java - Spring Security 预身份验证

Question

我被要求修改一个用 Java EE+Spring+Hibernate 构建的 Web 应用程序，以通过 Spring Security 提供安全性。

我以前没有使用过 Spring Security，我已经阅读了文档，但我没有看到使其工作的正确方法。

我们有一个登录页面，其中嵌入了另一个页面，其中包含实际的登录表单，该表单调用另一个 Web 应用程序的登录操作，并使用登录结果调用我们应用程序的方法之一。 (嵌入的登录页面是处理登录的“外国”Web 应用程序的一部分)。

我不明白应该如何配置 Spring Security。我想这是一个预身份验证场景，但我不知道如何让它发挥作用。

这是我的 security-context.xml

<?xml version="1.0" encoding="ISO-8859-1"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:security="http://www.springframework.org/schema/security"
       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-3.1.xsd
       http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd">

    <!-- preauthentication -->    
    <security:global-method-security pre-post-annotations="enabled">
    </security:global-method-security>

   <security:http auto-config="false" use-expressions="true" entry-point-ref="http403EntryPoint" access-denied-page="/403.jsp">
        <security:intercept-url pattern="/" access="permitAll"/>
        <security:intercept-url pattern="/403.jsp" access="permitAll"/>
        <security:intercept-url pattern="/gestion/**" access="permitAll"/>
        <security:intercept-url pattern="/consulta/**" access="hasRole('ROLE_ADMIN')"/>
        <security:intercept-url pattern="/importacion/**" access="hasRole('ROLE_ADMIN')"/>
        <!-- Allow non-secure access to static resources  -->
        <security:intercept-url pattern="/resources/**" access="permitAll"/>

        <security:logout logout-success-url="/"/>
    </security:http>

    <bean id="http403EntryPoint" class="org.springframework.security.web.authentication.Http403ForbiddenEntryPoint">
    </bean>

    <bean id="filterChainProxy" class="org.springframework.security.web.FilterChainProxy">
        <security:filter-chain-map path-type="ant">
            <security:filter-chain pattern="/**" filters="j2eePreAuthFilter"/>
        </security:filter-chain-map>
    </bean>


    <security:authentication-manager alias="authenticationManager">
        <security:authentication-provider ref='preAuthenticatedAuthenticationProvider'/>
    </security:authentication-manager>

    <bean id="preAuthenticatedAuthenticationProvider" class="org.springframework.security.web.authentication.preauth.PreAuthenticatedAuthenticationProvider">
        <property name="preAuthenticatedUserDetailsService" ref="preAuthenticatedUserDetailsService"/>
    </bean>

    <bean id="preAuthenticatedUserDetailsService"
            class="org.springframework.security.web.authentication.preauth.PreAuthenticatedGrantedAuthoritiesUserDetailsService"/>


    <bean id="j2eePreAuthFilter" class="es.myapp.security.MyUserJ2eePreAuthenticatedProcessingFilter">
        <property name="authenticationManager" ref="authenticationManager"/>
        <property name="authenticationDetailsSource" ref="authenticationDetailsSource"/>
        <property name="continueFilterChainOnUnsuccessfulAuthentication" value="false"/>
    </bean>

  <bean id="authenticationDetailsSource" class="org.springframework.security.web.authentication.preauth.j2ee.J2eeBasedPreAuthenticatedWebAuthenticationDetailsSource">
    <property name="mappableRolesRetriever" ref="j2eeMappableRolesRetriever"/>
    <property name="userRoles2GrantedAuthoritiesMapper" ref="j2eeUserRoles2GrantedAuthoritiesMapper"/>
  </bean>

  <bean id="j2eeMappableRolesRetriever" class="org.springframework.security.web.authentication.preauth.j2ee.WebXmlMappableAttributesRetriever">
  </bean>

   <bean id="j2eeUserRoles2GrantedAuthoritiesMapper" class="org.springframework.security.core.authority.mapping.SimpleAttributes2GrantedAuthoritiesMapper">
    <property name="attributePrefix" value="test"/>
  </bean>


</beans>

还有我的 web.xml

<web-app version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">

    <display-name>Aplicación Web</display-name>

    <!-- Define la localización de los ficheros de configuración de Spring -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
            /WEB-INF/classes/applicationContext.xml
        </param-value>
    </context-param>

    <!-- Reads request input using UTF-8 encoding -->
    <filter>
        <filter-name>characterEncodingFilter</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>characterEncodingFilter</filter-name>
        <url-pattern>/*</url-pattern>
    </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>myUserJ2eePreAuthenticatedProcessingFilter</filter-name>
        <filter-class>es.myapp.security.MyUserJ2eePreAuthenticatedProcessingFilter</filter-class>
    </filter>

    <filter-mapping>
        <filter-name>myUserJ2eePreAuthenticatedProcessingFilter</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>

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

    <!-- Handles all requests into the application -->
    <servlet>
        <servlet-name>Spring MVC Dispatcher Servlet</servlet-name>
        <servlet-class>es.myapp.controller.MyDispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/servlet-context.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>Spring MVC Dispatcher Servlet</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

    <!-- del. welcome files -->
    <!-- useful for Servlet 3 container (Tomcat 7 and Jetty 6) -->
    <welcome-file-list>
        <welcome-file></welcome-file>
    </welcome-file-list>

    <!-- Referencia a recursos jndi WAS -->
    <resource-ref id="ResourceRef_XISCO">
        <res-ref-name>jdbc/myapp</res-ref-name>
        <res-type>javax.sql.DataSource</res-type>
        <res-auth>Container</res-auth>
        <res-sharing-scope>Shareable</res-sharing-scope>
    </resource-ref>

</web-app>

Answer 1

在登录调用者应用程序中调用方法时，从另一个应用程序成功登录。您应该构建身份验证对象并设置安全上下文持有者。

    UserDetails user = userDetailsManager.loadUserByUsername(username);
    Authentication auth = new UsernamePasswordAuthenticationToken(user, null, user.getAuthorities());
    SecurityContextHolder.getContext().setAuthentication(auth);

不确定这是最好的方法，但这可能会解决您的问题。