Hibernate：引起：java.lang.ClassCastException：[Ljava.lang.Object;无法转换为<package.class> </package.class>

时间：2015-04-15 16:53:37

标签： java spring hibernate spring-security

我从数据库中获取数据并将其转换为我需要的类型时遇到问题。

我尝试使用Spring安全性从数据库中搜索已重置的用户。 它的工作方式如下：我插入userName和密码，方法loadUserByUserName(String userName)从UserDetailsServiceImpl类转到数据库，并且cheaks for用户存在于DB中，如果是（用户存在于DB中），则从DB获取它授权用户并为他呈现安全页面。但问题是在我的DAO impl类方法中无法将它转换为需要我的对象 - App1User我的实体类，我在登录页面上看到这样的消息（原因：[Ljava.lang.Object;不能被强制转换为app1.domain.App1User enter code here）：

我在DAOImpl类中的方法：

public App1User findUserByName(String userName) {


        Session session = sessionFactory.openSession();
        Transaction transaction = session.beginTransaction();
        List<App1User>  listUser = null;


        try {

            String stringSQL = "select au.userId, au.firstName, au.lastName, au.middleName, au.username, au.password, au.userPosition from App1User au where au.username = :userNameArg";
            Query query = session.createQuery(stringSQL);
            query.setParameter("userNameArg", userName);

            listUser = (List<App1User>)query.list();

            session.flush();    
            session.clear();
            transaction.commit();

        } catch(HibernateException e) {

            System.err.println(e);

            if(transaction != null)
                transaction.rollback();

        } finally {
            if(session != null)
                session.close();
        }

        return (App1User)listUser.get(0);
    }

我的hibernate.hbm.xml 

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration SYSTEM 
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">

<hibernate-configuration>

   <session-factory>

       <property name="hibernate.current_session_context_class">thread</property> 

        <property name="hibernate.default_schema">PUBLIC</property>

        <!-- JDBC connection pool (use the built-in) -->
        <property name="connection.pool_size">1</property>

        <property name="hibernate.transaction.auto_close_session">false</property>

        <!-- SQL dialect -->
        <property name="dialect">org.hibernate.dialect.H2Dialect</property>

        <!-- Disable the second-level cache  -->
        <property name="cache.provider_class">org.hibernate.cache.internal.NoCacheProvider</property>

        <!-- Echo all executed SQL to stdout -->
        <property name="show_sql">true</property>

        <!-- Drop and re-create the database schema on startup -->
        <property name="hbm2ddl.auto">create</property>

        <mapping resource="/WEB-INF/hibernateConf/App1User.hbm.xml"/>

    </session-factory>

</hibernate-configuration>

我现在的例外情况：

SEVERE: An internal error occurred while trying to authenticate the user.
org.springframework.security.authentication.InternalAuthenticationServiceException: [Ljava.lang.Object; cannot be cast to app1.domain.App1User
    at org.springframework.security.authentication.dao.DaoAuthenticationProvider.retrieveUser(DaoAuthenticationProvider.java:125)
    at org.springframework.security.authentication.dao.AbstractUserDetailsAuthenticationProvider.authenticate(AbstractUserDetailsAuthenticationProvider.java:143)
    at org.springframework.security.authentication.ProviderManager.authenticate(ProviderManager.java:167)
    at org.springframework.security.authentication.ProviderManager.authenticate(ProviderManager.java:192)
    at org.springframework.security.web.authentication.UsernamePasswordAuthenticationFilter.attemptAuthentication(UsernamePasswordAuthenticationFilter.java:93)
    at org.springframework.security.web.authentication.AbstractAuthenticationProcessingFilter.doFilter(AbstractAuthenticationProcessingFilter.java:217)
    at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:330)
    at org.springframework.security.web.authentication.logout.LogoutFilter.doFilter(LogoutFilter.java:120)
    at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:330)
    at org.springframework.security.web.csrf.CsrfFilter.doFilterInternal(CsrfFilter.java:120)
    at org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:107)
    at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:330)
    at org.springframework.security.web.header.HeaderWriterFilter.doFilterInternal(HeaderWriterFilter.java:64)
    at org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:107)
    at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:330)
    at org.springframework.security.web.context.request.async.WebAsyncManagerIntegrationFilter.doFilterInternal(WebAsyncManagerIntegrationFilter.java:53)
    at org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:107)
    at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:330)
    at org.springframework.security.web.context.SecurityContextPersistenceFilter.doFilter(SecurityContextPersistenceFilter.java:91)
    at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:330)
    at org.springframework.security.web.FilterChainProxy.doFilterInternal(FilterChainProxy.java:213)
    at org.springframework.security.web.FilterChainProxy.doFilter(FilterChainProxy.java:176)
    at org.springframework.web.filter.DelegatingFilterProxy.invokeDelegate(DelegatingFilterProxy.java:344)
    at org.springframework.web.filter.DelegatingFilterProxy.doFilter(DelegatingFilterProxy.java:261)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:239)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
    at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:219)
    at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:106)
    at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:502)
    at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:142)
    at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:79)
    at org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:610)
    at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:88)
    at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:518)
    at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1091)
    at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:668)
    at org.apache.coyote.http11.Http11NioProtocol$Http11ConnectionHandler.process(Http11NioProtocol.java:223)
    at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1517)
    at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.run(NioEndpoint.java:1474)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
    at org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
    at java.lang.Thread.run(Thread.java:745)
Caused by: java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to app1.domain.App1User
    at app1.persistance.UserManageDAOImpl.findUserByName(UserManageDAOImpl.java:102)
    at app1.web.authentification.UserDetailsServiceImpl.loadUserByUsername(UserDetailsServiceImpl.java:36)
    at org.springframework.security.authentication.dao.DaoAuthenticationProvider.retrieveUser(DaoAuthenticationProvider.java:114)
    ... 42 more

还有我的UserDetailServiceImpl类：

public UserDetails loadUserByUsername(String userName) 
                                                throws UsernameNotFoundException {


        App1User listUser = userManageDAOIF.findUserByName(userName);
        String password = null;
        Collection<GrantedAuthority> authorities = null;

        if(listUser != null) {

                password = (String)listUser.getPassword();

                authorities = new ArrayList<GrantedAuthority>();

                    authorities.add(new SimpleGrantedAuthority("ROLE_USER"));


                org.springframework.security.core.userdetails.User secureUser = new
                        org.springframework.security.core.userdetails.User(userName, password, authorities);

                return secureUser;

        } else {

            throw new UsernameNotFoundException("No user exist in datbase");
        }

    }

1 个答案:

答案 0 :(得分：3)

当您从DB中选择特定列时，您获得的结果不是实体对象，而是Object[]，其中每个数组元素都包含不同的列值。因此，您从query.list()方法获得的内容是：List<Object[]>。因此，当您将list.get(0)投射到App1User时，肯定会产生ClassCastException。

所以，你得到Object[]并构建App1User实体对象，如下所示：

// size of this array would be number of columns in select query
Object[] attr = listUser.get(0);
App1User appUser = new App1User(attr[0], attr[1], attr[2], ...);

当然不是最好的方法。更好的方法是从db获得List<App1User>。为此，只需更改查询：

String stringSQL = "select au.userId, au.firstName, au.lastName, au.middleName, au.username, au.password, au.userPosition from App1User au where au.username = :userNameArg";

为：

String stringSQL = "select au from App1User au where au.username = :userNameArg";
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