Hibernate/JPA中涉及继承时如何指定列名？

我想我想在这里鱼和熊掌兼得，但我们会看看是否有一个合理的解决方案来满足我正在寻找的问题。我有一个 Spring Boot/JPA/Hibernate 应用程序，它将与 MySQL 作为其后备存储进行通信。我有几种情况，从 OOP 的角度来看，我的实体类形成了一个父/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
}

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)
);

Ideally我想保持这个表的设计不变，“父车辆”列的名称与我上面的完全相同。但如果我正确理解 Hibernate/JPA API，那么我就不会think无需做出某种牺牲，这是可能的。我认为我要么需要牺牲应用程序层的继承，以便我可以完全按照数据库中的名称命名子类中的列：

@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
}

Or我认为我可以保留应用程序层继承，但随后需要重构我的数据库表，如下所示：

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：我是否可以保留我的应用程序层继承（并且有Motorcycle and Car继承这些属性Vehicle) and使用我的首选约定来命名我的数据库表列？

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.） 车辆类别：

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

摩托车小类

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