Question

我正在尝试使用Spring JPA（Hibernate）映射现有数据库表。例如，一个表（THING）具有包含另一个表（PLACE）的外键的复合主键。我试图通过为Thing创建一个单独的IdClass来映射这个，但是当我尝试通过Can not set cktest.Place field cktest.ThingId.place to java.lang.Long save Thing CrudRepository<Thing, ThingId>时我得到@Entity public class Place { @Id @GeneratedValue(strategy=GenerationType.AUTO) @Column(name = "ID") private long id; @Column(name = "NAME", nullable = false) private String name; @OneToMany(mappedBy = "place") Set<Thing> things; ... constructor, get/set methods, equals, hashCode ... }

我的地图不正确，还是这是JPA限制？有解决办法吗？

Place.java：

@Entity
@IdClass(ThingId.class)
public class Thing {
  @Id
  String name;

  @Id
  Place place;

  ... constructors, get/set methods, equals, hashCode ...
}

Thing.java：

@Embeddable
public class ThingId implements Serializable {
  @ManyToOne
  @JoinColumn(name = "PLACE_ID", referencedColumnName = "ID", nullable = false)
  private Place place;

  @Column(name="NAME")
  private String name;
}

ThingId.java：

... imports ...
public interface ThingRepository extends CrudRepository<Thing, ThingId> {}

ThingRepository.java：

@Configuration
@EnableAutoConfiguration
public class Main {

  public static void main(String[] args) {
    ConfigurableApplicationContext context = SpringApplication.run(Main.class);
    PlaceRepository placeRepository = context.getBean(PlaceRepository.class);
    ThingRepository thingRepository = context.getBean(ThingRepository.class);

    // save a place
    Place home = placeRepository.save(new Place("Home"));

    // save a thing
    Thing re = new Thing(home, "Fido");
    // Throws 

    thingRepository.save(re);  

    context.close();
  }
}

Main.java：

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Spring JPA：使用包含外键的复合键保存对象时出错

0 个答案: