JPA和Hibernate代理行为

Question

我试着在下面观察JPA2/Hibernate4代理行为,

//延迟加载的循环实体:

@Entity
public class Employee {

 @Id@Generated
 int id;
 String name;
 @OneToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
 Employee boss;

 public String toString() {
  return id + "|" + name + "|" + boss;
 }

 //getters and setters ...

}

//坚持实体:

// Outer entity:
Employee employee = new Employee();
employee.setName("engineer");
// Inner entity:
Employee boss = new Employee();
boss.setName("manager");
employee.setBoss(boss);

entityTransaction.begin();
entityManager.persist(employee);
entityTransaction.commit();
System.out.println(employee);

//输出:

Hibernate: insert into Employee (id, boss_id, name) values (default, ?, ?)
Hibernate: insert into Employee (id, boss_id, name) values (default, ?, ?)

2|engineer|1|manager|null

//加载外部实体:

String queryString = "select e from Employee e where e.id=" + employee.getId();
Query query = entityManager.createQuery(queryString);
Object loadedEmployee = query.getSingleResult();
System.out.println(loadedEmployee.getClass().getSimpleName());

//输出:

Hibernate: select employee0_.id as id2_, employee0_.boss_id as boss3_2_, employee0_.name as name2_ from Employee employee0_ where employee0_.id=2 limit ?

Employee

令我惊讶的是,上面加载的外部实体仍然是普通的实体,但我预计它将Hibernate proxy由此产生lazy loading.我可能在这里错过了一些东西,那么如何才能做到正确？一个简单但具体的例子非常感谢!

@编辑

根据@kostja我的回答我调整了代码并在下面以SE模式调试它,既不能LazyInitializationException生成也不能boss property代理.还有什么提示吗？

@EDIT 2

最后,我确认答案@kostja无疑是伟大的.

我在EE模式下测试,所以在proxied boss property下面观察,

// LazyInitializationException抛出:

public Employee retrieve(int id) {
 Employee employee = entityManager.find(Employee.class, id);
 // access to the proxied boss property outside of persistence/transaction ctx
 Employee boss = employee.getBoss();
 System.out.println(boss instanceof HibernateProxy);
 System.out.println(boss.getClass().getSimpleName());
 return boss;
}

//投入Spring Tx使用后的绿灯:

@Transactional
public Employee retrieve(int id) ...

//输出:

true
Employee_$$_javassist_0

另外,可以参考20.1.4.从Hibernate文档初始化集合和代理.

Answer 1

这是预期的JPA行为.您的查询中的实体没有理由被代理 - 这是查询的常规结果.boss但是,该实体的财产应该是代理人.它不会告诉我是否 - 当你对托管实体的延迟加载属性执行任何操作时,它将触发获取.

所以你应该访问交易之外的boss属性.如果没有提取,你会得到一个LazyInitializationException.

那你怎么去了解它取决于同类产品中EntityManager和PersistenceContext.

• 仅适用于JPA 2.0 - 调用em.detach(loadedEmployee)然后访问该boss属性.

对于JPA 1:

• 如果您在Java EE环境中,请使用方法标记@TransactionAttribute(TransactionAttributeType.NOT_SUPPORTED)以暂停事务.

• 在具有用户事务的SE环境中,transaction.commit()在访问boss属性之前调用.

• 如果使用的EXTENDED PersistenceContext会比交易更长,请致电em.clear().

EIDT:我认为你没有得到异常的原因是这FetchType.LAZY只是JPA提供者的一个提示,因此无法保证懒惰地加载该属性.与此相反,FetchType.EAGER保证渴望获取.我想,你的JPA提供商选择热切地加载.

我已经复制了这个例子,虽然有点不同,我可以重复地获取LazyInitializationException日志声明.该测试是在JBoss 7.1.1上运行的Arquillian测试,JPA 2.0基于Hibernate 4.0.1:

@RunWith(Arquillian.class)
public class CircularEmployeeTest {
    @Deployment
    public static Archive<?> createTestArchive() {
        return ShrinkWrap
                .create(WebArchive.class, "test.war")
                .addClasses(Employee.class, Resources.class)
                .addAsResource("META-INF/persistence.xml",
                        "META-INF/persistence.xml")
                .addAsResource("testSeeds/2CircularEmployees.sql", "import.sql")
                .addAsWebInfResource(EmptyAsset.INSTANCE, "beans.xml");
    }

    @PersistenceContext
    EntityManager em;

    @Inject
    UserTransaction tx;

    @Inject
    Logger log;

    @Test
    @TransactionAttribute(TransactionAttributeType.NOT_SUPPORTED)
    public void testConfirmLazyLoading() throws Exception {
        String query = "SELECT e FROM Employee e WHERE e.id = 1";

        tx.begin();
        Employee employee = em.createQuery(query,
                Employee.class).getSingleResult();
        tx.commit();
        log.info("retrieving the boss: {}", employee.getBoss());
    }
}