Spring + Hibernate =“手动”交易如何 [英] Spring + Hibernate = "manual" transactions how-to

问题描述

我的webapp（Spring3 + Hibernate3）总是与服务类一起使用，这些服务是用 @Transactional 和这个配置注释的：

 < tx：annotation-driven transaction-manager =transactionManager/> 
 
< bean id =transactionManagerclass =org.springframework.orm.hibernate3.HibernateTransactionManager> 
< property name =sessionFactoryref =mySessionFactory/> 
< / bean> 
  

现在...我在Google AppEngine上。由于我还不知道的一些令人讨厌的原因，@Transactional不起作用。它使用javax.naming中的某个类，该类没有列入白名单。结果是：
$ b

创建名为'mySessionFactory'的bean时出错：后处理
FactoryBean的对象失败;嵌套异常是
java.lang.SecurityException：无法获得类
的成员org.hibernate.impl.SessionFactoryImpl

请不要问我为什么....：-\

使用Spring的HibernateTemplate而不是我的dao（它使用原始会话工厂）解决了问题，但我知道它已经过时了。

因此，我想尝试使用手动旧式交易。问题：

• 哪里？我想保留服务层中的交易。


• 如何？

解决方案

SessionFactoryImpl 在Google App Engine白名单中。有很多Google点击讨论它。

至于做什么，您可以选择：

• 依赖于另一个JPA提供程序




• 根本不使用ORM， JdbcTemplate（我的最爱）

我不确定为什么你需要使用编程事务管理，因为Hibernate是你问题的根源，但如果你只是喜欢知道如何，这里是一个草案：

  public class SomeService implements SomeInterface {
 
 private SomeDao thisDaoWrapsJdbcTemplate; 
私人PlatformTransactionManager transactionManager; 
 
 public void setTransactionManager（PlatformTransactionManager transactionManager）{
 this.transactionManager = transactionManager; 
} 
 
 public void doBusiness（Business：business）{
 
 TransactionDefinition def = new DefaultTransactionDefinition（）; 
 TransactionStatus status = transactionManager.getTransaction（def）; 
 
尝试{
 
 //在这里做生意
金钱= Money.LOTS_OF 
 ... 
 //将钱汇入.. 
 thisDaoWrapsJdbcTemplate.depositLotsOfMoney（money）
 
 transactionManager.commit（status）; 
 $ b $ catch（DataAccessException dae）{
 
 transactionManager.rollback（status）; 
扔dae; 
} 
 return; 
} 
  

My webapp (Spring3 + Hibernate3) always worked with services class-annotated with @Transactional and this configuration:

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

    <bean id="transactionManager" class="org.springframework.orm.hibernate3.HibernateTransactionManager">
        <property name="sessionFactory" ref="mySessionFactory" />
    </bean>

Now... I'm on Google AppEngine. For some nasty reason I don't know yet, @Transactional does not work. It uses some class in javax.naming, which is not whitelisted. It ends up with:

Error creating bean with name 'mySessionFactory': Post-processing of the FactoryBean's object failed; nested exception is java.lang.SecurityException: Unable to get members for class org.hibernate.impl.SessionFactoryImpl

Please don't ask me why.... :-\

Using Spring's HibernateTemplate instead of my dao (which uses raw session factory) solved the problem, but I know it's a little obsolete.

So, I want to try using manual old style transactions. Questions:

• where? I'd like to keep the transactions in the service layer.
• how?

解决方案

SessionFactoryImpl dependency is not in Google App Engine whitelist. There is a number of Google hits discussing it.

As far as "what to do", you have options:

• Depend on on another JPA provider

• Don't use ORM at all, and go native with Spring's JdbcTemplate (my favorite)

• I am not sure why you need to use a programmatic transaction management since Hibernate is the root of your problem, but if you just like to know how, here is a draft:

public class SomeService implements SomeInterface {

   private SomeDao thisDaoWrapsJdbcTemplate;
   private PlatformTransactionManager transactionManager;

   public void setTransactionManager( PlatformTransactionManager transactionManager ) {
      this.transactionManager = transactionManager;
   }

   public void doBusiness( Business: business ) {

      TransactionDefinition def = new DefaultTransactionDefinition();
      TransactionStatus status = transactionManager.getTransaction( def );

      try {

         // do business here
         Money money = Money.LOTS_OF
         ...
         // wire the money in..
         thisDaoWrapsJdbcTemplate.depositLotsOfMoney( money )

         transactionManager.commit( status );

      } catch ( DataAccessException dae ) {

         transactionManager.rollback( status );
         throw dae;
      }
      return;
   }

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