4

'正在开发一个 JSF 2 项目。我已将 login.xhtml 页面定义为 web.xml 中的入口页面

    <welcome-file-list>
        <welcome-file>login.xhtml</welcome-file>
    </welcome-file-list>

我还有一个过滤器来检查用户是否登录

    @WebFilter(filterName = "loginCheckFilter", urlPatterns={"/*"})
    public class LoginCheckFilter implements Filter
    {
        @Inject
        private LoginStatus loginStatus;

        public void do Filter(...)
        {
            try{
            HttpServletRequest req = (HttpServletRequest) request;
            HttpServletResponse res = (HttpServletResponse) response;

            String path = req.getRequestURI();
            if(StringUtils.isNotBlank(path)
               && StringUtils.contains(path, ".xhtml")
               && !StringUtils.endsWith(path, "login.xhtml"))
             {
                    if(loginStatus == null
                       || !loginStatus.isLoggedIn())
                     {
                          res.sendRedirect(req.getContextPath() + "/login.xhtml");
                      }
                     else
                      {
                           chain.doFilter(request, response);
                       }
              }
              else
              {
                  chain.doFilter(request, response);
               }
            }catch (Exception ex)
             {
                  log.error(ex);
              }
           }

        .... ....
        }

我的 css 文件以以下样式引用:

    <link href="css/styles.css" rel="stylesheet" type="text/css"/>

一切正常,直到我将 css 引用样式更改为 JSF 2 资源处理程序 ( http://www.mkyong.com/jsf2/resources-library-in-jsf-2-0/ )。我已将所有 css 文件复制到资源文件夹下,并提供了库名称和版本号。所以现在我引用css如下:

    <h:outputStylesheet library="default" name="css/styles.css"/>

更改后,login.xhtml 不再呈现样式表。我在 login.xhtml 页面之后有一个 welcome.xhtml 页面,除了核心内容外,它的结构几乎相同,但这个页面呈现得非常好。我已经刷新了 login.xhtml 仍然没有呈现。但是一旦我登录,前进到下一页,然后回到 login.xhtml,然后刷新,就会呈现样式。此外,如果我取消 loginCheckFilter,login.xhtml 将呈现样式表。那么,如果有人遇到类似的情况并知道如何解决呢?谢谢!

4

1 回答 1

5
urlPatterns={"/*"}

您的过滤器还会阻止对 JSF 资源的请求。

您需要以允许 JSF 资源请求的方式重写您的过滤器。

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

    boolean loggedIn = loginStatus != null && loginStatus.isLoggedIn();
    boolean loginRequest = request.getRequestURI().startsWith(loginURL);
    boolean resourceRequest = request.getRequestURI().startsWith(request.getContextPath() + "/faces" + ResourceHandler.RESOURCE_IDENTIFIER);

    if (loggedIn || loginRequest || resourceRequest)) {
        if (!resourceRequest) { // Prevent restricted pages from being cached.
            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);
    } else {
        response.sendRedirect(loginURL);
    }
}
于 2013-02-27T00:50:58.057 回答