Hibernate扩展了同一个表的实体

Question

我有一个有两个字段的表我希望有两个对象.

第一个只有field1

@Entity(name = "simpleTableObject")
@Table(name = "someTable")
public class SimpleTableObject
{
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    protected long id;

    @Column(name = "field1")
    private String field1;

第二个有两个领域

@Entity(name = "tableObject")
@Table(name = "someTable")
public class TableObject
{
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    protected long id;

    @Column(name = "field1")
    private String field1;

    @Column(name = "field2")
    private String field2;

我使用加载每一个

@Transactional(readOnly = true)
@SuppressWarnings("unchecked")
public List get(Class aClass)
{
    ClassMetadata hibernateMetadata = sessionFactory.getClassMetadata(aClass);
    if (hibernateMetadata == null)
    {
        return null;
    }
    if (hibernateMetadata instanceof AbstractEntityPersister)
    {
        AbstractEntityPersister persister = (AbstractEntityPersister) hibernateMetadata;
        String                  tableName = persister.getTableName();
        if (tableName != null)
        {
            return sessionFactory.getCurrentSession().
                    createQuery("from " + tableName).list();
        }
    }
    return null;
}

我想做的是TableObject扩展SimpleTableObject.我该怎么做呢？

Answer 1

如果您想在所需表中保存公共字段意味着假设您有 A 类和 B 类，并且您有一些公共字段，如 created_by,updated_by 并且您想在两个实体中保存 field1,field2：IN 数据库级别：

query> select * from A;
+----++------------------------+
| id | created_by | updated_by |
+----+------------+------------+
|  3 |    xyz     | abc        |
+----+------------+------------+
query> select * from B;
 +----++------------------------+
| id | created_by | updated_by |
+----+------------+------------+
|  3 |    xyz     | abc        |
+----+------------+------------+

对于这种类型的结构，您应该像@Dragan Bozanovic 建议的那样使用@MappedSuperclass

但是，如果您想要父子关系并希望为每个类生成表，那么您可以使用 @Inheritance(strategy = InheritanceType.TABLE_PER_CLASS) 它将为每个类创建表，例如：假设您有 2 个类 Payment 和 CreditCard，Payment 是 CreditCard 的父类。

@Entity
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public class Payment {
    @Id
    @GeneratedValue(strategy = GenerationType.TABLE)
    private int id;
    @Column(nullable = false)
    private double amount;

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public double getAmount() {
        return amount;
    }

    public void setAmount(double amount) {
        this.amount = amount;
    }
}


@Entity
public class CreditCard extends Payment {
private String ccNumber;
private Date expireDate;

    public String getCcNumber() {
        return ccNumber;
    }

    public void setCcNumber(String ccNumber) {
        this.ccNumber = ccNumber;
    }

    public Date getExpireDate() {
        return expireDate;
    }

    public void setExpireDate(Date expireDate) {
        this.expireDate = expireDate;
    }
}

现在您将保存日期：

public class TestConcreteClasses {
    public static void main(String[] args) {
        Payment payment = new Payment();
        payment.setAmount(52.6);
        createData(payment);
        CreditCard creditCard = new CreditCard();
        creditCard.setAmount(10);
        creditCard.setCcNumber("2536985474561236");
        creditCard.setExpireDate(new Date());
        createData(creditCard);

    }

    private static void createData(Payment instance) {
        Session session = HibernateUtil.getSession();
        session.beginTransaction();
        session.save(instance);
        session.getTransaction().commit();
    }
}

然后数据将保存

query> select * from Payment;
+----+--------+
| id | amount |
+----+--------+
|  1 |   52.6 |
+----+--------+
1 row in set (0.00 sec)

 select * from CreditCard;
+----+--------+------------------+---------------------+
| id | amount | ccNumber         | expireDate          |
+----+--------+------------------+---------------------+
|  2 |     10 | 2536985474561236 | 2017-03-12 14:10:15 |
+----+--------+------------------+---------------------+
1 row in set (0.00 sec)

hibernate 中使用了 3 种类型的继承，这里是继承的 hibernate doc https://docs.jboss.org/hibernate/jpa/2.1/api/javax/persistence/InheritanceType.html，你应该根据你的要求。

Answer 2

我能够对你的问题做些什么.我定义了另一个类层次结构:UserWithRole扩展了User,它类似于你的.

将类定义User为具有继承策略的实体SINGLE_TABLE:

@Entity
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@Table(name = "USERS")
public class User {
 @Id
 @GeneratedValue(strategy = GenerationType.IDENTITY)
 protected Long id;
 @Column(nullable = false)
 protected String name;
 // toString(), default constructor, setters/getters, more constructors.
 ...
}

这种继承策略有一个相当大的缺点:

• 子类不能包含不可为空的列.

还有另一种策略JOINED允许在子类中创建不可为空的列.它为每个子类创建一个附加表,这些表具有FK到超类表.

定义UserWithRole类:

@Entity
public class UserWithRole extends User {
  private String role;
  // toString(), default constructor, setters/getters, more constructors.
 ...
}

添加Helper类以在数据库中创建用户并使用您的查询:

@Component
public class Helper {
 @Autowired
 EntityManager entityManager;
 @Transactional
 public void createUsers() {
  for (long i = 0; i < 10; i++) {
   User user;
   if (i % 2 == 0) {
    user = new UserWithRole("User-" + i, "Role-" + i);
   } else {
    user = new User("User-" + i);
   }
   entityManager.persist(user);
  }
  entityManager.flush();
 }

 @Transactional(readOnly = true)
 @SuppressWarnings("unchecked")
 public < T > List < T > get(Class < T > aClass) {
  SessionFactory sessionFactory = entityManager.getEntityManagerFactory().unwrap(SessionFactory.class);
  ClassMetadata hibernateMetadata = sessionFactory.getClassMetadata(aClass);
  if (hibernateMetadata == null) {
   return null;
  }
  if (hibernateMetadata instanceof AbstractEntityPersister) {
   AbstractEntityPersister persister = (AbstractEntityPersister) hibernateMetadata;
   String entityName = persister.getEntityName();

   if (entityName != null) {
    return sessionFactory.getCurrentSession().
    createQuery("from " + entityName).list();
   }
  }
  return null;
 }

}

如你所见,我改变了你的方法:

• 添加了泛型类型以避免不安全的类型转换;
• 使用的实体名称而不是表名,因为HQL需要实体名称.

我们开始测试了

获取所有User实例:

@Test
public void testQueryUsers() {
 helper.createUsers();
 for (User user: helper.get(User.class)) {
  System.out.println(user);
 }
}

输出(具有角色的用户仍然UserWithProfile是运行时的实例):

UserWithRole{id=1, name='User-0', role='Role-0'}
User{id=2, name='User-1'}
UserWithRole{id=3, name='User-2', role='Role-2'}
User{id=4, name='User-3'}
UserWithRole{id=5, name='User-4', role='Role-4'}
User{id=6, name='User-5'}
UserWithRole{id=7, name='User-6', role='Role-6'}
User{id=8, name='User-7'}
UserWithRole{id=9, name='User-8', role='Role-8'}
User{id=10, name='User-9'}

Hibernate发出的SQL查询:

select
user0_.id as id2_0_,
 user0_.name as name3_0_,
 user0_.role as role4_0_,
 user0_.dtype as dtype1_0_
from
users user0_

获取所有UserWithProfile实例:

@Test
public void testQueryUsersWithProfile() {
 helper.createUsers();
 for (User user: helper.get(UserWithRole.class)) {
  System.out.println(user);
 }
}

输出:

UserWithRole{id=1, name='User-0', role='Role-0'}
UserWithRole{id=3, name='User-2', role='Role-2'}
UserWithRole{id=5, name='User-4', role='Role-4'}
UserWithRole{id=7, name='User-6', role='Role-6'}
UserWithRole{id=9, name='User-8', role='Role-8'}

Hibernate发出的SQL查询:

select
userwithro0_.id as id2_0_,
 userwithro0_.name as name3_0_,
 userwithro0_.role as role4_0_
from
users userwithro0_
where
userwithro0_.dtype = 'UserWithRole'

请告诉我这是你要找的.

Answer 3

您可以为两个实体都有一个共同的超类:

@MappedSuperclass
public abstract class AbstractTableObject {
 // common mappings
}

@Entity
@Table(name = "someTable")
public class TableObject extends AbstractTableObject {
 // remaining mappings
}

@Entity
@Table(name = "someTable")
@Immutable
public class SimpleTableObject extends AbstractTableObject {
 // nothing here
}

此外,您可以将SimpleTableObject实体标记为@Immutable如上所示,以便不会意外地保留或更新它.