JSF 2.0用户会话(JSF 2.0 user session)

JSF 2.0用户会话(JSF 2.0 user session)

我早些时候问了我的问题，但我不清楚。 所以我再次问我的问题，希望能理解我。 我想为我的应用程序创建一个用户会话。 因此，用户必须登录才能访问我的主页。 为此，我创建了一个Bean和一个过滤器。 我用这个问题的答案JSF HTTP Session Login能够做到这一点。 但Filter中的userManager始终为null。 我不知道如何在这一行中输入连接用户的值。

UserManager userManager = (UserManager) req.getSession().getAttribute("userManager");

这是过滤器。

public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException { HttpServletRequest req = (HttpServletRequest) request; HttpServletResponse res = (HttpServletResponse) response; UserManager userManager = (UserManager) req.getSession().getAttribute("userManager"); String loginURL = req.getContextPath() + "/index.jsf"; if (userManager == null && !userManager.isLoggedIn() && !req.getRequestURI().equals(loginURL)) { res.sendRedirect(loginURL); } else { res.setHeader("Cache-Control", "no-cache, no-store, must-revalidate"); // HTTP 1.1. res.setHeader("Pragma", "no-cache"); // HTTP 1.0. res.setDateHeader("Expires", 0); // Proxies. chain.doFilter(request, response); }

当我把这个条件放在if !userManager.isLoggedIn() ，我有一个标记，上面写着：

空指针访问：变量userManager在此位置只能为空，1个快速修复可用，将@SuppressWarnings'null'添加到'doFilter（）'

我试图添加它并执行但是我收到了一个错误java.lang.NullPointerException 。 我是JSF的初学者。 拜托我需要你的帮忙。

ERROR [org.apache.catalina.core.ContainerBase.[jboss.web].[localhost].[/jsf-formulaire].[Faces Servlet]] "Servlet.service()" pour la servlet Faces Servlet a généré une exception: java.lang.NullPointerException at com.citech.pointage.fe.bean.filters.SessionFilter.doFilter(SessionFilter.java:61) [:] at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:274) [:6.1.0.Final] at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:242) [:6.1.0.Final] at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:275) [:6.1.0.Final] at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:161) [:6.1.0.Final] at org.jboss.web.tomcat.security.SecurityAssociationValve.invoke(SecurityAssociationValve.java:181) [:6.1.0.Final] at org.jboss.modcluster.catalina.CatalinaContext$RequestListenerValve.event(CatalinaContext.java:285) [:1.1.0.Final] at org.jboss.modcluster.catalina.CatalinaContext$RequestListenerValve.invoke(CatalinaContext.java:261) [:1.1.0.Final] at org.jboss.web.tomcat.security.JaccContextValve.invoke(JaccContextValve.java:88) [:6.1.0.Final] at org.jboss.web.tomcat.security.SecurityContextEstablishmentValve.invoke(SecurityContextEstablishmentValve.java:100) [:6.1.0.Final] at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:159) [:6.1.0.Final] at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102) [:6.1.0.Final] at org.jboss.web.tomcat.service.jca.CachedConnectionValve.invoke(CachedConnectionValve.java:158) [:6.1.0.Final] at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109) [:6.1.0.Final] at org.jboss.web.tomcat.service.request.ActiveRequestResponseCacheValve.invoke(ActiveRequestResponseCacheValve.java:53) [:6.1.0.Final] at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:362) [:6.1.0.Final] at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:877) [:6.1.0.Final] at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:654) [:6.1.0.Final] at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:951) [:6.1.0.Final] at java.lang.Thread.run(Unknown Source) [:1.6.0_03]

I asked my problem earlier but I was not clear. So I ask my question again hoping that one understands me. I would like to create a user session for my application. Thus, the user must log in to access my homepage. For this I create a Bean and a filter. I used the answer to this question JSF HTTP Session Login to be able to do it. But the userManager in the Filter is always null. I don't know how to put in this line the value of the user who is connected.

UserManager userManager = (UserManager) req.getSession().getAttribute("userManager");

Here is the filter.

public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException { HttpServletRequest req = (HttpServletRequest) request; HttpServletResponse res = (HttpServletResponse) response; UserManager userManager = (UserManager) req.getSession().getAttribute("userManager"); String loginURL = req.getContextPath() + "/index.jsf"; if (userManager == null && !userManager.isLoggedIn() && !req.getRequestURI().equals(loginURL)) { res.sendRedirect(loginURL); } else { res.setHeader("Cache-Control", "no-cache, no-store, must-revalidate"); // HTTP 1.1. res.setHeader("Pragma", "no-cache"); // HTTP 1.0. res.setDateHeader("Expires", 0); // Proxies. chain.doFilter(request, response); }

When I put this condition in the if !userManager.isLoggedIn(), I have a marker that says:

Null pointer access: The variable userManager can only be null at this location, 1 quick fix available, Add @SuppressWarnings 'null' to 'doFilter()'

I tried to add it and I executed but I got an error java.lang.NullPointerException. I'm beginner in JSF. Please, I need your help.

ERROR [org.apache.catalina.core.ContainerBase.[jboss.web].[localhost].[/jsf-formulaire].[Faces Servlet]] "Servlet.service()" pour la servlet Faces Servlet a généré une exception: java.lang.NullPointerException at com.citech.pointage.fe.bean.filters.SessionFilter.doFilter(SessionFilter.java:61) [:] at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:274) [:6.1.0.Final] at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:242) [:6.1.0.Final] at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:275) [:6.1.0.Final] at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:161) [:6.1.0.Final] at org.jboss.web.tomcat.security.SecurityAssociationValve.invoke(SecurityAssociationValve.java:181) [:6.1.0.Final] at org.jboss.modcluster.catalina.CatalinaContext$RequestListenerValve.event(CatalinaContext.java:285) [:1.1.0.Final] at org.jboss.modcluster.catalina.CatalinaContext$RequestListenerValve.invoke(CatalinaContext.java:261) [:1.1.0.Final] at org.jboss.web.tomcat.security.JaccContextValve.invoke(JaccContextValve.java:88) [:6.1.0.Final] at org.jboss.web.tomcat.security.SecurityContextEstablishmentValve.invoke(SecurityContextEstablishmentValve.java:100) [:6.1.0.Final] at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:159) [:6.1.0.Final] at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102) [:6.1.0.Final] at org.jboss.web.tomcat.service.jca.CachedConnectionValve.invoke(CachedConnectionValve.java:158) [:6.1.0.Final] at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109) [:6.1.0.Final] at org.jboss.web.tomcat.service.request.ActiveRequestResponseCacheValve.invoke(ActiveRequestResponseCacheValve.java:53) [:6.1.0.Final] at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:362) [:6.1.0.Final] at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:877) [:6.1.0.Final] at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:654) [:6.1.0.Final] at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:951) [:6.1.0.Final] at java.lang.Thread.run(Unknown Source) [:1.6.0_03]

最满意答案

你的病情应该是：

if ((userManager == null || !userManager.isLoggedIn()) && !req.getRequestURI().equals(loginURL)) // first && replaced by ||

现在，您可以继续使用JSF HTTP会话登录 。 验证用户的另一个选项是使用来自jsf元数据的PreRenderView事件，只需在页面中添加：

<f:metadata> <f:event type="preRenderView" listener="#{securityMB.checkLogIn}" /> </f:metadata>

而你的托管bean：

@ManagedBean(name = "securityMB") @RequestScoped public class SecurityBean { @ManagedProperty(value = "#{userManager}") private UserManager userManager; String loginURL = "/index.jsf"; public void checkLogIn() { if (userManager == null || !userManager.isLoggedIn()) { FacesContext.getCurrentInstance().getExternalContext().redirect(loginURL); } //getter and setter (userManager) }

Your condition should be:

if ((userManager == null || !userManager.isLoggedIn()) && !req.getRequestURI().equals(loginURL)) // first && replaced by ||

Now you can continue working with JSF HTTP Session Login. Another option to validate the user is with PreRenderView event from jsf metadata, just add this in your pages:

<f:metadata> <f:event type="preRenderView" listener="#{securityMB.checkLogIn}" /> </f:metadata>

And your managed bean:

@ManagedBean(name = "securityMB") @RequestScoped public class SecurityBean { @ManagedProperty(value = "#{userManager}") private UserManager userManager; String loginURL = "/index.jsf"; public void checkLogIn() { if (userManager == null || !userManager.isLoggedIn()) { FacesContext.getCurrentInstance().getExternalContext().redirect(loginURL); } //getter and setter (userManager) }