Question

我的超类是：

@Entity
@Table(name = "TEST_VEHICLE")
@ChangesListener
@AttributeOverride(name = "id", column = @Column(name = "VEHICLE_ID"))
@Inheritance(strategy = InheritanceType.JOINED)
@DiscriminatorColumn(name = "VEHICLE_TYPE_ID", discriminatorType = DiscriminatorType.INTEGER)
public abstract class Vehicle extends ParentEntity {
    @Column(name = "MAX_SPEED", nullable = false)
    private Integer maxSpeed;

    public Integer getMaxSpeed() {
        return maxSpeed;
    }

    public void setMaxSpeed(Integer maxSpeed) {
        this.maxSpeed = maxSpeed;
    }
}

和子类是：

@Entity
@Table(name = "TEST_BUS")
@DiscriminatorValue("2")
public class Bus extends Vehicle {
    @Column(name = "PASSENGER_NUMBER", nullable = false)
    private Short passengerNumber;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "FOO_OF_VEHICLE")
    private Foo foo;

    public Short getPassengerNumber() {
        return passengerNumber;
    }

    public void setPassengerNumber(Short passengerNumber) {
        this.passengerNumber = passengerNumber;
    }

    public Foo getFoo() {
        return foo;
    }

    public void setFoo(Foo foo) {
        this.foo = foo;
    }
}

在条件中使用foo上的Root<Vehicle>获取

root.fetch("foo", JoinType.LEFT);

导致此错误：

java.lang.IllegalArgumentException: Unable to locate Attribute  with the the given name [foo] on this ManagedType ...

如何从子类中获取字段？

更新

使用treat不能解决我的问题：

Root<Bus> busRoot = builder.treat(root, Bus.class);
busRoot.fetch("foo", JoinType.INNER);

我没有收到任何错误，但是foo还没有收到。生成的SQL是：

SELECT vehicle0_.VEHICLE_ID    AS VEHICLE_2_72_,
  vehicle0_.ATTACHMENT_COUNT   AS ATTACHME3_72_,
  vehicle0_.COMMENTS           AS COMMENTS4_72_,
  vehicle0_.CREATE_TIMESTAMP   AS CREATE_T5_72_,
  vehicle0_.CREATOR_USER_ID    AS CREATOR_8_72_,
  vehicle0_.MODIFIER_USER_ID   AS MODIFIER9_72_,
  vehicle0_.UPDATE_TIMESTAMP   AS UPDATE_T6_72_,
  vehicle0_.MAX_SPEED          AS MAX_SPEE7_72_,
  vehicle0_1_.FOO_OF_VEHICLE   AS FOO_OF_V3_70_,
  vehicle0_1_.PASSENGER_NUMBER AS PASSENGE1_70_,
  vehicle0_2_.ENGINE_TYPE      AS ENGINE_T1_71_,
  vehicle0_.VEHICLE_TYPE_ID    AS VEHICLE_1_72_
FROM TEST_VEHICLE vehicle0_
LEFT OUTER JOIN TEST_BUS vehicle0_1_
ON vehicle0_.VEHICLE_ID=vehicle0_1_.VEHICLE_ID
LEFT OUTER JOIN TEST_CAR vehicle0_2_
ON vehicle0_.VEHICLE_ID   =vehicle0_2_.VEHICLE_ID
WHERE vehicle0_.VEHICLE_ID=105

Answer 1

使用元模型可以解决这个问题。

public abstract class Bus_ extends com.rh.cores.architecture.tests.models.Vehicle_ {

    public static volatile SingularAttribute<Bus, Foo> foo;
    public static volatile SingularAttribute<Bus, Short> passengerNumber;
}

意味着：

root.fetch(Bus_.foo, JoinType.LEFT);

但由于JPA中的提取签名是这样的：

<Y> Fetch<X, Y> fetch(SingularAttribute<? super X, Y> attribute, JoinType jt);

上面的代码导致编译错误！改变这样的代码：

SingularAttribute attribute = Bus_.foo;
root.fetch(attribute, JoinType.LEFT);

我们可以在JPA标准中绕过泛型检查SingularAttribute<? super X, Y>，而Hibernate处理它！

JPA：在JOINED继承

1 个答案: