Question

我想我也想尝试一下我的蛋糕并在这里吃，但我们会看看我能找到合理的解决方案。我有一个Spring Boot / JPA / Hibernate应用程序，它将与MySQL作为其后备存储。我有几种情况，从OOP的角度来看，我的实体类形成了父/子层次结构，如下所示：

// Groovy pseudo-code!
class Vehicle {
  Long id
  Long maxSpeed
  String make
  String model
}

class Motorcycle extends Vehicle {
  Boolean isTwoStroke
}

class Car extends Vehicle {
  Boolean hasLeatherInterior
}

等。通常，在JPA之外，我可能会像这样设计各自的表：

CREATE TABLE motorcycles (
  motorcycle_id BIGINT UNSIGNED NOT NULL AUTO_INCREMENT,
  motorcycle_max_speed BIGINT UNSIGNED,
  motorcycle_make VARCHAR(50) NOT NULL,
  motorcycle_model VARCHAR(50) NOT NULL,
  motorcycle_is_two_speed BIT NOT NULL,

  # PK, FK, UC, index constraints down here (omitted for brevity)
);

CREATE TABLE cars (
  car_id BIGINT UNSIGNED NOT NULL AUTO_INCREMENT,
  car_max_speed BIGINT UNSIGNED,
  car_make VARCHAR(50) NOT NULL,
  car_model VARCHAR(50) NOT NULL,
  car_has_leather_interior BIT NOT NULL,

  # PK, FK, UC, index constraints down here (omitted for brevity)
);

理想情况我希望将此表设计保持原样，并使用＆＃34;父车辆的名称＆＃34;列完全像我上面那样。但是，如果我正确地理解了Hibernate / JPA API，那么我就不会在没有做出某种牺牲的情况下认为它的可能性。我想我要么在app层牺牲继承，这样我就可以像在DB中一样命名子类中的列：

@Entity
class Motorcycle {    // No longer extends Vehicle :-(
  @Id
  @GeneratedValue(strategy=GenerationType.IDENTITY)
  @Column(name = "motorcycle_id")
  Long id

  @Column(name = "motorcycle_max_speed")
  Long maxSpeed

  @Column(name = "motorcycle_make")
  String make

  @Column(name = "motorcycle_model")
  String model

  @Column(name = "motorcycle_is_two_speed")
  Boolean isTwoStroke
}

@Entity
class Car {    // No longer extends Vehicle :-(
  @Id
  @GeneratedValue(strategy=GenerationType.IDENTITY)
  @Column(name = "car_id")
  Long id

  @Column(name = "car_max_speed")
  Long maxSpeed

  @Column(name = "car_make")
  String make

  @Column(name = "car_model")
  String model

  @Column(name = "car_has_leather_interior")
  Boolean hasLeatherInterior
}

或我认为我可以保留app-layer继承，但之后需要像我这样重构我的数据库表：

CREATE TABLE motorcycles (
  id BIGINT UNSIGNED NOT NULL AUTO_INCREMENT,
  max_speed BIGINT UNSIGNED,
  make VARCHAR(50) NOT NULL,
  model VARCHAR(50) NOT NULL,
  motorcycle_is_two_speed BIT NOT NULL,

  # PK, FK, UC, index constraints down here (omitted for brevity)
);

CREATE TABLE cars (
  id BIGINT UNSIGNED NOT NULL AUTO_INCREMENT,
  max_speed BIGINT UNSIGNED,
  make VARCHAR(50) NOT NULL,
  model VARCHAR(50) NOT NULL,
  car_has_leather_interior BIT NOT NULL,

  # PK, FK, UC, index constraints down here (omitted for brevity)
);

所以我问：是否可以保留我的应用层继承（让Motorcycle和Car继承Vehicle）和使用我的首选约定命名我的数据库表列？

Answer 1

可以使用@MappedSuperclass（Designates a class whose mapping information is applied to the entities that inherit from it. A mapped superclass has no separate table defined for it.）车辆类：

@MappedSuperclass
public class Vehicle {
    @Id
    @GeneratedValue
    Long id;
    Long maxSpeed;
    String make;
    String model;
}

摩托车子类

 @Entity
    @AttributeOverrides({
            @AttributeOverride(name = "id", column = @Column(name = "motorcycle_id"))
    })
    public class Motorcycle extends Vehicle {
        Boolean isTwoStroke;
    }

当涉及继承时，如何在Hibernate / JPA中指定列名？

1 个答案: