postgresql 在子类的 Hibernate 中为每个表指定不同的序列
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Specifying distinct sequence per table in Hibernate on subclasses
提问by gutch
Is there a way to specify distinct sequences for each table in Hibernate, if the ID is defined on a mapped superclass?
如果 ID 是在映射的超类上定义的，有没有办法为 Hibernate 中的每个表指定不同的序列？
All entities in our application extend a superclass called DataObjectlike this:
我们应用程序中的所有实体都扩展了一个如下所示的超类DataObject：
@MappedSuperclass
public abstract class DataObject implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE)
@Column(name = "id")
private int id;
}
@Entity
@Table(name = "entity_a")
public class EntityA extends DataObject { ... }
@Entity
@Table(name = "entity_b")
public class EntityB extends DataObject { ... }
This causes all entities to use a shared sequence, the default hibernate_sequence.
这会导致所有实体使用共享序列，默认hibernate_sequence.
What I would like to do is use a separate sequence for each entity, for example entity_a_sequenceand entity_b_sequencein the example above. If the ID were specified on the subclasses then I could use the @SequenceGeneratorannotation to specify a sequence for each entity, but in this case the ID is on the superclass. Given that ID is in the superclass, is there a way I can use a separate sequence for each entity — and if so, how?
我想做的是为每个实体使用一个单独的序列，例如entity_a_sequence和entity_b_sequence上面的例子。如果在子类上指定了 ID，那么我可以使用@SequenceGenerator注释为每个实体指定一个序列，但在这种情况下，ID 位于超类上。鉴于 ID 在超类中，有没有一种方法可以为每个实体使用单独的序列——如果是这样，如何使用？
(We are using PostgreSQL 8.3, in case that's relevant)
（我们使用的是 PostgreSQL 8.3，以防万一）
回答by Fran?ois Cassin
Have you tried doing it this way ?
你试过这样做吗？
@MappedSuperclass
public abstract class DataObject implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "idgen")
@Column(name = "id")
private int id;
}
@Entity
@SequenceGenerator(initialValue = 1, name = "idgen", sequenceName = "entityaseq")
@Table(name = "entity_a")
public class EntityA extends DataObject {
}
@Entity
@SequenceGenerator(initialValue = 1, name = "idgen", sequenceName = "entitybseq")
@Table(name = "entity_b")
public class EntityB extends DataObject {
}
I'm sorry I don't have the required environment to test it right now but I'll try it later.
很抱歉，我现在没有测试所需的环境，但我稍后会尝试。
回答by Frans
We use this in the abstract superclass of all of our JPA entities:
我们在所有 JPA 实体的抽象超类中使用它：
@Id
@GeneratedValue(generator = "pooled")
@GenericGenerator(name = "pooled", strategy = "org.hibernate.id.enhanced.TableGenerator", parameters = {
@org.hibernate.annotations.Parameter(name = "value_column_name", value = "sequence_next_hi_value"),
@org.hibernate.annotations.Parameter(name = "prefer_entity_table_as_segment_value", value = "true"),
@org.hibernate.annotations.Parameter(name = "optimizer", value = "pooled-lo"),
@org.hibernate.annotations.Parameter(name = "increment_size", value = "100")})
private Long id;
It's a bit verbose, but it allows setting the prefer_entity_table_as_segment_valuewhich means you don't need to repeat the idfield or the generator annotations in the subclasses.
它有点冗长，但它允许设置prefer_entity_table_as_segment_value这意味着您不需要id在子类中重复字段或生成器注释。
回答by Gumovvy Steven
IHMO there is better way to do this:
IHMO 有更好的方法来做到这一点：
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
It works in my app.
它适用于我的应用程序。
回答by feel good and programming
TABLE generation stretergy uses separate db sequence for each tablebut it is little expensive operation
表生成策略为每个表使用单独的数据库序列，但它的操作代价很小
