Приетият отговор изглежда прекалено сложен с анотациите @JoinTable. Малко по-проста реализация се нуждае само от mappedBy. Използването на mappedBy показва притежаващия Entity или собственост, която вероятно трябва да бъде referencesTo, тъй като това ще се счита за "приятели". Връзката ManyToMany може да създаде много сложна графика. Използването на mappedBy прави кода така:
@Entity
public class Recursion {
@Id @GeneratedValue
private Integer id;
// what entities does this entity reference?
@ManyToMany
private Set<Recursion> referencesTo;
// what entities is this entity referenced from?
@ManyToMany(mappedBy="referencesTo")
private Set<Recursion> referencesFrom;
public Recursion init() {
referencesTo = new HashSet<>();
return this;
}
// getters, setters
}
И за да го използвате, трябва да имате предвид, че притежаваният имот е referencesTo. Трябва само да поставите връзки в това свойство, за да могат да бъдат препращани. Когато прочетете Entity обратно, ако приемем, че направите fetch join, JPA ще създаде колекциите за резултата. Когато изтриете обект, JPA ще изтрие всички препратки към него.
tx.begin();
Recursion r0 = new Recursion().init();
Recursion r1 = new Recursion().init();
Recursion r2 = new Recursion().init();
r0.getReferencesTo().add(r1);
r1.getReferencesTo().add(r2);
em.persist(r0);
em.persist(r1);
em.persist(r2);
tx.commit();
// required so that existing entities with null referencesFrom will be removed from cache.
em.clear();
for ( int i=1; i <= 3; ++i ) {
Recursion r = em.createQuery("select distinct r from Recursion r left join fetch r.referencesTo left join fetch r.referencesFrom where id = :id", Recursion.class).setParameter("id", i).getSingleResult();
System.out.println(r + " To=" + Arrays.toString(r.getReferencesTo().toArray()) + " From=" + Arrays.toString(r.getReferencesFrom().toArray()) );
}
tx.begin();
em.createQuery("delete from Recursion where id = 2").executeUpdate();
tx.commit();
// required so that existing entities with referencesTo will be removed from cache.
em.clear();
Recursion r = em.createQuery("select distinct r from Recursion r left join fetch r.referencesTo left join fetch r.referencesFrom where id = :id", Recursion.class).setParameter("id", 1).getSingleResult();
System.out.println(r + " To=" + Arrays.toString(r.getReferencesTo().toArray()) + " From=" + Arrays.toString(r.getReferencesFrom().toArray()) );
Което дава следния лог изход (винаги проверявайте генерираните SQL изрази):
Hibernate: create table Recursion (id integer not null, primary key (id))
Hibernate: create table Recursion_Recursion (referencesFrom_id integer not null, referencesTo_id integer not null, primary key (referencesFrom_id, referencesTo_id))
Hibernate: create sequence hibernate_sequence start with 1 increment by 1
Hibernate: alter table Recursion_Recursion add constraint FKsi0wfuwfs0bl19jjpofw4n8pt foreign key (referencesTo_id) references Recursion
Hibernate: alter table Recursion_Recursion add constraint FKarrkuyh2v1j5qnlui2vbpl7tk foreign key (referencesFrom_id) references Recursion
Hibernate: call next value for hibernate_sequence
Hibernate: call next value for hibernate_sequence
Hibernate: call next value for hibernate_sequence
Hibernate: insert into Recursion (id) values (?)
Hibernate: insert into Recursion (id) values (?)
Hibernate: insert into Recursion (id) values (?)
Hibernate: insert into Recursion_Recursion (referencesFrom_id, referencesTo_id) values (?, ?)
Hibernate: insert into Recursion_Recursion (referencesFrom_id, referencesTo_id) values (?, ?)
Hibernate: select distinct recursion0_.id as id1_2_0_, recursion2_.id as id1_2_1_, recursion4_.id as id1_2_2_, references1_.referencesFrom_id as referenc1_3_0__, references1_.referencesTo_id as referenc2_3_0__, references3_.referencesTo_id as referenc2_3_1__, references3_.referencesFrom_id as referenc1_3_1__ from Recursion recursion0_ left outer join Recursion_Recursion references1_ on recursion0_.id=references1_.referencesFrom_id left outer join Recursion recursion2_ on references1_.referencesTo_id=recursion2_.id left outer join Recursion_Recursion references3_ on recursion0_.id=references3_.referencesTo_id left outer join Recursion recursion4_ on references3_.referencesFrom_id=recursion4_.id where id=?
model.Recursion@7bdf6bb7 To=[model.Recursion@1bc53649] From=[]
Hibernate: select distinct recursion0_.id as id1_2_0_, recursion2_.id as id1_2_1_, recursion4_.id as id1_2_2_, references1_.referencesFrom_id as referenc1_3_0__, references1_.referencesTo_id as referenc2_3_0__, references3_.referencesTo_id as referenc2_3_1__, references3_.referencesFrom_id as referenc1_3_1__ from Recursion recursion0_ left outer join Recursion_Recursion references1_ on recursion0_.id=references1_.referencesFrom_id left outer join Recursion recursion2_ on references1_.referencesTo_id=recursion2_.id left outer join Recursion_Recursion references3_ on recursion0_.id=references3_.referencesTo_id left outer join Recursion recursion4_ on references3_.referencesFrom_id=recursion4_.id where id=?
model.Recursion@1bc53649 To=[model.Recursion@42deb43a] From=[model.Recursion@7bdf6bb7]
Hibernate: select distinct recursion0_.id as id1_2_0_, recursion2_.id as id1_2_1_, recursion4_.id as id1_2_2_, references1_.referencesFrom_id as referenc1_3_0__, references1_.referencesTo_id as referenc2_3_0__, references3_.referencesTo_id as referenc2_3_1__, references3_.referencesFrom_id as referenc1_3_1__ from Recursion recursion0_ left outer join Recursion_Recursion references1_ on recursion0_.id=references1_.referencesFrom_id left outer join Recursion recursion2_ on references1_.referencesTo_id=recursion2_.id left outer join Recursion_Recursion references3_ on recursion0_.id=references3_.referencesTo_id left outer join Recursion recursion4_ on references3_.referencesFrom_id=recursion4_.id where id=?
model.Recursion@42deb43a To=[] From=[model.Recursion@1bc53649]
Hibernate: delete from Recursion_Recursion where (referencesTo_id) in (select id from Recursion where id=2)
Hibernate: delete from Recursion_Recursion where (referencesFrom_id) in (select id from Recursion where id=2)
Hibernate: delete from Recursion where id=2
Hibernate: select distinct recursion0_.id as id1_2_0_, recursion2_.id as id1_2_1_, recursion4_.id as id1_2_2_, references1_.referencesFrom_id as referenc1_3_0__, references1_.referencesTo_id as referenc2_3_0__, references3_.referencesTo_id as referenc2_3_1__, references3_.referencesFrom_id as referenc1_3_1__ from Recursion recursion0_ left outer join Recursion_Recursion references1_ on recursion0_.id=references1_.referencesFrom_id left outer join Recursion recursion2_ on references1_.referencesTo_id=recursion2_.id left outer join Recursion_Recursion references3_ on recursion0_.id=references3_.referencesTo_id left outer join Recursion recursion4_ on references3_.referencesFrom_id=recursion4_.id where id=?
model.Recursion@6b739528 To=[] From=[]
person
K.Nicholas
schedule
13.06.2018
