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Hibernate EntityManagerImpl: Getting or setting up access to it

开发者 https://www.devze.com 2023-04-12 01:30 出处:网络
We are developing a Java EE application using Hibernate as a JPA provider. We now wish to use Hibernate Criterias but for that I need to get access to the HibernateEntityManagerImpl.

We are developing a Java EE application using Hibernate as a JPA provider. We now wish to use Hibernate Criterias but for that I need to get access to the HibernateEntityManagerImpl.

We have this currently in our componentContext.xml

  <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>

and in our classes:

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

Thanks!


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;


Use this code:

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

Note that JPA2 has a (different) Criteria API as well. It's harder to use IMHO, but it's more type-safe, and has the same advantages as the Hibernate Criteria API (be able to dynamically compose a query)


You can call em.getDelegate() and cast it to the Hibernate class

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