我有这种情况:
- 从IncomingMessage表中获取(读取和删除)一条记录
 - 读取记录内容
 - 在某些表中插入一些东西
 - 如果在步骤 1-3 中发生错误(任何异常),则将错误记录插入OutgoingMessage表
 - 否则,在OutgoingMessage表中插入一条成功记录
 
所以步骤 1,2,3,4 应该在一个事务中,或者步骤 1,2,3,5
我的流程从这里开始(这是一个计划任务):
public class ReceiveMessagesJob implements ScheduledJob {
// ...
    @Override
    public void run() {
        try {
            processMessageMediator.processNextRegistrationMessage();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
// ...
}
我在 ProcessMessageMediator 中的主要功能(processNextRegistrationMessage):
public class ProcessMessageMediatorImpl implements ProcessMessageMediator {
// ...
    @Override
    @Transactional
    public void processNextRegistrationMessage() throws ProcessIncomingMessageException {
        String refrenceId = null;
        MessageTypeEnum registrationMessageType = MessageTypeEnum.REGISTRATION;
        try {
            String messageContent = incomingMessageService.fetchNextMessageContent(registrationMessageType);
            if (messageContent == null) {
                return;
            }
            IncomingXmlModel incomingXmlModel = incomingXmlDeserializer.fromXml(messageContent);
            refrenceId = incomingXmlModel.getRefrenceId();
            if (!StringUtil.hasText(refrenceId)) {
                throw new ProcessIncomingMessageException(
                        "Can not proceed processing incoming-message. refrence-code field is null.");
            }
            sqlCommandHandlerService.persist(incomingXmlModel);
        } catch (Exception e) {
            if (e instanceof ProcessIncomingMessageException) {
                throw (ProcessIncomingMessageException) e;
            }
            e.printStackTrace();
            // send error outgoing-message
            OutgoingXmlModel outgoingXmlModel = new OutgoingXmlModel(refrenceId,
                    ProcessResultStateEnum.FAILED.getCode(), e.getMessage());
            saveOutgoingMessage(outgoingXmlModel, registrationMessageType);
            return;
        }
        // send success outgoing-message
        OutgoingXmlModel outgoingXmlModel = new OutgoingXmlModel(refrenceId, ProcessResultStateEnum.SUCCEED.getCode());
        saveOutgoingMessage(outgoingXmlModel, registrationMessageType);
    }
    private void saveOutgoingMessage(OutgoingXmlModel outgoingXmlModel, MessageTypeEnum messageType)
            throws ProcessIncomingMessageException {
        String xml = outgoingXmlSerializer.toXml(outgoingXmlModel, messageType);
        OutgoingMessageEntity entity = new OutgoingMessageEntity(messageType.getCode(), new Date());
        try {
            outgoingMessageService.save(entity, xml);
        } catch (SaveOutgoingMessageException e) {
            throw new ProcessIncomingMessageException("Can not proceed processing incoming-message.", e);
        }
    }
// ...
}
正如我所说,如果在步骤 1-3 中发生任何异常,我想插入一个错误记录:
catch (Exception e) {
    if (e instanceof ProcessIncomingMessageException) {
        throw (ProcessIncomingMessageException) e;
    }
    e.printStackTrace();
    //send error outgoing-message
    OutgoingXmlModel outgoingXmlModel = new OutgoingXmlModel(refrenceId,ProcessResultStateEnum.FAILED.getCode(), e.getMessage());
    saveOutgoingMessage(outgoingXmlModel, registrationMessageType);
    return;
}
这是 SqlCommandHandlerServiceImpl.persist() 方法:
public class SqlCommandHandlerServiceImpl implements SqlCommandHandlerService {
// ...
    @Override
    @Transactional
    public void persist(IncomingXmlModel incomingXmlModel) {
        Collections.sort(incomingXmlModel.getTables());
        List<ParametricQuery> queries = generateSqlQueries(incomingXmlModel.getTables());
        for (ParametricQuery query : queries) {
            queryExecuter.executeQuery(query);
        }
    }
// ...
}
但是当sqlCommandHandlerService.persist()抛出异常(这里是 org.hibernate.exception.ConstraintViolationException 异常)时,在 OutgoingMessage 表中插入错误记录后,当要提交事务时,我得到 UnexpectedRollbackException。我不知道我的问题出在哪里:
Exception in thread "null#0" org.springframework.transaction.UnexpectedRollbackException: Transaction rolled back because it has been marked as rollback-only
    at org.springframework.transaction.support.AbstractPlatformTransactionManager.commit(AbstractPlatformTransactionManager.java:717)
    at org.springframework.transaction.interceptor.TransactionAspectSupport.commitTransactionAfterReturning(TransactionAspectSupport.java:394)
    at org.springframework.transaction.interceptor.TransactionInterceptor.invoke(TransactionInterceptor.java:120)
    at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:172)
    at org.springframework.aop.framework.Cglib2AopProxy$DynamicAdvisedInterceptor.intercept(Cglib2AopProxy.java:622)
    at ir.tamin.branch.insuranceregistration.services.schedular.ReceiveMessagesJob$$EnhancerByCGLIB$$63524c6b.run(<generated>)
    at ir.asta.wise.core.util.timer.JobScheduler$ScheduledJobThread.run(JobScheduler.java:132)
我正在使用 hibernate-4.1.0-Final,我的数据库是 oracle,这是我的事务管理器 bean:
<bean id="transactionManager"
    class="org.springframework.orm.hibernate4.HibernateTransactionManager">
    <property name="sessionFactory" ref="sessionFactory" />
</bean>
<tx:annotation-driven transaction-manager="transactionManager"
    proxy-target-class="true" />
提前致谢。