如何在spring security中添加@secure注解

Question

如何在控制器的方法中添加@Secure注解并使其运行？ 现在，当我运行它时，出现如下异常：

org.springframework.beans.factory.BeanCreationException：创建文件中定义的名为“companyController”的 bean 时出错[C:\workspace\sts\springsource\vfabric-tc-server-developer-2.6.1.RELEASE\spring-insight-instance\wtpwebapps\BillingEngine\WEB-INF\classes\com\sesami\common\management\web\ [controller\CompanyController.class]: bean 初始化失败；嵌套异常是org.springframework.aop.framework.AopConfigException：意外的AOP异常；嵌套异常是org.springframework.beans.factory.BeanCreationException：创建名为“org.springframework.security.access.intercept.aopalliance.MethodSecurityInterceptor#0”的bean时出错：设置bean属性“accessDecisionManager”时无法解析对bean“accessDecisionManager”的引用';嵌套异常是 org.springframework.beans.factory.NoSuchBeanDefinitionException：未定义名为“accessDecisionManager”的 bean org.apache.catalina.core.StandardContext.listenerStart(StandardContext.java:4723) 在 org.apache.catalina.core.StandardContext$1.call(StandardContext.java 在 java.lang.Thread.run（来源未知） 引起原因：org.springframework.aop.framework.AopConfigException：意外的AOP异常；嵌套异常是org.springframework.beans.factory.BeanCreationException：创建名为“org.springframework.security.access.intercept.aopalliance.MethodSecurityInterceptor#0”的bean时出错：设置bean属性“accessDecisionManager”时无法解析对bean“accessDecisionManager”的引用';嵌套异常位于... 19 更多

我有 spring security .xml

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns:context="http://www.springframework.org/schema/context"
    xmlns="http://www.springframework.org/schema/security" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:beans="http://www.springframework.org/schema/beans"
    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">

<global-method-security secured-annotations="enabled">
<!-- 
<protect-pointcut access="ROLE_ADMIN"
        expression="execution(* com.sesami.common.management.web.controller.AdminController.*(..))" />
         -->
</global-method-security>

<!-- URL pattern based security -->
<http auto-config="false" entry-point-ref="authenticationEntryPoint"
    use-expressions="true">
    <custom-filter ref="authenticationFilter" position="FORM_LOGIN_FILTER" />
    <intercept-url access="hasRole('ROLE_ADMIN')" pattern="/common/admin/**" />
    <intercept-url pattern="/common/accounting/**" access="hasRole('ROLE_USER')" />
    <intercept-url pattern="/common/billing/**" access="hasRole('ROLE_COMPANY')" />
    <logout logout-success-url="/" logout-url="/logout"/>

</http>.........

在控制器中我添加这样的内容

@Secure("ROLE_ADMIN")
@RequestMapping(value = "/common/admin/addAdmin", method = RequestMethod.GET)
    public String add(ModelMap map) {
        map.addAttribute(new Administrator());
        return "/common/admin/addAdmin";
    }

我需要配置或导入一些类吗？

Answer 1

Cannot resolve reference to bean 'accessDecisionManager' while setting bean property 'accessDecisionManager'; nested exception is
org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'accessDecisionManager' is defined 

Spring 应该为您创建一个默认的 accessDecisionManager，但看起来这并没有发生，可能是由于某些配置问题。只是为了好玩，如果您在 http 配置中将 auto-config 设置为 true 会发生什么？

Answer 2

<bean id="accessDecisionManager" class="org.springframework.security.access.vote.UnanimousBased" xmlns="http://www.springframework.org/schema/beans">
    <constructor-arg>
        <list>
            <bean class="org.springframework.security.oauth2.provider.vote.ScopeVoter" />
            <bean class="org.springframework.security.access.vote.RoleVoter" />
            <bean class="org.springframework.security.access.vote.AuthenticatedVoter" />
        </list>
    </constructor-arg>
</bean>

你必须定义这个bean。