Question

我有以下场景，我需要子类才能指定MappedSuperClass中找到的实际属性类型。我使用hibernate作为提供程序，我不介意使用hibernate特定的注释来解决这个问题。

@MappedSuperclass
abstract class BaseA{
   ....

   @OneToMany(mappedBy = "baseA")
   public Set<? extends BaseB> getBaseB(){
      .....
   }
}

@MappedSuperclass
abstract class BaseB{
   .....

   @ManyToOne(optional = false)
   @JoinColumn(name = "basea_id")
   public BaseA getBaseA(){
      .....
   }
}

@Entity
class BaseAImpl extends BaseA{
    public Set<BaseBImpl> getBaseB(){
      .....
   }
}

@Entity    
class BaseBImpl{

   public BaseAImpl getBaseA(){
      .....
   }
}

Answer 1

在这种情况下，

@AssociationOverride会解雇你。请参阅documentation（当然它是JPA注释）。您可以将它与@AttributeOverrides注释结合使用以覆盖基本类型。示例（摘自示例）：

@MappedSuperclass
public class Employee {
    ...
    @ManyToOne
    protected Address address;
    ...
}

@Entity 
@AssociationOverride(name="address", 
                         joinColumns=@JoinColumn(name="ADDR_ID"))
    // address field mapping overridden to ADDR_ID foreign key
public class PartTimeEmployee extends Employee {
    ...
}

覆盖JPA MappedSuperClass通用属性

1 个答案: