5
我想爲Web應用程序實現自定義錯誤頁面。我用下面的方法:Spring MVC:錯誤頁面和主要信息
的web.xml
<error-page>
<error-code>404</error-code>
<location>/404/</location>
</error-page>
彈簧security.xml文件
<http use-expressions="true">
<form-login ... />
<access-denied-handler error-page="/403/" />
....
</http>
兩個網頁通過適當的控制器處理。但似乎principal
在這種情況下是不可接受的,即我無法獲得有關當前登錄用戶的任何信息。
它是默認行爲還是我在代碼中有錯誤?
謝謝
UPD#1: 我的配置:
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/spring-service.xml
/WEB-INF/spring-security.xml
/WEB-INF/spring-data.xml
/WEB-INF/spring-mail.xml
</param-value>
</context-param>
<filter>
<filter-name>springSecurityFilterChain</filter-name>
<filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>
<filter-mapping>
<filter-name>springSecurityFilterChain</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<filter>
<filter-name>hibernateFilter</filter-name>
<filter-class>org.springframework.orm.hibernate3.support.OpenSessionInViewFilter</filter-class>
<init-param>
<param-name>sessionFactoryBeanName</param-name>
<param-value>sessionFactory</param-value>
</init-param>
</filter>
<filter-mapping>
<filter-name>hibernateFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<servlet>
<servlet-name>spring</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>0</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
如何你在web.xml文件中映射安全過濾器嗎? – Pastur
@AbelPastur查看更新 – nKognito