2

我无法让 sec:authorize hasRole() 使用角色层次结构。如果我有一个角色 ROLE_BOSS 的用户是 ROLE_WORKER 的父级，那么由于某种原因是错误的。但是，在我的服务类中，@PreAuthorize("hasRole('ROLE_WORKER')") 确实有效。我假设他们都使用相同的评估器，那么为什么 taglib 不起作用？谢谢您的帮助。

JSP：

<sec:authorize access="hasRole('ROLE_BOSS')">
  <p>This shows up.</p>
</sec:authorize>
<sec:authorize access="hasRole('ROLE_WORKER')">
  <p>This does not show up, but should.</p>
</sec:authorize>

-config.xml 安全性：

<bean id="expressionHandler" class="org.springframework.security.access.expression.method.DefaultMethodSecurityExpressionHandler">
  <property name="permissionEvaluator" ref="permissionEvaluator"/>
  <property name="roleHierarchy" ref="roleHierarchy"/>
</bean>

<sec:global-method-security pre-post-annotations="enabled">
  <sec:expression-handler ref="expressionHandler"/>
</sec:global-method-security>

<bean id="permissionEvaluator" class="com.myapp.security.MyPermissionEvaluator">
  <constructor-arg index="0">
    <map key-type="java.lang.String" value-type="com.myapp.security.Permission">
      <entry key="contractReadAccess" value-ref="contractReadPermission"/>
      <entry key="contractWriteAccess" value-ref="contractWritePermission"/>
    </map>
  </constructor-arg>
</bean>

<bean id="contractReadPermission" class="com.myapp.security.ContractReadPermission"/>
<bean id="contractWritePermission" class="com.myapp.security.ContractWritePermission"/>

<sec:http use-expressions="true" access-decision-manager-ref="accessDecisionManager">
  <sec:intercept-url pattern="/worker/**" access="isAuthenticated()" requires-channel="https"/>
  <sec:intercept-url pattern="/boss/**" access="hasRole('ROLE_BOSS')" requires-channel="https"/>

  <sec:form-login login-page="/login" authentication-failure-url="/login?login_error=1" authentication-success-handler-ref="successHandler"/>
  <sec:logout logout-url="/logout" logout-success-url="/login" invalidate-session="true"/>
<sec:remember-me/>
</sec:http>

<bean id="accessDecisionManager" class="org.springframework.security.access.vote.AffirmativeBased">
  <constructor-arg>
    <list>
      <ref bean="roleVoter" />
      <bean class="org.springframework.security.web.access.expression.WebExpressionVoter">
        <property name="expressionHandler">
          <bean class="org.springframework.security.web.access.expression.DefaultWebSecurityExpressionHandler">
            <property name="roleHierarchy" ref="roleHierarchy"/>
          </bean>
        </property>
      </bean>
      <bean class="org.springframework.security.access.vote.AuthenticatedVoter"/>
    </list>
  </constructor-arg>
</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">
    <value>
      ROLE_BOSS > ROLE_WORKER
    </value>
  </property>
</bean>

<sec:authentication-manager alias="authenticationManager">
  <sec:authentication-provider user-service-ref="myUserDetailsService"/>
</sec:authentication-manager>
4

3 回答 3

3

对于像我这样使用 Java Config 的人。这是一个非常简单的解决方案，只需在扩展类中添加以下代码WebSecurityConfigurerAdapter：

@Bean
    public RoleHierarchyVoter roleVoter() {
        return new RoleHierarchyVoter(roleHierarchy());
    }

    @Bean
    public RoleHierarchy roleHierarchy() {
        RoleHierarchyImpl roleHierarchy = new RoleHierarchyImpl();
        roleHierarchy.setHierarchy("ROLE_BOSS > ROLE_WORKER");
        return roleHierarchy;
    }

    private SecurityExpressionHandler<FilterInvocation> webExpressionHandler() {
        DefaultWebSecurityExpressionHandler defaultWebSecurityExpressionHandler = new DefaultWebSecurityExpressionHandler();
        defaultWebSecurityExpressionHandler.setRoleHierarchy(roleHierarchy());
        return defaultWebSecurityExpressionHandler;
    }

    @Override
    public void init(WebSecurity web) throws Exception {
        web.expressionHandler(webExpressionHandler());
        super.init(web);
    }
于 2017-11-10T22:01:57.233 回答
1

很奇怪，我认为这是不正确的，但它似乎有效。我开始深入研究 Spring 源代码，我认为我可以通过将 DefaultWebSecurityExpressionHandler 从 accessDecisionManager 中取出并将其放在我所有安全配置的最顶部来使其工作。所以在我的 -config.xml 顶部我有这个：

<bean id="webExpressionHandler" class="org.springframework.security.web.access.expression.DefaultWebSecurityExpressionHandler">
  <property name="permissionEvaluator" ref="permissionEvaluator"/>
  <property name="roleHierarchy" ref="roleHierarchy"/>
</bean>

我的 accessDecisionManager 现在是：

<bean id="accessDecisionManager" class="org.springframework.security.access.vote.AffirmativeBased">
  <constructor-arg>
    <list>
      <ref bean="roleVoter" />
      <bean class="org.springframework.security.web.access.expression.WebExpressionVoter">
        <property name="expressionHandler" ref="webExpressionHandler"/>
      </bean>
      <bean class="org.springframework.security.access.vote.AuthenticatedVoter"/>
    </list>
  </constructor-arg>
</bean>
于 2012-10-30T05:29:15.013 回答
0

你有没有尝试过？

<%@ taglib prefix='sec' uri='http://www.springframework.org/security/tags' %> 

<sec:authorize ifAnyGranted='ROLE_BOSS,ROLE_WORKER'> 
  <h1>ROLE_BOSS and ROLE_WORKER can see this</h1><br/> 
</sec:authorize>

或者

<sec:authorize access="hasAnyRole('ROLE_BOSS','ROLE_WORKER')">
  <h1>ROLE_BOSS and ROLE_WORKER can see this</h1><br/> 
</sec:authorize>
于 2013-06-28T10:56:54.337 回答