如何在JSF中实现登录过滤器

Question

我想阻止某些页面的访问，即使用户知道某些页面的url。例如，/localhost:8080/user/home.xhtml(需要先登录)如果没有登录，则重定向到/index.xhtml。

如何在JSF中实现?我在谷歌上看到需要一个过滤器，但我不知道怎么做

Answer 1

您需要实现javax.servlet.Filter类，在doFilter()方法中做所需的工作，并将其映射到覆盖受限页面的URL模式上，/user/*可能?在doFilter()中，您应该以某种方式检查会话中是否存在登录用户。此外，您还需要考虑JSF ajax和资源请求。JSF ajax请求需要一个特殊的XML响应来让JavaScript执行重定向。需要跳过JSF资源请求，否则您的登录页面将不再有任何CSS/JS/图像。

假设您有一个/login.xhtml页面，该页面通过externalContext.getSessionMap().put("user", user)将登录的用户存储在JSF管理的bean中，那么您可以通过session.getAttribute("user")以通常的方式获得它，如下所示:

@WebFilter("/user/*")
public class AuthorizationFilter implements Filter {
    private static final String AJAX_REDIRECT_XML = "<?xml version="1.0" encoding="UTF-8"?>"
        + "<partial-response><redirect url="%s"></redirect></partial-response>";
    @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 loginURL = request.getContextPath() + "/login.xhtml";
        boolean loggedIn = (session != null) && (session.getAttribute("user") != null);
        boolean loginRequest = request.getRequestURI().equals(loginURL);
        boolean resourceRequest = request.getRequestURI().startsWith(request.getContextPath() + ResourceHandler.RESOURCE_IDENTIFIER + "/");
        boolean ajaxRequest = "partial/ajax".equals(request.getHeader("Faces-Request"));
        if (loggedIn || loginRequest || resourceRequest) {
            if (!resourceRequest) { // Prevent browser from caching restricted resources. See also https://stackoverflow.com/q/4194207/157882
                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.
            }
            chain.doFilter(request, response); // So, just continue request.
        }
        else if (ajaxRequest) {
            response.setContentType("text/xml");
            response.setCharacterEncoding("UTF-8");
            response.getWriter().printf(AJAX_REDIRECT_XML, loginURL); // So, return special XML response instructing JSF ajax to send a redirect.
        }
        else {
            response.sendRedirect(loginURL); // So, just perform standard synchronous redirect.
        }
    }

    // You need to override init() and destroy() as well, but they can be kept empty.
}

此外，过滤器还禁用了安全页面上的浏览器缓存，因此浏览器的后退按钮将不再显示它们。

如果您碰巧使用了JSF实用程序库OmniFaces，那么上面的代码可以简化如下:

@WebFilter("/user/*")
public class AuthorizationFilter extends HttpFilter {
    @Override
    public void doFilter(HttpServletRequest request, HttpServletResponse response, HttpSession session, FilterChain chain) throws ServletException, IOException {
        String loginURL = request.getContextPath() + "/login.xhtml";
        boolean loggedIn = (session != null) && (session.getAttribute("user") != null);
        boolean loginRequest = request.getRequestURI().equals(loginURL);
        boolean resourceRequest = Servlets.isFacesResourceRequest(request);
        if (loggedIn || loginRequest || resourceRequest) {
            if (!resourceRequest) { // Prevent browser from caching restricted resources. See also https://stackoverflow.com/q/4194207/157882
                Servlets.setNoCacheHeaders(response);
            }
            chain.doFilter(request, response); // So, just continue request.
        }
        else {
            Servlets.facesRedirect(request, response, loginURL);
        }
    }
}

参见:

• 我们的Servlet过滤器wiki页面
• 如何在数据库中处理用户的认证/授权?
• 使用JSF 2.0/Facelets，是否有办法将全局侦听器附加到所有AJAX调用?
• 避免JSF web应用程序的后退按钮
JSF:如何在JSF中控制访问和权限?

Answer 2

虽然使用简单的Servlet过滤器当然是合法的，但也有其他选择，如

• Spring Security
• Java EE Security
Apache Shiro