我有一个奇怪的问题,即hibernate没有在多对一的关系中创建预期的实体类型。我们有子类层次结构下面的实体(简化):Hibernate在关系中创建错误的实体子类型
@Entity
@Table(name = "A")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "DISCRIMINATOR", discriminatorType = DiscriminatorType.STRING, length = 1)
public abstract class A {
@Id
...
public Long getId() { ... }
...
}
@Entity
@DiscriminatorValue("1")
public class A1 extends A {
...
}
@Entity
@DiscriminatorValue("2")
public class A2 extends A {
...
}
@Entity
@Table(name = "B")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "DISCRIMINATOR", discriminatorType = DiscriminatorType.STRING, length = 1)
public abstract class B<AClass extends A> {
protected AClass a;
@Id
...
public Long getId() { ... }
...
public abstract AClass getA();
public void setA(AClass a) { ... }
}
@Entity
@DiscriminatorValue("1")
public class B1 extends B<A1> {
...
@Override
@ManyToOne(fetch = EAGER)
@JoinColumn(name = "A_ID")
public A1 getA() { ... }
}
@Entity
@DiscriminatorValue("2")
public class B2 extends B<A2> {
...
@Override
@ManyToOne(fetch = EAGER)
@JoinColumn(name = "A_ID")
public A2 getA() { ... }
}
在persistence.xml
两个实体在
A2
A1
B2
B1
现在我在DB创建A1的实例和B1的顺序声明:
A1 a1 = new A1();
entityManager.persist(a1);
B1 b1 = new B1();
b1.setA(a1);
entityManager.persist(b1);
我可以看到实例保存到数据库的每个ID都是1,DISCRIMINATOR也是1,B中的A_ID也是1.
现在当我试图让B(在另一个Hibernate的Session):
B b = entityManager.find(B.class, 1L);
我得到异常:
org.hibernate.PropertyAccessException: Exception occurred inside getter of B
Caused by: java.lang.ClassCastException: A2 cannot be cast to A1
at B1.getA(B1.java:61)
... 108 more
调试,我发现,Hibernate是创造正确的实体键入B1并为与A的关系创建类型A2的不正确实体。如果persistence.xml
中的顺序发生更改,则会创建正确的类型A1。在这种情况下,hibernate似乎并未将A表的DISCRIMINATOR列考虑在内,但总是创建在配置中声明的第一个子类型。这怎么解决?注释有问题吗?
(我也有其在超B中第一注释的具体实施方法getA()
的,但是这会导致类似的问题。)
谢谢你的回答。昨天我有机会测试这个,它在问题的场景中工作:) – Gandalf