Hibernate CriteriaBuilder：当它是实体时不是实体

Question

我有一个包含 30 个左右实体的工作应用程序。主要使用自动生成查询的 CRUD 存储库。现在，有些查询没有优化，我想使用 Entity Graph 系统来连接其中的一些表。

到目前为止一切顺利，我遵循了几个教程并编写了代码。剩下的唯一问题是，当我运行它时，我收到了这个错误：

Not an entity: class com.stingray.syncmanager.models.StationRoleModel

这很奇怪，因为就像我之前所说的，我的应用程序已经使用了这个实体。在我看来，实体管理器或会话没有“意识到”这些实体的存在，就像它们在单独的“上下文”中一样，如果这有意义的话。

谁能指出我正确的方向？这是我的代码：

CustomSqlRepository.java

public CustomSqlRepository(HibernateConfig hibernateConfig, EntityManager entityManager) {
    Map<String, String> settings = new HashMap<>();
    settings.put("connection.driver_class", hibernateConfig.getDriverClassName());
    settings.put("hibernate.connection.url", hibernateConfig.getConnectionUrl());
    settings.put("hibernate.connection.username", hibernateConfig.getUsername());
    settings.put("hibernate.connection.password", hibernateConfig.getPassword());
    settings.put("hibernate.dialect", "org.hibernate.dialect.MySQLDialect");
    settings.put("hibernate.current_session_context_class", "thread");

    ServiceRegistry serviceRegistry = new StandardServiceRegistryBuilder()
            .applySettings(settings).build();

    MetadataSources metadataSources = new MetadataSources(serviceRegistry);
    Metadata metadata = metadataSources.buildMetadata();

    sessionFactory = metadata.getSessionFactoryBuilder().build();
    this.entityManager = entityManager;
}

public Session getCurrentSession() {
    return sessionFactory.getCurrentSession();
}

@Override
public List<StationRoleModel> findAllByRole(RoleModel role) {
    EntityGraph<?> graph = entityManager.getEntityGraph("stationrole-entity-graph-with-station");

    try (Session session = getCurrentSession()) {
        Transaction transaction = session.beginTransaction();

        CriteriaBuilder builder = session.getCriteriaBuilder();

        //===================================================================
        //This line crashes with "Not an entity" which doesn't make sense at all because it IS an entity...
        //===================================================================
        CriteriaQuery<StationRoleModel> criteria = builder.createQuery(StationRoleModel.class);
            Root<StationRoleModel> from = criteria.from(StationRoleModel.class);
        criteria.select(from);
        criteria.where(builder.equal(from.get("role"), role));
        TypedQuery<StationRoleModel> typed = entityManager
                    .createQuery(criteria)
                    .setHint("javax.persistence.fetchgraph", graph);

        try {
            return typed.getResultList();
        } catch (final NoResultException nre) {
            return new ArrayList<>();
        } finally {
            transaction.commit();
        }
    }
}

StationRoleModel.java

@Entity(name = "StationRoleModel")
@Table(name = "station_role")
@Getter
@Setter
@AllArgsConstructor(access = AccessLevel.PRIVATE)
@NoArgsConstructor(access = AccessLevel.PUBLIC)
@NamedEntityGraph(name = "stationrole-entity-graph-with-station",
        attributeNodes = @NamedAttributeNode("station"))
public class StationRoleModel extends BaseEntity {
    ...
}

Answer 1

在我看来，实体管理器或会话没有“意识到”这些实体的存在，就像它们在单独的“上下文”中一样，如果这有意义的话。

那是因为情况如此。您传入了一个EntityManager，它与现有的EntityManagerFactory 相关联，因此也与SessionFactory 相关联，但在存储库的构造函数中，您正在创建一个新的SessionFactory。 SessionFactory/Session 和 EntityManagerFactory/EntityManager 的管理应该由容器（在你的情况下是 Spring）来完成，所以你不应该乱构建任何一个。只需将EntityManager 与@Autowired 一起注入即可。