如何在Hibernate中删除子项的ID-s项？

Question

我正在尝试编写一个JPQL查询,它会删除所有PlaylistItem-s,ArtContent通过ArtContentID 来引用特定的-s .

我试过这个:

public int deleteItemsByContentIds(Long[] contentIds) {
    EntityManager em = getEntityManager();
    int result = em.createQuery(
    "delete from PlaylistItem where artContent.id in (:idsArray) ")
    .setParameter("idsArray", contentIds).executeUpdate();

    return result;
}

但它引发了一个例外:

Servlet.service() for servlet RemoveContentServlet threw exception: 
javax.ejb.EJBException: java.lang.IllegalArgumentException: 
Encountered array-valued parameter binding, but was expecting [java.lang.Long]

什么是可以理解的,因为没有setParameter方法将数组作为参数.那么解决这个问题的最佳方法是什么？

简化的类定义:

@Entity
public class PlaylistItem implements Serializable {

@Id
@GeneratedValue
private Long id;

private int position;

@ManyToOne(optional = false)
@JoinColumn(name = "PLAYLIST_ID")
private Playlist playlist;

@ManyToOne(optional = false)
@JoinColumn(name = "ART_CONTENT_ID")
private ArtContent artContent;

...

}

@Entity
public class ArtContent implements Serializable {

@Id
@GeneratedValue
private Long id;
...
}

Answer 1

您可以继续使用,.setParameter但需要使值扩展Collection(如ArrayList)而不是使用数组类型.也许只需改为:

public int deleteItemsByContentIds(Long[] contentIds) {
    EntityManager em = getEntityManager();
    int result = em.createQuery(
    "delete from PlaylistItem where artContent.id in (:idsArray) ")
    .setParameter("idsArray", Arrays.asList(contentIds)).executeUpdate();

    return result;
}