我试图按照不同的文章来介绍如何在GlassFish 3.1的Mojarra 2.1.0(带有RichFaces 4)中处理ViewExpiredException
。但是我在web.xml
中定义的内容对接缝没有任何作用。我使用表单基础安全性,用户必须登录才能访问内容。我只想让Glassfish(catalina)在 session 超时时将用户重定向到一个简单的JSF页面,并提供指向登录页面的链接。
我总是收到以下错误消息;
javax.faces.application.ViewExpiredException: viewId:/app_user/activity/List.xhtml - View /app_user/activity/List.xhtml could not be restored.
at com.sun.faces.lifecycle.RestoreViewPhase.execute(RestoreViewPhase.java:202)
at com.sun.faces.lifecycle.Phase.doPhase(Phase.java:101)
at com.sun.faces.lifecycle.RestoreViewPhase.doPhase(RestoreViewPhase.java:113)
当用户 session 超时并在服务器日志中捕获异常时,谁可以重定向用户?
克里斯,问候。
layout.xhtml
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml"
xmlns:ui="http://java.sun.com/jsf/facelets"
xmlns:f="http://java.sun.com/jsf/core"
xmlns:h="http://java.sun.com/jsf/html"
xmlns:rich="http://richfaces.org/rich">
<h:head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<!-- Had no effect
<meta http-equiv="Cache-Control" content="no-cache, no-store, must-revalidate" />
<meta http-equiv="Pragma" content="no-cache" />
<meta http-equiv="Expires" content="0" />
-->
<h:outputStylesheet name="css/default.css"/>
<h:outputStylesheet name="css/cssLayout.css"/>
<title>
<h:outputText value="Partner Tapestry - " /> <ui:insert name="title">Browser Title</ui:insert>
</title>
</h:head>
<h:body>
<div id="top" >
<ui:insert name="top">Top Default</ui:insert>
</div>
<div id="messages">
<rich:messages id="messagePanel"/>
</div>
<div id="content">
<ui:insert name="content">Content Default</ui:insert>
</div>
</h:body>
</html>
web.xml
...
<servlet-mapping>
<servlet-name>Faces Servlet</servlet-name>
<url-pattern>/faces/*</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>
2
</session-timeout>
</session-config>
<error-page>
<exception-type>javax.faces.application.ViewExpiredException</exception-type>
<location>/faces/resources/sessionExpired.xhtml</location>
</error-page>
...
login.xhtml
<ui:composition template="/resources/masterLayout.xhtml"
xmlns="http://www.w3.org/1999/xhtml"
xmlns:ui="http://java.sun.com/jsf/facelets"
xmlns:h="http://java.sun.com/jsf/html"
xmlns:a4j="http://richfaces.org/a4j">
<ui:define name="title">
<h:outputText value="Login"></h:outputText>
</ui:define>
<ui:define name="top">
<h:outputText value="Welcome" />
</ui:define>
<ui:define name="content">
<h:form>
<h:panelGrid columns="3">
<h:outputLabel for="username" value="Username" />
<h:inputText id="username" value="#{userController.userAuthentication}" />
<br />
<h:outputLabel for="password" value="Password" />
<h:inputSecret id="password" value="#{userController.passwordAuthentication}" />
<br />
<h:outputText value=" " />
<a4j:commandButton action="#{userController.login()}" value="Login"/>
</h:panelGrid>
</h:form>
</ui:define>
</ui:composition>
最后 sessionExpire.xhtml
<ui:composition template="/resources/masterLayout.xhtml"
xmlns="http://www.w3.org/1999/xhtml"
xmlns:ui="http://java.sun.com/jsf/facelets"
xmlns:h="http://java.sun.com/jsf/html"
xmlns:a4j="http://richfaces.org/a4j">
<ui:define name="title">
<h:outputText value="Session Exired"></h:outputText>
</ui:define>
<ui:define name="top">
<h:outputText value="Session expired, please login again" />
</ui:define>
<ui:define name="content">
<h:form>
<a4j:commandLink action="/resources/login?faces-redirect=true" value="Login" />
</h:form>
</ui:define>
</ui:composition>
最佳答案
对于同步POST请求,您的方法应该可以很好地工作。但是,您正在使用异步(ajax)POST请求。 Ajax请求的异常需要由JavaScript中的jsf.ajax.addOnError
init函数或自定义 ExceptionHandler
处理。
也可以看看
关于session - ViewExpiredException,无法还原页面,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/7512708/