如何在spring data rest / HATEOAS中创建一个引用已经存在的子实体的新父实体

Question

在我的项目中，我有两个域模型。一个父实体和一个子实体。父级引用子实体的列表。 （例如帖子和评论）两个实体都有它们的弹簧数据 JPA CrudRepository<Long, ModelClass> 接口，它们公开为 @RepositoryRestResource

HTTP GET 和 PUT 操作工作正常，并返回这些模型的漂亮 HATEOS 表示。

现在我需要一个特殊的 REST 端点“创建一个引用一个或多个已经存在子实体的新父级”。我想将对孩子的引用发布为我在请求正文中传递的文本/uri-list，如下所示：

POST http://localhost:8080/api/v1/createNewParent
HEADER
  Content-Type: text/uri-list
HTTP REQUEST BODY:
   http://localhost:8080/api/v1/Child/4711
   http://localhost:8080/api/v1/Child/4712
   http://localhost:8080/api/v1/Child/4713

如何实现这个休息端点？这是我迄今为止尝试过的：

 @Autowired
 ParentRepo parentRepo  // Spring Data JPA repository for "parent" entity


 @RequestMapping(value = "/createNewParent", method = RequestMethod.POST)
 public @ResponseBody String createNewParentWithChildren(
    @RequestBody Resources<ChildModel> childList,                         
    PersistentEntityResourceAssembler resourceAssembler
) 
{
   Collection<ChildModel> childrenObjects = childList.getContent()

   // Ok, this gives me the URIs I've posted
   List<Link> links = proposalResource.getLinks();

   // But now how to convert these URIs to domain objects???
   List<ChildModel> listOfChildren = ... ???? ...

   ParentModel newParnet = new ParentModel(listOfChildren)
   parentRepo.save(newParent)

}

参考/相关 https://github.com/spring-projects/spring-hateoas/issues/292

Answer 1

我遇到了同样的问题。这个问题有点老了，但我找到了另一个解决方案：

实际上，您必须在父级上强制合并，但在创建时会调用 persis。您可以通过使用空子列表保存父级来绕过，将子级添加到列表并再次保存：

List<ChildModel> listOfChildren = ... ???? ...

ParentModel newParnet = new ParentModel()
parent = parentRepo.save(newParent)
parent.getChilds().addAll(listOfChildren)
parentRepo.save(parent)

要获得合并权限，您必须编写自定义 repo 代码：

public interface PollModelRepositoryCustom {
    public PollModel merge(PollModel poll);
}

及其实现

@Repository
public class PollModelRepositoryCustomImpl implements PollModelRepositoryCustom {
    @PersistenceContext
    private EntityManager entityManager;

    public PollModel merge(PollModel poll) {
        return entityManager.merge(poll);
    }
}

然后你可以打电话：parentRepo.(newParent) 而不是parentRepo.save(newParent)

Answer 2

附注中有一个相关问题，也需要考虑：当我想保存父实体时，我不想以任何方式触摸、保存或更改已经存在的子实体.这在 JPA 中并不容易。因为 JPA 还将（尝试）持久化依赖的子实体。这失败了，但有例外：

javax.persistence.PersistenceException: org.hibernate.PersistentObjectException: detached entity passed to persist:

为了避免这种情况，您必须将子实体合并到 JPA save() 调用的事务中。我发现在一个事务中拥有两个实体的唯一方法是创建一个单独的@Services，它被标记为@Transactional。似乎完全是过度杀伤和过度工程。

这是我的代码：

PollController.java // 父实体的自定义 REST 端点

@BasePathAwareController
public class PollController {

@RequestMapping(value = "/createNewPoll", method = RequestMethod.POST)
public @ResponseBody Resource createNewPoll(
    @RequestBody Resource<PollModel> pollResource, 
    PersistentEntityResourceAssembler resourceAssembler
) throws LiquidoRestException
{
  PollModel pollFromRequest = pollResource.getContent();
  LawModel proposalFromRequest = pollFromRequest.getProposals().iterator().next();             // This propsal is a "detached entity". Cannot simply be saved.
  //jpaContext.getEntityManagerByManagedType(PollModel.class).merge(proposal);      // DOES NOT WORK IN SPRING.  Must handle transaction via a seperate PollService class and @Transactional annotation there.

  PollModel createdPoll;
  try {
    createdPoll = pollService.createPoll(proposalFromRequest, resourceAssembler);
  } catch (LiquidoException e) {
    log.warn("Cannot /createNewPoll: "+e.getMessage());
    throw new LiquidoRestException(e.getMessage(), e);
  }

  PersistentEntityResource persistentEntityResource = resourceAssembler.toFullResource(createdPoll);

  log.trace("<= POST /createNewPoll: created Poll "+persistentEntityResource.getLink("self").getHref());

  return persistentEntityResource;   // This nicely returns the HAL representation of the created poll
}

PollService.java // 用于事务处理

@Service
public class PollService {

    @Transactional    // This should run inside a transaction (all or nothing)
    public PollModel createPoll(@NotNull LawModel proposal, PersistentEntityResourceAssembler resourceAssembler) throws LiquidoException {
    //===== some functional/business checks on the passed enties (must not be null etc)
    //[...]

    //===== create new Poll with one initial proposal
    log.debug("Will create new poll. InitialProposal (id={}): {}", proposal.getId(), proposal.getTitle());
    PollModel poll = new PollModel();
    LawModel proposalInDB = lawRepo.findByTitle(proposal.getTitle());  // I have to lookup the proposal that I already have

    Set<LawModel> linkedProposals = new HashSet<>();
    linkedProposals.add(proposalInDB);
    poll.setProposals(linkedProposals);

    PollModel savedPoll = pollRepo.save(poll);

    return savedPoll;
}