我正在尝试通过复制的实体获取实体字段

Posted 2023-02-26

【中文标题】我正在尝试通过复制的实体获取实体字段

【英文标题】：I'm trying to get an entity field through a copied entity

【发布时间】：2019-08-28 16:18:11

【问题描述】：

我有 2 个实体：歌手和专辑。歌手有几张专辑。我复制了歌手的精华，可以得到他的ID和名字，但是我无法得到他的专辑列表。 我有这个例外

线程“主”org.hibernate.LazyInitializationException 中的异常：无法延迟初始化角色集合：Entity.SingerEntity.albums，无法初始化代理 - 没有会话

@Entity
@Table(name = "\"singer\"")
public class SingerEntity 
    @Id
    private int singer_id;
    private String name;

    @OneToMany(mappedBy = "singer_id", cascade = CascadeType.ALL, orphanRemoval = true)
    private List<AlbumEntity> albums;

    public List<AlbumEntity> getAlbums() 
        return albums;
    

    public void setAlbums(List<AlbumEntity> albums) 
        this.albums = albums;
    

@Entity
@Table(name = "album")
public class AlbumEntity 
    @Id
    private int album_id;
    private String album_title;
    private String genre;
    private int singer_id;

主要方法：

public static void main(String[] args) 

    SingerDAO sDAO = new SingerDAO();
    AlbumDAO aDAO = new AlbumDAO();

    SingerEntity s1 = new SingerEntity(1, "Singer");
    sDAO.insert(s1);

    AlbumEntity a1 = new AlbumEntity(1, "a1", "g1", s1);
    AlbumEntity a2 = new AlbumEntity(2, "a2", "g1", s1);
    AlbumEntity a3 = new AlbumEntity(3, "a3", "g2", s1);
    AlbumEntity a4 = new AlbumEntity(4, "a4", "g2", s1);

    aDAO.insert(a1);
    aDAO.insert(a2);
    aDAO.insert(a3);
    aDAO.insert(a4);

    s1 = sDAO.findById(1);
    System.out.println(s1.getId() + " , " + s1.getName());

    List<AlbumEntity> albums = s1.getAlbums();
    System.out.println(albums.get(0).getAlbum_title());

类 SingerDAO：

public class SingerDAO 

public SingerEntity findById(int id) 
    Session session = null;
    SingerEntity singer = null;
    try 
        session = HibernateUtil.getSessionFactory().openSession();
        singer = (SingerEntity) session.get(SingerEntity.class, id);
        Hibernate.initialize(singer);
     catch (Exception e)
        e.printStackTrace();
     finally 
        if (session != null && session.isOpen())
            session.close();
        
    
    return singer;

public void select()
    Session session = HibernateUtil.getSessionFactory().openSession();
    session.beginTransaction();
    Query query= session.createQuery("from SingerEntity");
    List<SingerEntity> list = query.list();
    for (SingerEntity s:list) 
        System.out.println(s.getName());
    
    session.getTransaction().commit();
    session.close();


public void insert(SingerEntity singerEntity)
    Session session = HibernateUtil.getSessionFactory().openSession();
    session.beginTransaction();
    session.save(singerEntity);
    session.getTransaction().commit();
    session.close();


public void update(SingerEntity sE)
    Session session = HibernateUtil.getSessionFactory().openSession();
    session.beginTransaction();

    session.update(sE);
    session.getTransaction().commit();
    session.close();


public void delete(SingerEntity sE)
    Session session = HibernateUtil.getSessionFactory().openSession();
    session.beginTransaction();
    session.delete(sE);
    session.getTransaction().commit();
    session.close();

【参考方案1】：

首先正确映射 OneToMany 双向关系。你应该有

@Entity
@Table(name = "singer")
public class SingerEntity 
    ...
    @OneToMany(mappedBy = "singer", cascade = CascadeType.ALL, orphanRemoval = true)
    private List<AlbumEntity> albums;
    ...

@Entity
@Table(name = "album")
public class AlbumEntity 
    ...
    @ManyToOne
    @JoinColumn(name = "singer_id)
    private SingerEntity singer;
    ....

如果您想更好地将专辑添加到实体，请在 Singer 实体上使用自定义方法来保持引用正确，例如：

public void addAlbum(AlbumEntity a) 
    albums.add(a);
    a.setSinger(this);

之后，您可以保存专辑，只保留歌手，因为您有级联选项。

然后，由于 OneToMany 关系集合默认是惰性的，您应该将它加载到您获得 SingerEntity 的同一事务中。因此 SingerEntity 将附加到 Hibernate 会话。例如，您可以在方法上使用 Spring @Transactional 注释。

我没有看到你的 DAO 代码，但 main 可以看：

SingerDAO sDAO = new SingerDAO();
AlbumDAO aDAO = new AlbumDAO();

SingerEntity s1 = new SingerEntity(1,"Singer");
s1.addAlbum(new AlbumEntity(1,"a1","g1",1)); 
s1.addAlbum(new AlbumEntity(2,"a2","g2",1)); 
s1.addAlbum(new AlbumEntity(3,"a3","g3",1)); 
s1.addAlbum(new AlbumEntity(4,"a4","g4",1)); 
sDAO.insert(s1);

更改 findById 以使用一个事务进行延迟加载：

public SingerEntity findById(int id) 
    Session session = null;
    SingerEntity singer = null;
    try 
        session = HibernateUtil.getSessionFactory().openSession();
        //start transaction
        singer = (SingerEntity) session.get(SingerEntity.class, id);
        Hibernate.initialize(singer);
        //end transaction 
     catch (Exception e)
        e.printStackTrace();
     finally 
        if (session != null && session.isOpen())
            session.close();
        
    
    return singer;

【讨论】：

感谢您的回答，但我不太明白您的意思。你能写出main方法的代码吗？

我只需要在控制台中显示所有艺术家专辑的名称。

我得到了这个例外：“实体映射中的重复列：Entity.AlbumEntity 列：singer_id（应使用 insert="false" update="false"）"

你不能有私有的intsinger_id；和私人 SingerEntity 歌手；在一个实体中。你应该删除intsinger_id，异常就会消失。

我不能删除singer_id，因为这是实体的主键