使用Spring CrudRepository的Hibernate LazyInitializationException
我总是得到例外：
org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: com.knapp.vk.domain.Business.businessCategorySet, could not initialize proxy - no Session 我不想把fetch设置为渴望。 什么是其他好的解决方案？
商业实体类：
@Entity public class Business { @Id @GeneratedValue private int pk; @ManyToMany private Set businessCategorySet = new HashSet(); ... }
BusinessRepository接口：
import org.springframework.data.repository.CrudRepository; import org.springframework.transaction.annotation.Transactional; @Transactional public interface BusinessRepository extends CrudRepository { }
配置：
import javax.persistence.EntityManager; import javax.persistence.EntityManagerFactory; import javax.persistence.Persistence; import org.springframework.context.annotation.Bean; import org.springframework.context.annotation.Configuration; import org.springframework.data.jpa.repository.config.EnableJpaRepositories; import org.springframework.orm.hibernate4.HibernateExceptionTranslator; import org.springframework.orm.jpa.JpaTransactionManager; import org.springframework.transaction.PlatformTransactionManager; import org.springframework.transaction.annotation.EnableTransactionManagement; @Configuration @EnableJpaRepositories(basePackages = "com.knapp.vk.repositorynew") @EnableTransactionManagement public class JPAConfiguration { @Bean public EntityManagerFactory entityManagerFactory() throws SQLException { return Persistence.createEntityManagerFactory("standardManager"); } @Bean public EntityManager entityManager(EntityManagerFactory entityManagerFactory) { return entityManagerFactory.createEntityManager(); } @Bean public PlatformTransactionManager transactionManager() throws SQLException { JpaTransactionManager txManager = new JpaTransactionManager(); txManager.setEntityManagerFactory(entityManagerFactory()); return txManager; } @Bean public HibernateExceptionTranslator hibernateExceptionTranslator() { return new HibernateExceptionTranslator(); } }
所以你可以通过在Service中初始化businessCategoruSet来实现它，而不是在Repository中使用@Transactional而是在Service中使用它。 让我们在存储库中编写一些像id或业务列表获取业务的方法，并从服务中访问该方法。
@Transactional public Class BusinessService{ @resource BusinessRepository businessRepository; public Business getBusiness(){ Business business = businessRepository.getBusiness(); //Here you should initialize BusinessCategorySet Object object = business.getBusinessCategoriesSet().size(); } }
有关@Transactional更多详细信息， 请参阅此链接
您还可以在存储库中添加方法：
@Transactional(readOnly = true) public List findAllEagerly() { CriteriaBuilder builder = em.getCriteriaBuilder(); CriteriaQuery query = builder.createQuery(Business.class); Root root = query.from(Business.class); root.fetch(Business_.businessCategorySet); return em.createQuery(query).getResultList(); }
或者更容易在JpaRepository中使用@Query：
@Query("SELECT b FROM Business b JOIN FETCH b.businessCategorySet") public List findAll();
我认为好的解决方案是使用@NamedEntityGraph
@Entity @NamedEntityGraph(name = "Business.detail", attributeNodes = @NamedAttributeNode("businessCategorySet")) public class Business {...}
并在存储库中
public interface BusinessRepository extends CrudRepository { @EntityGraph(value = "Business.detail", type = EntityGraphType.LOAD) List findAll(); }
也可以看看：
Spring Data存储库的自定义实现
配置Fetch-和LoadGraphs
Hibernate会话的范围在事务中。 因此，如果您在服务层开始交易，您的hibernate会话将可用到服务层。 如果你在另一个对象中调用任何对象，例如（business.getBusinessCategorySet（））需要加载延迟，将导致抛出“org.hibernate.LazyInitializationException”。 要解决此问题，您有两个解决方案：1）在服务层2中加载所有相关对象。使用Open Session In View（OSIV）配置，它将使您的hibernate会话打开，直到查看图层。