JPA整数判别器值

Question

我正在尝试遵循JPA教程并设计一个类层次结构来表示下表:

CREATE TABLE Entities (
  entity_id INT PRIMARY KEY,
  entity_type INT CHECK (0 <= entity_type AND entity_type <= 2)
  entity_name VARCHAR(255)
);

此表映射到类层次结构:

@Entity
@Inheritance
@Table(schema="myschema", name="Entities")
@DiscriminatorColumn(name="entity_type")
@SequenceGenerator(schema="myschema", name="entity_id_seq", sequenceName="entity_id_seq", allocationSize=100)
public abstract class LOTEntity {
  @Id
  @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="entity_id_seq")
  @Column(name="entity_id")
  protected long entityId;

  @Column(name="entity_name")
  protected String entityName = "";

  public LOTEntity() {}
  public LOTEntity(String name) { this.entityName = name; }
}

和

@Entity
@DiscriminatorValue("1")
class LOTClass extends LOTEntity {
  public LOTClass() {}
  public LOTClass(String name) { super(name); }
}

这不起作用,但是,因为entity_type是INT不是String:

内部异常:org.postgresql.util.PSQLException:错误:列"entity_type"的类型为整数但表达式的类型为字符不同

但是,如果我@DiscriminatorValue("1")改为@DiscriminatorValue(1),我得到一个编译错误:

类型不匹配:无法从int转换为String

我需要一个整数.有什么快速建议？

Answer 1

在您的@DiscriminatorColumn注释上,您需要指定discriminatorType:

@DiscriminatorColumn(name="entity_type", discriminatorType=DiscriminatorType.INTEGER)