当我用AJAX像服务器发送一个登录请求后，服务器验证数据是否正确，是则跳到另一个
页面，否则返回一个数据错误的信息给AJAX，页面不刷新。
当验证正确时我用的是servlet的
request.getRequestDispatcher("../index.jsp").forward(request, response);跳转方式
错误就直接out.print();可是不管怎样都是跳到了那个请求页面
请问要怎样才能跳到另一个页面，还能带着数据过去
ajax代码
//声明XMLHttpRequest对象
var xmlrequest;
//初始化XMLHttpRequest
function createXMLHttpRequest(){
if(window.XMLHttpRequest){
xmlrequest = new XMLHttpRequest();
}else if(window.ActiveXObject){
try{
xmlrequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch(e){
try{
xmlrequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch(e){
}
}
}
}
//定义回调函数
function processResponse(){
//响应完成
if(xmlrequest.readyState == 4){
//响应正常
if(xmlrequest.status == 200){
var head = xmlrequest.responseText;
alert(head);
}
}
}
servlet代码
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
System.out.println("post被触发了");
request.setCharacterEncoding("UTF-8");
response.setContentType("text/html;charset=UTF-8");
String account = request.getParameter("account");
String password = request.getParameter("password");
if (account == "admin" & password == "admin"){
request.getRequestDispatcher("../index.jsp").forward(request, response);
}else{
PrintWriter out = response.getWriter();
out.print("错误");
}
