如何在Java Web应用程序中过期会话时重定向到“登录”页面？

Question

I'm running a web application in JBoss AS 5. I also have a servlet filter which intercepts all the requests to the server. Now, I want to redirect the users to the login page, if the session has expired. I need to do this 'isSessionExpired()' check in the filter and need to redirect the user accordingly. How do I do it? I'm setting my session time limit in web.xml, as below:

<session-config>
    <session-timeout>15</session-timeout>
</session-config>

Answer 1

You could use a Filter and do the following test:

HttpSession session = request.getSession(false);// don't create if it doesn't exist
if(session != null && !session.isNew()) {
    chain.doFilter(request, response);
} else {
    response.sendRedirect("/login.jsp");
}

The above code is untested .

This isn't the most extensive solution however. You should also test that some domain-specific object or flag is available in the session before assuming that because a session isn't new the user must've logged in. Be paranoid !

Answer 2

How to redirect to Login page when Session is expired in Java web application?

This is a wrong question. You should differentiate between the cases "User is not logged in" and "Session is expired". You basically want to redirect to login page when user is not logged in. Not when session is expired. The currently accepted answer only checks HttpSession#isNew() . But this obviously fails when the user has sent more than one request in the same session when the session is implicitly created by the JSP or what not. Eg when just pressing F5 on the login page.

As said, you should instead be checking if the user is logged in or not. Given the fact that you're asking this kind of question while standard authentication frameworks like j_security_check , Shiro, Spring Security, etc already transparently manage this (and thus there would be no need to ask this kind of question on them), that can only mean that you're using a homegrown authentication approach.

Assuming that you're storing the logged-in user in the session in some login servlet like below:

@WebServlet("/login")
public class LoginServlet extends HttpServlet {

    @EJB
    private UserService userService;

    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        request.getRequestDispatcher("/WEB-INF/login.jsp").forward(request, response);
    }

    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        String username = request.getParameter("username");
        String password = request.getParameter("password");
        User user = userService.find(username, password);

        if (user != null) {
            request.getSession().setAttribute("user", user);
            response.sendRedirect(request.getContextPath() + "/home");
        } else {
            request.setAttribute("error", "Unknown login, try again");
            doGet(request, response);
        }
    }

}

Then you can check for that in a login filter like below:

@WebFilter("/*")
public class LoginFilter implements Filter {

    @Override
    public void doFilter(ServletRequest req, ServletResponse res, FilterChain chain) throws ServletException, IOException {    
        HttpServletRequest request = (HttpServletRequest) req;
        HttpServletResponse response = (HttpServletResponse) res;
        HttpSession session = request.getSession(false);
        String loginURI = request.getContextPath() + "/login";

        boolean loggedIn = session != null && session.getAttribute("user") != null;
        boolean loginRequest = request.getRequestURI().equals(loginURI);

        if (loggedIn || loginRequest) {
            chain.doFilter(request, response);
        } else {
            response.sendRedirect(loginURI);
        }
    }

    // ...
}

No need to fiddle around with brittle HttpSession#isNew() checks.

Answer 3

you can also do it with a filter like this:

public class RedirectFilter implements Filter {

public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException {
    HttpServletRequest req=(HttpServletRequest)request;

    //check if "role" attribute is null
    if(req.getSession().getAttribute("role")==null) {
        //forward request to login.jsp
        req.getRequestDispatcher("/login.jsp").forward(request, response);
    } else {
        chain.doFilter(request, response);
    }
}
}

Answer 4

Check for session is new.

HttpSession session = request.getSession(false);
if (!session.isNew()) {
  // Session is valid
}
else {
  //Session has expired - redirect to login.jsp
}

Answer 5

Inside the filter inject this JavaScript which will bring the login page like this. If you don't do this then in your AJAX call you will get login page and the contents of login page will be appended.

Inside your filter or redirect insert this script in response:

String scr = "<script>window.location=\""+request.getContextPath()+"/login.do\"</script>";
response.getWriter().write(scr);

Answer 6

You need to implement the HttpSessionListener interface, server will notify session time outs.

like this;

import javax.servlet.http.HttpSessionEvent;
import javax.servlet.http.HttpSessionListener;

public class ApplicationSessionListener implements HttpSessionListener {

public void sessionCreated(HttpSessionEvent event) {
   System.out.println("Session Created");
 }

public void sessionDestroyed(HttpSessionEvent event) {
   //write your logic
   System.out.println("Session Destroyed");
  }
 }

Check this example for better understanding

http://www.myjavarecipes.com/how-to-catch-session-timeouts/

Answer 7

Until the session timeout we get a normal request, after which we get an Ajax request. We can identify it the following way:

String ajaxRequestHeader = request.getHeader("X-Requested-With");
if ("XMLHttpRequest".equals(ajaxRequestHeader)) {
    response.sendRedirect("/login.jsp");
}

Answer 8

i found this posible solution:

public void logout() {
    ExternalContext ctx = FacesContext.getCurrentInstance().getExternalContext();
    String ctxPath = ((ServletContext) ctx.getContext()).getContextPath();
    try {
        //Use the context of JSF for invalidate the session,
        //without servlet
        ((HttpSession) ctx.getSession(false)).invalidate();
        //redirect with JSF context.
        ctx.redirect(ctxPath + "absolute/path/index.jsp");
    } catch (IOException ex) {
        System.out.println(ex.getMessage());
    }
}

Answer 9

When the use logs in, put its username in the session:

`session.setAttribute("USER", username);`

At the beginning of each page you can do this:

<%
String username = (String)session.getAttribute("USER");
if(username==null) 
// if session is expired, forward it to login page
%>
<jsp:forward page="Login.jsp" />
<% { } %>