Java:Spring security 3角色层次结构

Java:Spring security 3角色层次结构

问题描述:

我正在使用Spring框架mvc 3 + spring security 3. 我想在我的spring安全中启用角色层次结构。 据http://static.springsource.org/spring-security/site/docs/3.1.x/reference/authz-arch.html我应该写Java:Spring security 3角色层次结构

<bean id="roleVoter" class="org.springframework.security.access.vote.RoleHierarchyVoter"> 
    <constructor-arg ref="roleHierarchy" /> 
</bean> 
<bean id="roleHierarchy" 
    class="org.springframework.security.access.hierarchicalroles.RoleHierarchyImpl"> 
<property name="hierarchy"> 
    ROLE_ADMIN > ROLE_STAFF 
    ROLE_STAFF > ROLE_USER 
    ROLE_USER > ROLE_GUEST 
</property> 
</bean> 

但我应该在哪里放呢?我试图把它变成我的应用程序-security.xml文件:

<beans:beans xmlns="http://www.springframework.org/schema/security" 
xmlns:beans="http://www.springframework.org/schema/beans" 
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
xsi:schemaLocation="http://www.springframework.org/schema/beans 
    http://www.springframework.org/schema/beans/spring-beans-3.0.xsd 
    http://www.springframework.org/schema/security 
    http://www.springframework.org/schema/security/spring-security-3.0.xsd"> 
    <http> 
     <intercept-url pattern="/entryPost/**" access="ROLE_USER" requires-channel="https"/> 
     <intercept-url pattern="/entryDelete/**" access="ROLE_ADMIN" requires-channel="https"/> 
     <intercept-url pattern="/commentDelete/**" access="ROLE_ADMIN" requires-channel="https"/> 
     <intercept-url pattern="/login" access="ROLE_ANONYMOUS" requires-channel="https"/> 
     <form-login login-page="/login" default-target-url="/entryList/1" authentication-failure-url="/login?error=true" /> 
     <logout logout-success-url="/login" /> 
     <session-management> 
      <concurrency-control max-sessions="1" /> 
     </session-management> 
     <access-denied-handler error-page="/accessDenied"/> 
    </http> 
    <authentication-manager> 
     <authentication-provider> 
      <jdbc-user-service data-source-ref="dataSource" 
      users-by-username-query="SELECT username,password,'true' as enabled FROM member WHERE username=?" 
      authorities-by-username-query="SELECT member.username,role FROM member,memberRole WHERE member.username=? AND member.id=memberRole.member_id"/> 
     </authentication-provider> 
    </authentication-manager> 
<bean id="roleVoter" class="org.springframework.security.access.vote.RoleHierarchyVoter"> 
    <constructor-arg ref="roleHierarchy" /> 
</bean> 
<bean id="roleHierarchy" class="org.springframework.security.access.hierarchicalroles.RoleHierarchyImpl"> 
    <property name="hierarchy"> 
     ROLE_ADMIN > ROLE_STAFF 
     ROLE_STAFF > ROLE_USER 
     ROLE_USER > ROLE_GUEST 
    </property> 
</bean> 

但它不工作:HTTP状态404

当我把它变成APP-servlet.xml中:

<?xml version="1.0" encoding="UTF-8"?> 
<beans xmlns="http://www.springframework.org/schema/beans" 
xmlns:mvc="http://www.springframework.org/schema/mvc" 
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
xmlns:context="http://www.springframework.org/schema/context" 
xsi:schemaLocation="http://www.springframework.org/schema/beans 
    http://www.springframework.org/schema/beans/spring-beans-3.0.xsd 
    http://www.springframework.org/schema/context 
    http://www.springframework.org/schema/context/spring-context-3.0.xsd 
    http://www.springframework.org/schema/mvc 
    http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd"> 
    <context:component-scan base-package="rus.web"/> 
    <bean id="entryValidator" class="rus.domain.EntryValidator"/> 
    <bean id="commentValidator" class="rus.domain.CommentValidator"/> 
    <mvc:annotation-driven/> 
    <mvc:resources mapping="/resources/**" location="/resources/"/> 
    <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver"> 
     <property name="prefix" value="/WEB-INF/jsp/"/> 
     <property name="suffix" value=".jsp"/> 
    </bean> 
    <bean id="messageSource" class="org.springframework.context.support.ResourceBundleMessageSource"> 
     <property name="basename" value="messages"/> 
    </bean> 
    <!--<bean class="org.springframework.web.servlet.handler.SimpleMappingExceptionResolver"> 
     <property name="defaultErrorView" value="error"/> 
    </bean> --> 

<bean id="roleVoter" class="org.springframework.security.access.vote.RoleHierarchyVoter"> 
    <constructor-arg ref="roleHierarchy" /> 
</bean> 
<bean id="roleHierarchy" class="org.springframework.security.access.hierarchicalroles.RoleHierarchyImpl"> 
    <property name="hierarchy"> 
     ROLE_ADMIN > ROLE_STAFF 
     ROLE_STAFF > ROLE_USER 
     ROLE_USER > ROLE_GUEST 
    </property> 
</bean> 
</beans> 

它抛出异常:

org.springfram ework.beans.factory.xml.XmlBeanDefinitionStoreException:来自ServletContext资源的XML文档中的第35行[/WEB-INF/rus-servlet.xml]无效;嵌套的异常是org.xml.sax.SAXParseException:cvc-complex-type.2.3:元素的'属性'不能包含字符[子],因为该类型的内容类型仅为元素。

org.xml.sax.SAXParseException:cvc-complex-type.2.3:元素的'属性'不能包含character [children],因为该类型的内容类型是仅包含元素的。

我该怎么办才能解决这个问题?

+0

我有同样的问题。我按照这里的说明解决了问题:http://*.com/questions/7809313/accessdeniedexception-if-using-rolehierarchyimpl – 2012-08-14 06:13:01

的文件是错误的,这是无效的:

<property name="hierarchy"> 
    ROLE_ADMIN > ROLE_STAFF 
    ROLE_STAFF > ROLE_USER 
    ROLE_USER > ROLE_GUEST 
</property> 

您需要包装内<value>内容:

<property name="hierarchy"> 
    <value> 
     ROLE_ADMIN > ROLE_STAFF 
     ROLE_STAFF > ROLE_USER 
     ROLE_USER > ROLE_GUEST 
    </value> 
</property> 

我建议在SpringSource JIRA提交的问题,要求他们修复文档。

+2

好吧,它帮助,但不是那么多...当我尝试与ROLE_ADMIN去到页面条目邮政/(访问=“ROLE_USER”)它说:“访问被拒绝”。为什么这样? – Twisty