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带有 404 错误的 Spring 安全问题?

转载 作者:IT老高 更新时间:2023-10-28 13:46:32 25 4
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大家好,我使用的是 spring security 3.0.2,urlRewrite 3.1.0,我有一个spring security的问题,我有一个规则,应用程序中的所有页面都需要身份验证,除了某些页面,所以我的security.xml是:

<http use-expressions="true" > 
<intercept-url pattern="/" access="permitAll" />
<intercept-url pattern="/error" filter="none" />
<intercept-url pattern="/**" access="isAuthenticated()" />
.
.
.</http>

在 web.xml 中我定义了错误页面

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

问题是,如果我不是登录用户,并且输入了一些应用程序中不存在的 URL,例如 app/notFoundUrl,spring security 会将此页面与需要身份验证的模式/** 匹配,所以用户没有按预期重定向到错误页面,而是重定向到登录页面,然后重定向到错误页面

我希望如果用户在登录与否时输入了错误的 url,他会直接重定向到错误页面。

我认为问题与web.xml有关,这里是:

<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

<!-- Beans in these files will makeup the configuration of the root web application context -->
<!-- Bootstraps the root web application context before servlet initialization-->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<!-- Deploys the 'projects' dispatcher servlet whose configuration resides in /WEB-INF/servlet-config.xml-->
<servlet>
<servlet-name>p</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/servlet-config.xml
</param-value>
</init-param>
</servlet>

<!-- Maps all /p URLs to the 'p' servlet -->
<servlet-mapping>
<servlet-name>p</servlet-name>
<url-pattern>/p/*</url-pattern>
</servlet-mapping>

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


<!-- force encoding on the requests -->
<filter>
<filter-name>encoding-filter</filter-name>
<filter-class>org.springframework.web.filter.CharacterEncodingFilter</filter-class>
<init-param>
<param-name>encoding</param-name>
<param-value>utf-8</param-value>
</init-param>
<init-param>
<param-name>forceEncoding</param-name>
<param-value>true</param-value>
</init-param>
</filter>
<filter-mapping>
<filter-name>encoding-filter</filter-name>
<url-pattern>/*</url-pattern>
<dispatcher>REQUEST</dispatcher>
<dispatcher>FORWARD</dispatcher>
</filter-mapping>

<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>UrlRewriteFilter</filter-name>
<filter-class>org.tuckey.web.filters.urlrewrite.UrlRewriteFilter</filter-class>

</filter>
<filter-mapping>
<filter-name>UrlRewriteFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>





<!-- Security -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/application-config.xml
/WEB-INF/app-security.xml
/WEB-INF/mvc-config.xml
</param-value>
</context-param>


<session-config>
<session-timeout>1</session-timeout>
</session-config>


</web-app>

任何想法如何解决这个问题?

最佳答案

你说过:

i want that if the user typed a bad url if he's logged in or not, he's redirected to the error page directly

Spring security 会在它知道其 url 是否有效之前拦截每个请求,因此获取它的一种方法是拦截具有某些模式的所有有效 url,并在最后添加一个任何人都可以访问的通用模式.

<intercept-url pattern="/" access="permitAll" />
<intercept-url pattern="/validUrl1Pattern" access="permitAll" />
<intercept-url pattern="/validUrl2Pattern" access="permitAll" />
<intercept-url pattern="/validUrl2Pattern" access="permitAll" />
...
<intercept-url pattern="/**" access="ROLE_ANONYMOUS" />

这种配置的问题是,如果您的应用程序很复杂,可能很难找到所有有效 url 的模式。

关于带有 404 错误的 Spring 安全问题?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4153438/

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