spring-projects / spring-projects/spring-data-relational
Make sure referenced immutable Entities get updated in collections even if the immutable parent doesn't have an explicit id [DATAJDBC-433]
Aperta
@schauder ci sta già lavorando.
Dal 31/12/2020.
type: bug
- Lingua principale
- Java
- Stelle
- 827
- Fork
- 394
- Metriche di merge delle PR
- Nessuna PR unita negli ultimi 30g
Descrizione
Jens Schauder opened DATAJDBC-433 and commented
If an entity A contains a collection List<C> with A and C both being immutable, A needs to get updated even though it might not have an id if C has or any of the classes it references directly or indirectly.
This was ensured by DATAJDBC-291 by the test below but it is not entirely sure it is still handled properly by the variant about to get merged.
@Test
public void updatingCollectionInEntityWithoutId() {
CollectionOwner owner = new CollectionOwner(null, "owner", singletonList(new IdLessIntermediate("intermediate", singletonList(new Element(null, "Lovecraft")))));
CollectionOwner saved = template.save(owner);
Long id = saved.id;
assertThat(id).isNotNull();
Element savedLovecraft = saved.list.get(0).list.get(0);
assertThat(savedLovecraft.id).isNotNull();
CollectionOwner changedOwner = new CollectionOwner(id, "owner", singletonList(new IdLessIntermediate("intermediate", asList(savedLovecraft, new Element(null, "Sakurai")))));
CollectionOwner savedAgain = template.save(changedOwner);
SoftAssertions softly = new SoftAssertions();
softly.assertThat(savedAgain.id).isEqualTo(id);
softly.assertThat(savedAgain.list.get(0).list.get(0)).isEqualTo(savedLovecraft);
softly.assertThat(savedAgain.list.get(0).list.get(1).id).describedAs("Element has id set and that element is part of the aggregate").isNotNull();
softly.assertAll();
}
Issue Links:
- DATAJDBC-291 Nested data structure with Spring Data JDBC
Guida per i contributori
Apri la guida per i contributori
Come iniziare
- Leggi tutta la issue e poi la guida ai contributi del progetto.
- Commenta sulla issue per dire che te ne occupi tu — evita che due persone facciano lo stesso lavoro.
- Fai un fork del repository e lavora su un branch.
- Apri una pull request che faccia riferimento al numero della issue.
Valutazione
Questa issue non è ancora stata valutata.