Question

我们正在使用Hibernate作为JPA提供程序开发Java EE应用程序。我们现在希望使用Hibernate Criterias，但为此我需要访问HibernateEntityManagerImpl。

我们目前在componentContext.xml

中有这个

  <context:component-scan base-package="com.volvo.it.lsm"/>

<!-- JPA EntityManagerFactory -->
<bean id="EmployeeDomainEntityManagerFactory" parent="hibernateEntityManagerFactory"      class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean"> 
     <property name="persistenceUnitName" value="EmployeeDomainPU" /> 
</bean>

在我们的课程中：

@PersistenceContext(unitName = "EmployeeDomainPU")
public void setEntityManager(EntityManager em) {
    this.em = em;
}

谢谢！

Answer 1

EntityManager em ...

//if you want Session with JPA 1.0
org.hibernate.Session session = (org.hibernate.Session) em.getDelegate();

//If you use JPA 2.0 this is preferred way
org.hibernate.Session session2 = em.unwrap(org.hibernate.Session.class);

//if you really want Hibernates implementation of EntityManager, just cast
//it (this is not needed for Criteria queries though). Actual implementing class
//is of course Hibernates business and it can vary
org.hibernate.ejb.EntityManagerImpl emi = 
 (org.hibernate.ejb.EntityManagerImpl) em;

Answer 2

使用此代码：

Session session = (Session) entityManager.getDelegate();
session.createCriteria(...);

请注意，JPA2也有（不同的）Criteria API。使用IMHO更难，但它更安全，并且具有与Hibernate Criteria API相同的优点（能够动态组合查询）

Answer 3

您可以调用em.getDelegate()并将其强制转换为Hibernate类

Hibernate EntityManagerImpl：获取或设置对它的访问权限

3 个答案: