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copied!<p>You can solve this the following way:</p> <ul> <li><p>Tell the browser to not cache the login page, so that it will always send a fullworthy GET request on it instead of loading from the browser cache.</p></li> <li><p>When the login page is requested while the user is already logged-in, then send a redirect to the desired page (which you've remembered in the session when the user logs in).</p></li> </ul> <p>This could be done in one and same <a href="https://stackoverflow.com/tags/servlet-filters/info">filter</a> (although I'd personally prefer to disable browser caching on all change-sensitive dynamic pages, so that would then be done in another filter on a more generic URL pattern covering those pages).</p> <p>Here's a kickoff example assuming that you're using container managed authentication (for homegrown authentication wherein you manually put the user in the session, just check the presence of the session attribute instead):</p> <pre><code>@WebFilter("/login.xhtml") public class LoginPageFilter implements Filter { @Override public void doFilter(ServletRequest req, ServletResponse res, FilterChain chain) throws IOException, ServletException { HttpServletRequest request = (HttpServletRequest) req; HttpServletResponse response = (HttpServletResponse) res; HttpSession session = request.getSession(); // Tell browser to not cache this page. response.setHeader("Cache-Control", "no-cache, no-store, must-revalidate"); // HTTP 1.1. response.setHeader("Pragma", "no-cache"); // HTTP 1.0. response.setDateHeader("Expires", 0); // Proxies. // Check if user is logged in. if (request.getRemoteUser() == null) { // Not logged in, so remember previous URL and continue request. session.setAttribute("referrer", request.getHeader("referer")); // Yes, the misspelling is "correct"! chain.doFilter(request, response); } else { // Logged in, so redirect to initial referring URL, if any. String referrer = (String) session.getAttribute("referrer"); if (referrer != null) { response.sendRedirect(referrer); } else { // There was no referring page. Just continue request? chain.doFilter(request, response); } } } // ... } </code></pre>

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