在Hibernate中映射具有多个ID的扩展实体

Question

I have implemented a db which is consisted of Article and Fruit tables, with the following specifications:

create table Fruit
(
ART_ID            bigint,
FRU_ID          bigint not null auto_increment,
FRU_FROZEN   varchar(15),
primary key(FRU_ID)
);

# Implemented

create table Article
(
ART_ID            bigint,
ART_NAME         varchar(10) not null,
ART_COST   varchar(10) not null,
primary key(ART_ID)
);

alter table Fruits add constraint FK_FRUIT_ARTICLE foreign key (ART_ID) references Article (ART_ID) on delete restrict on update restrict;

and with following class entities : Article.java

@Entity      
@Table(name = "Article")
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public class Article implements Serializable
 {
@Id @GeneratedValue(strategy=GenerationType.AUTO)
@Basic(optional = false)
@Column(name = "ART_ID")
private Long id;

@Basic(optional = false)
@Column(name = "ART_NAME")
private String name;

@Basic(optional = true)
@Column(name = "ART_COST")
private String cost; 

// constructors, getters and setters

Fruit.java

@Entity
@Table(name="Fruit")
public class Fruit extends Article{


@Id @GeneratedValue(strategy=GenerationType.IDENTITY)
@Basic(optional = false)
@Column(name = "FRU_ID")
private long fruitID;

@Basic(optional = false)
@Column(name = "FRU_FROZEN")
private String fruitFrozen;

// constructors, getters and setters

Now, my problem is how to have ID in Article and in Fruit, if I keep it this way, it throws me exception that sub-class must not contain IDClass because it will result with multiple ID's. Any help is appreciated. Thanks in advance.

Answer 1

It means what it says, an Entity cannot have two IDs.

Consider the operation

entityManager.find(Article.class, 1l);

How is it supposed to know which one to return if there are two articles (an article and a fruit) that both have id 1?

If your tables need to each have IDs, entity inheritance is not an appropriate solution. Consider putting the common elements in an @MappedSuperclass instead.