java - @EmbeddedId 和 @Id 异常

Question

我有两个 SQL 表，如下:

CREATE TABLE lost_travelers
(
    id BIGINT PRIMARY KEY DEFAULT nextval('global_seq'),
    /* a lot of other columns */
);

CREATE TABLE lost_travelers_locations
(
    lost_traveler_id BIGINT NOT NULL,
    latitude REAL NOT NULL,
    longitude REAL NOT NULL,
    location_type VARCHAR NOT NULL,

    FOREIGN KEY (lost_traveler_id) REFERENCES travelers (id) ON DELETE CASCADE
);

我希望它位于单独的表中的原因是因为 lost_travelers 表确实有很多属性。

我遇到的问题与 JPA/Hibernate 映射有关。基本上，我不希望 lost_travelers_locations 成为一个实体(有 id)。但是，当我尝试使用 @Embeddable 注释时，出现以下错误。

Caused by: org.hibernate.AnnotationException: model.location.LostTravelerLocation must not have @Id properties when used as an @EmbeddedId: model.traveler.LostTraveler.lostTravelerLocation

我的类(class)分别是:

迷失旅行者位置:

@Embeddable
@Table(name = "lost_travelers_locations")
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public class LostTravelerLocation extends Location
{
    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "lost_traveler_id")
    private LostTraveler lostTraveler;

    @Enumerated(EnumType.STRING)
    @Column(name = "location_type")
    private LocationType locationType;

    public LostTraveler getLostTraveler()
    {
        return lostTraveler;
    }

    public void setLostTraveler(LostTraveler lostTraveler)
    {
        this.lostTraveler = lostTraveler;
    }

    public LocationType getLocationType()
    {
        return locationType;
    }

    public void setLocationType(LocationType locationType)
    {
        this.locationType = locationType;
    }
}

位置类别:

@MappedSuperclass
public abstract class Location
{
    @Column(name = "latitude")
    @NotNull
    private float longitude;

    @Column(name = "longitude")
    @NotNull
    private float latitude;

    public float getLongitude()
    {
        return longitude;
    }

    public void setLongitude(float longitude)
    {
        this.longitude = longitude;
    }

    public float getLatitude()
    {
        return latitude;
    }

    public void setLatitude(float latitude)
    {
        this.latitude = latitude;
    }
}

迷失的旅行者:

@Entity
@Table(name = "lost_travelers")
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public class LostTraveler extends Traveler
{    
    @EmbeddedId
    private LostTravelerLocation lostTravelerLocation;

    /* A lot of other properties */

   public LostTravelerLocation getLostTravelerLocation()
   {
        return lostTravelerLocation;
   }

   public void setLostTravelerLocation(LostTravelerLocation lostTravelerLocation)
   {
        this.lostTravelerLocation = lostTravelerLocation;
   }

}

抽象类旅行者:

@MappedSuperclass
public abstract class Traveler extends EntityWithId
{
    /* A lot of properties as well */
}

EntityWithId:

@MappedSuperclass
public class EntityWithId
{
    @Id
    @SequenceGenerator(name = "global_seq", sequenceName = "global_seq", allocationSize = 1)
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "global_seq")
    private Long id;

    public Long getId()
    {
        return id;
    }

    public void setId(Long id)
    {
        this.id = id;
    }
}

我现在不知道问题是什么。我只是坚持 LostTraveler 是一个实体，而 LostTravelerLocation 不是。提前致谢。

Answer 1

在实体上使用 @Embeddable 时，不能声明 @Table 注解，因为它会导致冲突。一方面你说这可以嵌入到任何表中，另一方面你说它有一个独立的表，JPA 会提示。我注意到的另一件事是，您没有嵌入可嵌入的(我没有看到 @Embedded )，而是使用 @EmbeddedId ，这主要是用于复合键 ID。

也许 @Embeddable/@Embedded 不是适合您想要执行的方法，特别是因为您想要创建两个具有一对一关系的表一个映射。使用一对一映射或将 LostTravelerLocation 正确嵌入到 LostTraveler 中