我想为 Web 应用程序实现自定义错误页面。我使用以下方式:
web.xml
<error-page>
<error-code>404</error-code>
<location>/404/</location>
</error-page>
弹簧安全.xml
<http use-expressions="true">
<form-login ... />
<access-denied-handler error-page="/403/" />
....
</http>
两个页面都由适当的控制器处理。但在这种情况下似乎principal
无法访问,即我无法获得有关当前登录用户的任何信息。
是默认行为还是我的代码有错误?
谢谢
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>