考虑以下实体。我打算让Child类引用两个衍生食品类(Local Food或Foreign)中的任何一个。这是一个人为的例子,我的真实域对象非常复杂,因此组合和使用FoodType列不是一个选项,因为Food子类只在少数特征中相似。
@MappedSuperclass
public abstract class Food {
}
@Entity
public class LocalFood extends Food {
private long id;
private String name;
}
@Entity
public class ForeignFood extends Food {
private long id;
private String name;
}
@Entity
public class Child {
private Food food; //Base Class needed here
@ManyToOne()
public Food getFood() {
return food;
}
}
Caused by: org.hibernate.AnnotationException: @OneToOne or @ManyToOne on com.sample.Child.food references an unknown entity: com.sample.Food
也没有使用继承和鉴别器。
@Entity
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
public abstract class Food {
private long id; // set , get (Auto gen)
}
是否可以使这种映射工作?
答案 0 :(得分:2)
JB Nizet是对的。现在食品类看起来像这样。
@Entity
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class Food {
private long id;
public void setId(long id) {
this.id = id;
}
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
public long getId() {
return id;
}
}
从子类中删除了id。