休眠 - WrongClassException

Question

这是我正在构建的音乐系统领域模型的一部分：

@Entity
@Table(name="library_entities", schema="myschema")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name="entity_type", discriminatorType = DiscriminatorType.STRING)
public abstract class LibraryEntity {

    public LibraryEntity() {}
    
    // ... some fields and their getters/setters ...

}

@Entity
public abstract class Artist extends LibraryEntity {
    
    public Artist() {}

    @OneToMany(mappedBy="artist")
    private Set<Album> albums = new HashSet<>();

    @OneToMany(mappedBy="artist")
    private Set<Track> tracks = new HashSet<>();

    // ... some Artist-specific fields and their getters/setters

    public Set<Album> getAlbums() {
        return this.albums;
    }

    public Set<Track> getTracks() {
        return this.tracks;
    }

}

@Entity
@DiscriminatorValue("B")
public class Band extends Artist {

    public Band() {}

    // ...        
    private Set<Singer> members = new HashSet<>();

}

@Entity
@DiscriminatorValue("S")
public class Singer extends Artist {

    public Singer() {}

    @ManyToMany(mappedBy="members")
    private Set<Band> bands;

}

@Entity
@DiscriminatorValue("A")
public class Album extends LibraryEntity {

    public Album() {}

    @ManyToOne
    @JoinColumn(name="artist_id")
    private Artist artist;

    @OneToMany(mappedBy="album")
    private Set<Track> tracks = new HashSet<>();

    public Set<Track> getTracks() {
        return this.tracks;
    }

}

@Entity
@DiscriminatorValue("T")
public class Track extends LibraryEntity {

    public Track() {}

    @ManyToOne
    @JoinColumn(name="artist_id")
    private Artist artist;

    @ManyToOne
    @JoinColumn(name="album_id")
    private Album album;

    public Artist getArtist() {
        return this.artist;
    }

    public Album getAlbum() {
        return this.album;
    }

}

生成的架构（如预期）：

--------------------------------------------
| entity_type | ... | album_id | artist_id |
--------------------------------------------

一切正常，但是当我运行以下代码时：

public class AppMain {

    public static void main(String[] args) {

        final List<Artist> results =
            entityManager
            .createQuery("SELECT artist FROM Artist artist")
            .getResultList();

        try {

            for(Artist artist : results) {
                artist.getTracks(); // works fine
                artist.getAlbums(); // throws WrongClassException
            }
            
        } catch(Exception e){
            e.printStackTrack();
        }            

    }

}

通过跟踪 Hibernate 生成的 SQL 查询，很明显 Hibernate 错误地将 Track 获取为 Album，从而抛出了 WrongClassException。

org.hibernate.WrongClassException: Object [id=93c348c2-fb21-4239-9e47-63ef08ae3040] 
was not of the specified subclass [com.database.Models.Album] : loaded object was of 
wrong class class com.database.Models.Track

我在这里做错了什么？请帮忙。

[更新]
我意识到通过第一次调用artist.getAlbums()，第二次调用（artist.getTracks()）失败了。第一次调用 artist.getTracks()，第二次调用 (artist.getAlbums()) 失败。

[更新 2]
当我删除时问题消失了：

public class Album extends LibraryEntity {

    @ManyToOne                      // removed
    @JoinColumn(name="artist_id")   // removed
    private Artist artist;          // removed

}
public class Artist extends LibraryEntity {

    @OneToMany(mappedBy="artist")   // removed
    private Set<AlbumModel> albums = new HashSet<AlbumModel>(); //removed

}

但是，这种关系是绝对需要的，因为 Track 可以是单个轨道（因此需要对 Artist 的引用）。而Album 肯定也需要这种关系。

Answer 1

当您在Album 和Track 上使用@DiscriminatorOptions(force = true) 来强制为关联连接添加类型鉴别器约束时，这应该可以工作。

Answer 2

我似乎找到了解决办法。
根据this，两个类（在本例中为Album 和Track）不应该使用相同的连接列（尽管我认为应该有这样的能力）。 所以通过创建album_artist_id 和track_artist_id，问题就解决了。

编辑：
我找到了更好的解决方案。 我添加了另一个名为 Artwork 的抽象类，其中包含对类 Artist 的引用。
Album 和 Track 都派生自 Artwork。
Artwork 也标有 @MappedSuperclass 注释。
并且Album 和Track 已更改为：

@Entity
@DiscriminatorValue("A")
@AssociationOverride(name="artist", 
    joinColumns=@JoinColumn(name="album_artist_id"))
public class Album extends Artwork {
    ...
}

@Entity
@DiscriminatorValue("T")
@AssociationOverride(name="artist", 
    joinColumns=@JoinColumn(name="track_artist_id"))
public class Track extends Artwork {
    ...
}

最后我们仍然有两个外键，但从 OOP 的角度来看，它是固定的。
让我知道你的意见。