Question

这是我的代码段：

public class Object1 implements Serializable {
    @Id
    @Column(length = 36)
    protected String id;

    @Column(length = 36, insertable = false, updatable = false)
    protected String parentID;

    @OneToMany(fetch = FetchType.LAZY)
    @JoinColumn(name = "object2ID", referencedColumnName = "parentID")
    protected List<Object2> parents = new ArrayList<>();

    public List<Object2> getParents() {
        return parents;
    }
}

public class Object2 implements Serializable {
    @Id
    @Column(length = 36)
    protected String id;

    @Column(length = 36, insertable = false, updatable = false)
    protected String object2ID;

    @Column(length = 36, insertable = false, updatable = false)
    protected String parentID;

    @OneToMany(fetch = FetchType.LAZY)
    @JoinColumn(name = "object2ID", referencedColumnName = "parentID")
    protected List<Object2> parents = new ArrayList<>();

    public List<Object2> getParents() {
        return parents;
    }
}

和应用程序类：

public class Application {
    public static Logger logger = LoggerFactory.getLogger(Application.class);

    public static void main(String[] args) {
        Path HIBERNATE_CONFIGURATION = Paths.get("");
        Configuration configuration = new Configuration().configure(HIBERNATE_CONFIGURATION.toFile());
        ServiceRegistry serviceRegistry = new StandardServiceRegistryBuilder().applySettings(
                configuration.getProperties()).build();
        SessionFactory sessionFactory = configuration.buildSessionFactory(serviceRegistry);

        Session session = sessionFactory.openSession();
        Object1 object1 = (Object1) session.get(Object1.class, "1");
        logger.info(object1.toString());

        Object2 object2 = object1.getParents().get(0);
        logger.info(object2.toString());

        while (!object2.getParents().isEmpty()) {
            object2 = object2.getParents().get(0);
            logger.info(object2.toString());
        }

        session.close();
    }

}

我按预期获得了Object1，但是object2抛出异常org.hibernate.LazyInitializationException: could not initialize proxy - no Session

会话未关闭，为什么我收到此错误？

我使用的是Hibernate核心：4.3.7.Final

解决：

您好。谢谢大家。我发现解决了我的问题。我尝试获取OneToMany，但在db引用中实际是类型ManyToMany。我为db和model创建了一些小改动。

我重命名对象。这是新的代码段：

@Entity
@Table(name = "Houses")
public class House implements Serializable {
    @Id
    @Column(length = 36)
    protected String id;

    @Column(length = 36)
    protected String parentGUID;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "parentguid", referencedColumnName = "guid")
    protected AddressObject address;

    public AddressObject getAddress() {
        return address;
    }

    @Override
    public String toString() {
        return "Object1{" +
                "id='" + id + '\'' +
                ", parentGUID='" + parentGUID + '\'' +
                '}';
    }
}

@Entity
@Table(name = "AddressObjects")
public class AddressObject implements Serializable {
    @Id
    @Column(length = 36)
    protected String id;

    @Column(length = 36, unique = true)
    protected String guid;

    @Column(length = 36, nullable = true)
    protected String parentGUID;

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "address")
    protected List<House> houses = new ArrayList<>();

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "upHierarchicObject")
    protected List<AddressObject> downHierarchicObject = new ArrayList<>();

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "parentguid", referencedColumnName = "guid")
    protected AddressObject upHierarchicObject;

    public List<House> getHouses() {
        return houses;
    }

    public List<AddressObject> getDownHierarchicObject() {
        return downHierarchicObject;
    }

    public AddressObject getUpHierarchicObject() {
        return upHierarchicObject;
    }

    @Override
    public String toString() {
        return "Object2{" +
                "id='" + id + '\'' +
                ", guid='" + guid + '\'' +
                ", parentGUID='" + parentGUID + '\'' +
                '}';
    }
}

和应用程序类：

public class Application {
    public static Logger logger = LoggerFactory.getLogger(Application.class);

    public static void main(String[] args) {
        ArrayList<Object> objects = new ArrayList<>();
        Path HIBERNATE_CONFIGURATION = Paths.get("config/hibernate.test.cfg.xml");
        Configuration configuration = new Configuration().configure(HIBERNATE_CONFIGURATION.toFile());
        ServiceRegistry serviceRegistry = new StandardServiceRegistryBuilder().applySettings(
                configuration.getProperties()).build();
        SessionFactory sessionFactory = configuration.buildSessionFactory(serviceRegistry);

        Session session = sessionFactory.openSession();
        House house = (House) session.get(House.class, "1");
        objects.add(house);

        AddressObject addressObject = house.getAddress();
        objects.add(addressObject);

        while (addressObject.getUpHierarchicObject() != null) {
            addressObject = addressObject.getUpHierarchicObject();
            objects.add(addressObject);
        }

        for (Object obj : objects) {
           logger.info("Object: {}", obj);
        }
        session.close();
    }
}

但是，除了LazyInitializationException，我不知道为什么。这是hibernate的一个错误吗？

Answer 1

懒惰就是你忘了在查询中进行提取（这里是为了父母），你试图在关闭会话后检索java对象在代码的这一部分中，您如何知道来自object1的object2是通过Parents获取的？这里我们只测试null

object1.getParents() != null

但是在惰性的情况下，对象不是null而是虚拟

如果要使用它，则需要打开会话但在此步骤中会话已经关闭因为我们在从数据库中重新获取对象后立即关闭会话

示例：

Session s = sessions.openSession();
Transaction tx = s.beginTransaction();

Employee e = (Employee) s.createQuery("from Employee e where e.name=:empName").setString("empName", eName).uniqueResult();
List roles = u.getRoles();
tx.commit();
s.close();
String role = roles.get(0); //  This line will throw error

 Easy Solution

 Use lazy=false in your Entity class.

注意

此处会话未关闭。这可以做到这一点

获取条件可以在Entity类的多对一条件中执行 fetch =＆＃34; join＆＃34; 或 FetchMode.JOIN 的技巧映射中定义的获取策略会影响：

retrieval via get()

希望这有帮助。

Answer 2

您必须提供虚拟循环才能首先获取所有对象。

例如

 for (Object2 obj:object1.getParents()) {
          obj.getName();
 }

如何为惰性对象获取hibernate层次结构

2 个答案: