涉及继承时如何在 Hibernate/JPA 中指定列名？

Question

我想我想在这里吃我的蛋糕，但我们会看看是否有一个可行的解决方案来解决我正在寻找的问题。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，我的实体类形成了这样的父/子层次结构：

// 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
}

等等。通常，在 JPA 之外，我可能会像这样设计它们各自的表：

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)
);

理想情况下我想保持这个表格设计原样，“父车辆”列的名称与我上面的一样。但是，如果我正确理解了 Hibernate/JPA API，那么我不认为在不做出某种牺牲的情况下是不可能的。我认为我要么需要牺牲应用层的继承，以便我可以完全按照我在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
}

或者我认为我可以保留应用层继承，但随后需要像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：我是否可以保留我的应用层继承（并让 Motorcycle 和 Car 从 Vehicle 继承这些属性）并保留我的数据库表列使用我的首选约定命名？

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.) 车辆等级：

@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;
    }