在我的项目中，我有两个领域模型。父实体和子实体。父引用子实体列表。 (例如 Post 和 Comments)两个实体都有它们的 spring 数据 JPA CrudRepository<Long, ModelClass>公开为 @RepositoryRestResource 的接口(interface)

HTTP GET 和 PUT 操作工作正常，并返回这些模型的良好 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存在子实体的新父级”。我想将对 child 的引用作为我在请求正文中传递的文本/uri 列表发布，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

我如何实现这个休息端点？到目前为止，这是我尝试过的:

 @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)

}

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

最佳答案

旁注中有一个相关问题，还需要考虑: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)保留依赖的子实体。这失败了异常(exception):

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() 调用的事务处理中。我发现在一个事务中同时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。似乎完全矫枉过正和过度工程。

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;
}

关于java - 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，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/47357297/