Question

Please find below entity code,

@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...
}   

if we want to persist parent and child both at the same time then below code works perfectly fine

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();
}

but here in this case i have parent object in the database and want to persist child object what could be the possible solution…?

Answer 1

You get the parent from the database, do the attachments, and persist the two children:

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);