急,急,急 差点砸电脑了 !一个最简单的AJAX例子 怎么跑不起来
JSP部分:
<%@ page language= "java " import= "java.util.* " pageEncoding= "utf-8 "%>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN " "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd ">
<html>
<head>
<title> 无标题文档 </title>
<script language= "javascript ">
var XMLHttpReq = new ActiveXObject( "MSXML2.XMLHTTP.3.0 ");
function callServer() {
XMLHttpReq.open( "GET ", "servlet/ajax_sevelet ", "true ");
XMLHttpReq.onreadystatechange = updatePage;
XMLHttpReq.send(null);
}
function send_msg(){
callServer();
}
</script>
</head>
<body>
<form id= "form1 " name= "form1 " method= "post " action= " ">
姓名: <label>
<input type= "text " ondblclick= "send_msg() " name= "name " />
</label> <p>
密码: <label>
<input type= "text " name= "password " />
</label>
</form>
<a href= "servlet/ajax_sevelet "> ei </a> /* 点这里可以调用sevlet,证明路径没错啊
</body>
</html>
sevlet部分:
package shiyan;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileReader;
import java.io.FileWriter;
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 ajax_sevelet extends HttpServlet {
/**
* Destruction of the servlet. <br>
*/
public void destroy() {
super.destroy(); // Just puts "destroy " string in log
// Put your code here
}
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws
ServletException,
IOException {
doPost(request,response);
}
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
int i=1;
response.setContentType( "text/html ");
PrintWriter out = response.getWriter();
String name=request.getParameter(