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在Spring Security 3中动态创建新角色和权限

我在Struts 2 + Spring IOC项目中使用Spring Security 3.

我在我的项目中使用过滤器,身份验证提供程序等.

你可以在这里看到我的security.xml

<?xml version="1.0" encoding="UTF-8"?>
<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"
  xmlns:p="http://www.springframework.org/schema/p"
   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-3.1.xsd">


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

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

<beans:bean class="code.permission.MyCustomPermissionEvaluator" id="customPermissionEvaluator" />

<!-- User Login -->

  <http auto-config="true" use-expressions="true" pattern="/user/*" >
<intercept-url pattern="/index.jsp" access="permitAll"/>
<intercept-url pattern="/user/showLoginPage.action" access="permitAll"/>
<intercept-url pattern="/user/showFirstPage" access="hasRole('ROLE_USER') or hasRole('ROLE_VISIT')"/>
<intercept-url pattern="/user/showSecondUserPage" access="hasRole('ROLE_USER')"/>
<intercept-url pattern="/user/showThirdUserPage" access="hasRole('ROLE_VISIT')"/>
<intercept-url pattern="/user/showFirstPage" access="hasRole('ROLE_USER') or hasRole('ROLE_VISIT')"/>
<form-login login-page="/user/showLoginPage.action" />
<logout invalidate-session="true"
        logout-success-url="/"
        logout-url="/user/j_spring_security_logout"/>
<access-denied-handler ref="myAccessDeniedHandler" …
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spring spring-security

33
推荐指数
3
解决办法
4万
查看次数

Spring安全方法级别安全注释无效

我正在使用Struts 2 + Spring Security 3创建一个简单的Web应用程序.我想使用Pre-Post Annotations进行方法级安全性.

但是注释不起作用.

这是我的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"   xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
  <display-name>MyCustomSpringSecurity</display-name>

  <context-param>
<param-name>contextConfiguration</param-name>
<param-value>/WEB-INF/applicationContext.xml</param-value>
  </context-param>

  <listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

  <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>struts2</filter-name>
    <filter-class>org.apache.struts2.dispatcher.FilterDispatcher</filter-class>
</filter>

  <filter-mapping>
    <filter-name>struts2</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>
  <welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
</web-app>
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这是我的struts.xml

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE struts PUBLIC "-//Apache Software Foundation//DTD Struts Configuration 2.1//EN" "http://struts.apache.org/dtds/struts-2.1.dtd">
<struts>
<package name="default" extends="struts-default">
    <action name="firstPage" class="code.action.MyAction" method="showPage">
        <result name="success">/firstPage.jsp</result>
    </action>
</package>
</struts>
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这是我的applicationContext.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans …
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spring annotations struts2 spring-security

3
推荐指数
1
解决办法
2092
查看次数

Spring + Spring Security + hasPermission不工作

我正在尝试使用Spring Security 3和Struts 2以及Spring DI.我在我的代码中放了hasRole()注释,它工作正常,但hasPermission()注释不起作用.我做了表达式处理程序,自定义权限评估程序等.

这是代码

applicationContext.xml中

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE beans PUBLIC "-//SPRING//DTD BEAN//EN" "http://www.springframework.org/dtd/spring-beans.dtd">
<beans> 
 <import resource="mySecurity.xml"/>
<bean id="myAction" name="myAction" class="code.action.MyAction" autowire="byName">
</bean>
<bean id="permission" name="permission" class="code.permission.Permission" autowire="byName">
</bean>
<bean id="myEntity" name="myEntity" class="code.entities.MyEntity" autowire="byName">
</bean>
<bean id="expressionHandler" class="org.springframework.security.access.expression.method.DefaultMethodSecurityExpressionHandler" 
    name="expressionHandler" autowire="byName">
</bean>
<bean class="code.permission.MyCustomPermissionEvaluator" id="customPermissionEvaluator" name="customPermissionEvaluator" autowire="byName" />
</beans>
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这是我的acl-context.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
 xmlns:p="http://www.springframework.org/schema/p"
 xmlns:util="http://www.springframework.org/schema/util"
 xsi:schemaLocation="http://www.springframework.org/schema/beans
      http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
   http://www.springframework.org/schema/util
   http://www.springframework.org/schema/util/spring-util-3.0.xsd">

    <!-- Declare a simple map containing all our roles -->
    <util:map id="permissionsMap">
         <entry key="ROLE_USER" value-ref="user"/>
     </util:map>

 <!-- …
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spring spring-security

2
推荐指数
1
解决办法
5239
查看次数

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spring ×3

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annotations ×1

struts2 ×1