Question

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

我不想将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<BusinessCategory> businessCategorySet = new HashSet<BusinessCategory>();

...
}

BusinessRepository接口：

import org.springframework.data.repository.CrudRepository;
import org.springframework.transaction.annotation.Transactional;

@Transactional
public interface BusinessRepository extends CrudRepository<Business, Integer>
{

}

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();
    }
}

Answer 1

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。让我们编写一些方法，比如通过id或业务列表获取业务，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 Refer this link

的详细信息

Answer 2

您还可以在存储库中添加方法：

@Transactional(readOnly = true)
public List<Business> findAllEagerly() {
    CriteriaBuilder builder = em.getCriteriaBuilder();
    CriteriaQuery<Business> query = builder.createQuery(Business.class);
    Root<Business> root = query.from(Business.class);
    root.fetch(Business_.businessCategorySet);
    return em.createQuery(query).getResultList();
}

或更容易在JpaRepository中使用@Query：

@Query("SELECT b FROM Business b JOIN FETCH b.businessCategorySet")
public List<Business> 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 {...}

和存储库

public interface BusinessRepository extends CrudRepository<Business, Integer> {
    @EntityGraph(value = "Business.detail", type = EntityGraphType.LOAD)
    List<Business> findAll();
}

另见：

Custom implementations for Spring Data repositories

Configuring Fetch- and LoadGraphs

Answer 3

Hibernate会话的范围在事务中。因此，如果您在服务层开始交易，您的休眠会话将可用到服务层。如果你正在调用另一个对象内的任何对象，例如（business.getBusinessCategorySet（））需要加载lazy，会导致抛出＆＃34; org.hibernate.LazyInitializationException＆＃34;。要解决此问题，您有两个解决方案 1）加载服务层内的所有相关对象 2）使用Open Session In View（OSIV）配置，它将使您的休眠会话打开，直到查看图层。

使用Spring CrudRepository的Hibernate LazyInitializationException

3 个答案: