最近小弟在学习ajax，就随便做了一个ajax请求servlet并返回结果的小测试，但是一直调不通，一直是“网页上有错误”，看了一下XMLHttpRequest的值，一直是0。这是怎么回事啊？那位高手请指点一下，不胜感激！！！！！
希望实现功能：通过URL请求把参数传给ajaxServlet类处理，并显示处理结果！
index.jsp
<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%>
<%
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 'index.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">
<script type="text/javascript">
var xmlHttp;
function createXMLHttpRequest()
{
try{
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
}
catch(e)
{
try{
xmlHttp = new XMLHttpRequest();
}
catch(e)
{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
}
}
}
function startRequest()
{
createXMLHttpRequest();
xmlHttp.onreadystatechange = handleXMLHttpRequest;
xmlHttp.open("GET",basePath+"servlet/ajaxServlet?username=pangbo&password=19871201",true);
send(null);
}
function handleXMLHttpRequest()
{
if(xmlHttp.readyState==4)
{
if(xmlHttp.state==200)
{
alert(xmlHttp.responseText());
}
}
}
</script>
</head>
<body>
<input name="submit" value="提交" type="submit" onclick="startRequest()"/>
</body>
</html>
ajaxServlet类：
package com.ac.servlet;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.*;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class ajaxServlet extends HttpServlet{
/**
*
*/
private static final long serialVersionUID = 1L;
public void init(ServletConfig config) throws ServletException
{
super.init(config);
}
public void doGet(HttpServletRequest request,HttpServletResponse response) throws IOException,ServletException
{
String username = request.getParameter("username");
String password = request.getParameter("password");
String responseText = username+"的密码是"+password+"请牢记！！！！";
PrintWriter out = response.getWriter() ;
out.println(responseText);
out.close();
}
}
web.xml(应该没有配错)
<?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">
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>ajaxServlet</servlet-name>
<servlet-class>com.ac.servlet.ajaxServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ajaxServlet</servlet-name>
<url-pattern>/servlet/ajaxServlet</url-pattern>
</servlet-mapping>
</web-app>
