Question

@Embeddable
class Y {
    protected Y(){}
    @Column(nullable = false)
    int i;
    @Column(nullable = false)
    int j;
}
@Entity
class X {
    protected X(){}
    @Id
    @Column(name = "id")
    int id;

    @ElementCollection(fetch = FetchType.EAGER)
    Set<Y> s = new HashSet<>();
}

Session session = sessionFactory.openSession();
Transaction transaction = session.beginTransaction();
X a = new X();
a.s.add(new Y());

session.save(a);
transaction.commit();
session.close();

休眠映射表x_s包含3列，分别是x，i和j的ID。但是，如果通过更新这样使分离的对象再次持久化：

Session session = sessionFactory.openSession();
Transaction transaction = session.beginTransaction();
X x = session.get(X.class, 0);
transaction.commit();
session.close();

Session session1 = sessionFactory.openSession();
Transaction transaction1 = session1.beginTransaction();
session1.saveOrUpdate(x); // and select or update should execute
transaction1.commit();
session1.close();

奇怪的是，除了select之外，hibernate会删除并插入，上面的代码会生成：

Hibernate: select x0_.id as id1_0_0_, s1_.X_id as X_id1_1_1_, s1_.s_i as s_i2_1_1_, s1_.s_j as s_j3_1_1_ from X x0_ left outer join X_s s1_ on x0_.id=s1_.X_id where x0_.id=?
Hibernate: delete from X_s where X_id=? and s_i=? and s_j=?
Hibernate: insert into X_s (X_id, s_i, s_j) values (?, ?, ?)
Hibernate: select x_.id from X x_ where x_.id=?
Hibernate: delete from X_s where X_id=? and s_i=? and s_j=?
Hibernate: insert into X_s (X_id, s_i, s_j) values (?, ?, ?)

根据此answer，它说原因是表x_s没有pk。但是，休眠生成的表x_s已经将所有3列用作组合的pk。

消除这些插入和删除命令的唯一方法似乎是将Set<Y>更改为List<Y>，hibernate会自动添加列顺序并将命令和x_id组合为pk。但是从数据库架构视图来看，没有实质性差异

P.S 我使用休眠5.3

Answer 1

很明显，JPA看到对象不相等是清除并重新插入的。请尝试覆盖equals()对象的hashcode()和Embeddable。使用默认实现时，仅通过引用就可以认为对象是相等的，在您的情况下不会如此。

休眠-删除有关ElementCollection

1 个答案: