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我已经被这个问题困扰了一段时间,我需要你的帮助。我会尽量提供尽可能多的细节,并会根据需要进行编辑。

我正在尝试在 Spring 安全性的自定义 UserDetailsS​​ervice 中的数据库中创建用户。场景是:当用户登录我的应用程序时,如果用户存在,我会检查我的数据库;如果不存在,我会调用 Web 服务来检查该用户是否存在并导入他的所有信息,以便将它们保存到我自己的数据库中并授予他对 Web 应用程序的访问权限。

我可以从数据库中读取信息(SELECT 查询),但是如果没有得到以下(美丽的)错误,我就无法进行任何插入:

org.springframework.security.authentication.AuthenticationServiceException: Could not commit JPA transaction; nested exception is javax.persistence.RollbackException: Error while committing the transaction
org.springframework.security.authentication.dao.DaoAuthenticationProvider.retrieveUser(DaoAuthenticationProvider.java:85)

在应用程序的其他部分(一旦经过身份验证),我可以使用来自控制器的任何服务。插入+选择工作正常。

关于我的配置,我有以下文件: spring security config (Spring security context) is in "applicationContext-security.xml" spring beans (Root Context) in "applicationContext.xml" spring mvc (DispatcherServlet) in webmvc-config.xml和一个 applicationContext-jpa.xml + persistence.xml

UserDetailsS​​ervice 的实现:

@Transactional(propagation = Propagation.REQUIRES_NEW)
public class CustomUserDetailsService implements UserDetailsService {


    @PersistenceContext
    private EntityManager em;

    [...Other Declarations...]

    @Autowired
    private userService userService;

    @Override
    public UserDetails loadUserByUsername(String login) throws UsernameNotFoundException {
        User user = null; //extends spring security UserDetails 

            //If I try to find existing users in database here, it works perfectly
        User user = userService.findUserByLogin(login);

        List<GrantedAuthority> authorities = ...
        Set<String> permissions = ...

        if(user == null){
             //Call webservice and get User if exists => newUser

           //Save in user table
                userService.saveUser(newUser);  
               //No error here, I can even try a userService.findUser(newUser.getId()); and I get the results. the commit is after returning the UserDetails from this method.


             //I also tried the following solution... the entitymanager is well Autowired and I can access everything as usual
           //   em.getTransaction().begin();  // => Throw here 'nested exception is java.lang.IllegalStateException: Not allowed to create transaction on shared EntityManager - use Spring transactions or EJB CMT instead'
           //   em.persist(operator); 
           //   em.getTransaction().commit();
           //   em.close();
                    user = newUser

            }
            else{
                      //Other things
            }

        }
        return user;  //if I tried a saveUser, commit is done after the return and the exception is thrown
    }
}

我在类和方法本身上尝试了 @Transactional 的所有组合。

现在所有的配置文件[只有配置文件的有用部分]

网页.xml:

 <context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>classpath*:META-INF/spring/applicationContext*.xml</param-value>
  </context-param>
  <listener>
     <listener-class>org.springframework.security.web.session.HttpSessionEventPublisher</listener-class>
  </listener>
  <listener>
     <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
  </listener>
  <servlet>
    <servlet-name>myApp</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
      <param-name>contextConfigLocation</param-name>
      <param-value>WEB-INF/spring/webmvc-config.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
  </servlet>

应用上下文.xml:

<context:spring-configured/>

<context:component-scan base-package="com.company.myApp">
    <context:exclude-filter expression="org.springframework.stereotype.Controller" type="annotation"/>
</context:component-scan>


<bean class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close" id="dataSource">
    <property name="driverClassName" value="${database.driverClassName}"/>
    <property name="url" value="${database.url}"/>
    <property name="username" value="${database.username}"/>
    <property name="password" value="${database.password}"/>
    <property name="testOnBorrow" value="true"/>
    <property name="testOnReturn" value="true"/>
    <property name="testWhileIdle" value="true"/>
    <property name="timeBetweenEvictionRunsMillis" value="1800000"/>
    <property name="numTestsPerEvictionRun" value="3"/>
    <property name="minEvictableIdleTimeMillis" value="1800000"/>
    <property name="validationQuery" value="SELECT 1 FROM DUAL"/>
</bean>

<bean class="org.springframework.orm.jpa.JpaTransactionManager" id="transactionManager">
    <property name="entityManagerFactory" ref="entityManagerFactory"/>
</bean>

<tx:annotation-driven mode="aspectj" transaction-manager="transactionManager"/>

<bean class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean" id="entityManagerFactory">
    <property name="persistenceUnitName" value="persistenceUnit"/>
    <property name="dataSource" ref="dataSource"/>
</bean>

我的事务管理器处于模式 =“aspectj”(我试过没有它,代理)我的所有 Aspectj 库都运行良好。加载时编织等等。我还在我的应用程序中使用 *.aj 文件。我还尝试在我的 customUserDetailsS​​ervice 类上添加一个 aop:advisor

applicationContext-security.xml:

<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-3.1.xsd
        http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd" profile="local,dev,int">
    <global-method-security pre-post-annotations="enabled">
        <expression-handler ref="expressionHandler"/>
    </global-method-security>
    <!-- HTTP security configurations -->
    <http auto-config="true" use-expressions="true">
        <expression-handler ref="webExpressionHandler"/>
        <form-login login-processing-url="/resources/j_spring_security_check" login-page="/login" authentication-failure-url="/login?login_error=t" />
        <logout logout-url="/resources/j_spring_security_logout" />
        <!-- Configure these elements to secure URIs in your application -->
        <intercept-url pattern="/resources/**" access="permitAll" />
        <intercept-url pattern="/login/**" access="permitAll" />
        <intercept-url pattern="/**" access="isAuthenticated()" />
        <session-management session-fixation-protection="migrateSession">
            <concurrency-control max-sessions="1"/>
        </session-management>
        <x509 subject-principal-regex="CN=[^,]* ([^,]*),.*$" user-service-ref="customUserDetailsService"  />
        <logout delete-cookies="JSESSIONID" />
    </http>
    <!-- Configure Authentication mechanism -->
    <authentication-manager alias="authenticationManager">
        <!-- SHA-256 values can be produced using 'echo -n your_desired_password | sha256sum' (using normal *nix environments) -->
        <authentication-provider  user-service-ref="customUserDetailsService">
            <password-encoder hash="sha-256" />
        </authentication-provider>
    </authentication-manager>    
</beans:beans>

applicationContext-jpa.xml 只有基本的存储库声明:

<beans:beans xmlns:beans="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://www.springframework.org/schema/data/jpa"
  xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd http://www.springframework.org/schema/data/jpa http://www.springframework.org/schema/data/jpa/spring-jpa.xsd">

      <repositories base-package="com.company.myApp.common.repository" />

</beans>

webmvc-config 有 context:component-scan 用于控制器和其他仅适用于 webapp 的东西。with<aop:aspectj-autoproxy/>和其他 aop:advisors。

我正在测试嵌入在 Eclipse STS 中的 ApacheTomcat 服务

编辑:完整的堆栈跟踪:

Could not commit JPA transaction; nested exception is javax.persistence.RollbackException: Error while committing the transaction
org.springframework.security.authentication.dao.DaoAuthenticationProvider.retrieveUser(DaoAuthenticationProvider.java:85)
org.springframework.security.authentication.dao.AbstractUserDetailsAuthenticationProvider.authenticate(AbstractUserDetailsAuthenticationProvider.java:132)
org.springframework.security.authentication.ProviderManager.authenticate(ProviderManager.java:156)
org.springframework.security.authentication.ProviderManager.authenticate(ProviderManager.java:174)
org.springframework.security.web.authentication.UsernamePasswordAuthenticationFilter.attemptAuthentication(UsernamePasswordAuthenticationFilter.java:94)
org.springframework.security.web.authentication.AbstractAuthenticationProcessingFilter.doFilter(AbstractAuthenticationProcessingFilter.java:194)
org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:323)
org.springframework.security.web.authentication.preauth.AbstractPreAuthenticatedProcessingFilter.doFilter(AbstractPreAuthenticatedProcessingFilter.java:88)
org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:323)
org.springframework.security.web.authentication.logout.LogoutFilter.doFilter(LogoutFilter.java:105)
org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:323)
org.springframework.security.web.context.SecurityContextPersistenceFilter.doFilter(SecurityContextPersistenceFilter.java:87)
org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:323)
org.springframework.security.web.session.ConcurrentSessionFilter.doFilter(ConcurrentSessionFilter.java:125)
org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:323)
org.springframework.security.web.FilterChainProxy.doFilter(FilterChainProxy.java:173)
org.springframework.web.filter.DelegatingFilterProxy.invokeDelegate(DelegatingFilterProxy.java:346)
org.springframework.web.filter.DelegatingFilterProxy.doFilter(DelegatingFilterProxy.java:259)
org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235)
org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
org.springframework.web.filter.HiddenHttpMethodFilter.doFilterInternal(HiddenHttpMethodFilter.java:77)
org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:76)
org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235)
org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
org.springframework.web.filter.CharacterEncodingFilter.doFilterInternal(CharacterEncodingFilter.java:88)
org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:76)
org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235)
org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233)
org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:615)
org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:293)
org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:859)
org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:602)
org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:409)
java.util.concurrent.ThreadPoolExecutor$Worker.runTask(ThreadPoolExecutor.java:886)
java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:908)
java.lang.Thread.run(Thread.java:619)
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1 回答 1

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我终于找到了解决方案......这是一个愚蠢的错误,但我接近解决方案。上面的配置其实很好。RootContext 和 SecurityContext 共享事务和 bean。

通过从实体管理器手动调用persist() 方法而不开始事务,给了我另一个堆栈跟踪。问题实际上是 JSR303 验证失败......(来自网络服务的错误 RegExp)

因此,对于将来的使用,如果您想将“数据源”共享给 Spring SecurityContext,则此配置(在问题中)有效。为了从我的问题中获取堆栈跟踪,我使用了以下内容:

   em.persist(user); 
   em.flush();  
   em.close();

无论如何,谢谢你的帮助:)

于 2013-03-22T14:49:48.210 回答