Question

我想为Web应用程序实现自定义错误页面。我使用以下方式：

的web.xml

<error-page>
  <error-code>404</error-code>
  <location>/404/</location>
</error-page>

弹簧security.xml文件

<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。但似乎principal在这种情况下无法获取，即我无法获得有关当前登录用户的任何信息。

是默认行为还是代码中有错误？

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>

Answer 1

要从错误页面访问主要数据，您需要将spring安全过滤器映射为：

<filter-mapping>
    <filter-name>springSecurityFilterChain</filter-name>
    <url-pattern>/*</url-pattern>
    <dispatcher>FORWARD</dispatcher>
    <dispatcher>REQUEST</dispatcher>
    <dispatcher>ERROR</dispatcher>
</filter-mapping>

Spring MVC：错误页面和主要信息

1 个答案: