Question

我有这个自我关系注释

public class Estoria implements Serializable {

 @EmbeddedId 
 protected EstoriaPK estoriaPK; 

@ManyToOne
@JoinColumns({@JoinColumn(name = "id_estoria", referencedColumnName = "id", updatable = false, insertable = false, nullable = true)
          ,@JoinColumn(name = "id_projeto" , referencedColumnName = "id_projeto" ,     insertable = false, updatable = false)})
private Estoria subtask;

@OneToMany(mappedBy = "subtask",cascade = CascadeType.ALL) 
private Collection<Estoria> subtasks; 

}

我试着坚持这个集合::

public void persistSubtasks(int idEstoria, Collection<Estoria> subtasks) {
    Estoria rootEstoria = entityManager.createNamedQuery("Estoria.findById", Estoria.class)
                    .setParameter("id", idEstoria)
                    .getSingleResult();

        rootEstoria.setSubtasks(subtasks);
}

但是数据库持久性没有持久化根引用ID：

+----+------------+------------+--------------------------+-----------     -------------------------------------------------------------------+------   ------+--------+--------------+
| id | id_projeto | id_estoria | nome                     | descricao                                                                      | estimativa | status | data_criacao |
+----+------------+------------+--------------------------+------------------------------------------------------------------------------+------------+--------+--------------+
|  1 |          7 |       NULL | Check all old activities | No one know if every feature was checked before builded                      |       NULL |      0 | 2015-01-02   |
|  2 |          7 |       "supost to be 1" | Register all users       | Everyone needs to be registered and identified by profile                    |       NULL |      0 | 2015-01-02   |

有人能给我一些帮助吗？

好消息这篇文章＆gt; Understanding annotations and JPA(hibernate)解决了我的问题，但我真的不知道我需要引用root＆gt;＆gt;叶/叶＆gt;＆gt;根

Answer 1

您必须在Child ID中设置父ID。在子任务集合中设置rootEstoria id。

public void persistSubtasks(int idEstoria, Collection<Estoria> subtasks) {
    Estoria rootEstoria = entityManager.createNamedQuery("Estoria.findById", Estoria.class)
                .setParameter("id", idEstoria)
                .getSingleResult();
    for(Estoria subtask:subtasks){
        subtask.setSubtask(rootEstoria);
    }
    rootEstoria.setSubtasks(subtasks);
}

坚持JPA集合oneToMany不起作用

1 个答案: