Is it possible for Hibernate / JPA to fill in the reference itself or do I first have to persist entity A so that I receive an ID that I can set to entity B?
I have the following entities for this example:
@Entity
class A(uuid: UUID? = null,
@OneToMany(
mappedBy = "aUUID",
cascade = [CascadeType.ALL],
fetch = FetchType.LAZY,
orphanRemoval = true)
val b: List<B>
) : BaseEntity(uuid)
@Entity
class B(uuid: UUID? = null,
@Column(nullable = false) val aUUID: UUID,
) : BaseEntity(uuid)
B is referencing with a foreign key to A
You can persist both entities together (not exactly for this case but same logic):
A a = new A();
a.setBList(new ArrayList());
B b = new B();
a.getBList().add(b);
yourJpaRepository.save(a); // will persist both entities and add reference
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