JPA合并（）分离实体插入时，它存在

我使用JPA与休眠提供程序。我的继承人部分代码：JPA合并（）分离实体插入时，它存在

@Entity 
@IdClass(xxxPK.class) 
@NamedQueries({ 
... 
}) 
@Inheritance(strategy = InheritanceType.JOINED) 
public class xxx extends AbstractEntity { 

public static final String GET_TRADES = "getInstruments"; 
public static final String GET_STRUCTURE = "getStructure"; 
public static final String GET_ALL = "getAllTrades"; 

@Column(nullable = false) 
@Id 
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "xxx") 
@SequenceGenerator(name = "xxxx", sequenceName = "xxxxS", allocationSize = 1) 
private Long systemId; 

@Id 
@Column(nullable = false) 
private Integer version; 
//////

而一个后代：

@Entity 
@PrimaryKeyJoinColumns(
    { 
      @PrimaryKeyJoinColumn(name = "systemId"), 
      @PrimaryKeyJoinColumn(name = "version") 
    } 
) 
public class YYY extends xxx { 
//////

我创建一个YYY实例和persiti是如我所料。之后，我在每个表中的每一行db：

fx = Persistence.invoke(new Persistence.Invoke<YYY>() { 
      @Override 
      public YYY run(EntityManager em) { 
       BufferedReader reader = null; 
       try { 
        reader = new BufferedReader(new FileReader(new ClassPathResource("json/XXX_from.json").getFile())); 
       } catch (IOException ex) { 
        System.out.print("File is not found"); 
       } 
       YYYfx = (YYY) EntityConverter.getConverter(YYY.class).fromJson(reader); 
       em.persist(fx); 
       return fx; 
      } 
     });

所以，之后，fx被分离，并作为SPEC。赛斯我可以使它在另一个会话通过调用合并（）持久的，所以叫它但：

Persistence.invoke(new Persistence.Invoke<Object>() { 
    @Override 
    public Object run(EntityManager em) { 
     fx.setZZZ("A"); 
     em.merge(fx); 
     return null; 
    } 
});

之后，我有另一行XXX和YYY表与新的主。它看起来像提供程序将fx作为TRANSIENT而不是作为DETACHED然后它就好了 - 它是一个新的对象，我会插入它。我不知道该怎么做我已经在这样愚蠢的情况下失去了4个小时。请帮忙！

public class Persistence { 

    private static EntityManagerFactory emf; 

    public static EntityManagerFactory getEmf() { 
     return emf; 
    } 

    public void setEmf(EntityManagerFactory emf) { 
     Persistence.emf = emf; 
    } 

    public static <T> T invoke(Invoke<T> inv) { 
     T result = null; 
     EntityManager em = getEmf().createEntityManager(); 
     EntityTransaction transaction = em.getTransaction(); 
     transaction.begin(); 
      try { 
      result = inv.run(em); 
      transaction.commit(); 
     } catch (Exception e) { 
      throw new RuntimeException(e); 
     } finally { 
      em.close(); 
     } 
     return result; 
    } 

    public interface Invoke<T> { 
     public T run(EntityManager em); 
    } 

}

来源

2014-11-14 user2709715