onclick=function(){
//1.创建XmlHttpRequest对象
var xmlhttp;
if(window.XMLHttpRequest){
xmlhttp=new XMLHttpRequest();
}else{
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP")
}
//console.log(xmlhttp);
//2.发送Ajax
xmlhttp.open("GET","/ajax/content",true);
xmlhttp.send();
//3.处理服务器响应
xmlhttp.onreadystatechange=function(){
if(xmlhttp.readState==4 && xmlhttp.status==200){
var tt=xmlhttp.responseText;
document.getElementById("divContent").innerHTML=tt;
}
}
}
</script>
</body>
</html>
1回答
-
同学你好，在处理服务器响应的if判断中，readyStatus的单词拼写错误，少了y，如:
建议同学修改一下再试试。
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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