1

请在下面找到实体代码,

@Entity
public class A implements Serializable {

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private Long id;

    @OneToMany(cascade={CascadeType.PERSIST, CascadeType.MERGE}, fetch = FetchType.LAZY, mappedBy="parentActivity")        
    private Set<A> subActivities;

    @ManyToOne(cascade = CascadeType.REFRESH, fetch = FetchType.LAZY)
    @JoinColumn(name = "PARENTACTIVITYID", insertable = true, updatable = true)
    private A parentActivity;

    // Getters, Setters, serialVersionUID, etc...
}   

如果我们想同时保留父母和孩子,那么下面的代码可以正常工作

public static void main(String[] args) {

    EntityManager em = ... // from EntityManagerFactory, injection, etc.

    em.getTransaction().begin();

    A parentActivuty   = new A();
    A subActivity1      = new A();
    A subActivity2 = new A();

    son.setParentActivity(parent);
    daughter.setParentActivity(parent);
    parent.setSubActivity(Arrays.asList(subActivity1, subActivity2));

    em.persist(parent);
    em.persist(son);
    em.persist(daughter);

    em.getTransaction().commit();
}

但在这种情况下,我在数据库中有父对象并希望保留子对象可能的解决方案是什么......?

4

1 回答 1

3

您从数据库中获取父级,执行附件,并持久化两个子级:

A parent = em.get(A.class, parentId);
A son = new A();
A daughter = new A();
son.setParentActivity(parent);
daughter.setParentActivity(parent);
em.persist(son);
em.persist(daughter);
parent.getSubActivities().add(son);
parent.getSubActivities().add(daughter);
于 2012-09-14T07:11:24.850 回答