ajax 为什么不能进入readyState方法
这是servlet类package com.ajax.servlet;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class AjaxServlet extends HttpServlet {
protected void doGet(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
PrintWriter out=resp.getWriter();
System.out.println("doGet");
out.println("Hello World");
out.flush();
}
}
这是jsp页面
<%@ page language="java" import="java.util.*" pageEncoding="GBK"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<base href="<%=basePath%>">
<title>My JSP 'ajax.jsp' starting page</title>
<meta http-equiv="pragma" content="no-cache">
<meta http-equiv="cache-control" content="no-cache">
<meta http-equiv="expires" content="0">
<meta http-equiv="keywords" content="keyword1,keyword2,keyword3">
<meta http-equiv="description" content="This is my page">
<!--
<link rel="stylesheet" type="text/css" href="styles.css">
-->
<script type="text/javascript">
function ajaxSubmit(){
var XMLHttpRequest=null;
if (window.ActiveXObject){
XMLHttpRequest = new ActiveXObject("Microsoft.XMLHTTP");
}else if(window.XMLHttpRequest){
XMLHttpRequest=new XMLHttpRequest();
}
if(XMLHttpRequest!=null){
XMLHttpRequest.open("GET","AjaxServlet",true);
XMLHttpRequest.onreadystatechange=ajaxCallback;
XMLHttpRequest.send(null);
}
}
function ajaxCallback(){
if (XMLHttpRequest.readyState == 4){
if (XMLHttpRequest.status == 200){
//document.getElementById("chatArea").value = XMLHttpRequest.responseText;
var respText= XMLHttpRequest.responseText;
document.getElementById("div1").innerHTML=respText;
}
else{
window.alert("您所请求的页面有异常。");
}
}
}
</script>
</head>
<body>
<input type="button" value="get content from server" onclick="ajaxSubmit();">
<div id="div1"></div>
</body>
</html>
这是我的web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<display-name></display-name>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<description></description>
<display-name>AjaxServlet</display-name>
<servlet-name>AjaxServlet</servlet-name>
<servlet-class>com.ajax.servlet.AjaxServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>AjaxServlet</servlet-name>
<url-pattern>/AjaxServlet</url-pattern>
</servlet-mapping>
</web-app>
求给位大神帮忙 ajax java servlet web.xml javascript --------------------编程问答-------------------- 看下你的ajax 返回方法成功没? 看下具体状态码
out.write("Hello World");
--------------------编程问答-------------------- ajax返回的方法没有问题,但是我在jsp用alert(XMLHttpRequest.readyState)结果显示的是undefined --------------------编程问答--------------------
ajax返回的方法没有问题,但是我在jsp用alert(XMLHttpRequest.readyState)结果显示的是undefined --------------------编程问答--------------------
response.setContentType("text/xml; charset=UTF-8");--------------------编程问答-------------------- 你debug 看下 XMLHttpRequest 的属性 --------------------编程问答-------------------- 我记得貌似以前有人遇到这类问题,后面的做法是
response.setHeader("Cache-Control", "no-cache");
response.setHeader("Pragma", "no-cache");
PrintWriter out=resp.getWriter();
System.out.println("doGet");
out.write("return ajax value");
out.close();
var XMLHttpRequest=null;这段代码另外放在一个function里面的
if (window.ActiveXObject){
XMLHttpRequest = new ActiveXObject("Microsoft.XMLHTTP");
}else if(window.XMLHttpRequest){
XMLHttpRequest=new XMLHttpRequest();
}
补充:Java , Web 开发