相关疑难解决方法(0)

我正在使用Spring 4.0.0.RELEASE,Spring Data Commons 1.7.0.M1,Spring Hateoas 0.8.0.RELEASE

我的资源是一个简单的POJO:

public class UserResource extends ResourceSupport { ... }

我的资源汇编程序将User对象转换为UserResource对象:

@Component
public class UserResourceAssembler extends ResourceAssemblerSupport<User, UserResource> { 
    public UserResourceAssembler() {
        super(UserController.class, UserResource.class);
    }

    @Override
    public UserResource toResource(User entity) {
        // map User to UserResource
    }
}

在我的UserController中,我想Page<User>从我的服务中检索然后将其转换为PagedResources<UserResource>使用PagedResourcesAssembler,如下所示:https://stackoverflow.com/a/16794740/1321564

@RequestMapping(value="", method=RequestMethod.GET)
PagedResources<UserResource> get(@PageableDefault Pageable p, PagedResourcesAssembler assembler) {
    Page<User> u = service.get(p)
    return assembler.toResource(u);
}

这不会调用UserResourceAssembler,只User返回内容而不是我的自定义UserResource.

返回单个资源有效:

@Autowired
UserResourceAssembler …

我正在尝试做一些我认为应该非常简单的事情.我有一个Question对象,设置有spring-boot,spring-data-rest和spring-hateoas.所有的基础工作都很好.我想添加一个自定义控制器,它返回一个List<Question>与我Repository的/questions网址完全相同的格式,以便两者之间的响应兼容.

这是我的控制器:

@Controller
public class QuestionListController {

    @Autowired private QuestionRepository questionRepository;

    @Autowired private PagedResourcesAssembler<Question> pagedResourcesAssembler;

    @Autowired private QuestionResourceAssembler questionResourceAssembler;

    @RequestMapping(
            value = "/api/questions/filter", method = RequestMethod.GET,
            consumes = MediaType.APPLICATION_JSON_VALUE,
            produces = MediaType.APPLICATION_JSON_VALUE)
    public @ResponseBody PagedResources<QuestionResource> filter(
            @RequestParam(value = "filter", required = false) String filter,
            Pageable p) {

        // Using queryDSL here to get a paged list of Questions
        Page<Question> page = 
            questionRepository.findAll(
                QuestionPredicate.findWithFilter(filter), p);

        // Option 1 - default resource assembler
        return …

我正在尝试使用spring-boot-starter-data-rest使用Spring Boot构建RESTful API.有一些实体:帐户,交易,类别和用户 - 只是通常的东西.

当我通过默认生成的API 检索http:// localhost:8080/transactions中的对象时,一切顺利,我得到一个包含所有事务的列表作为JSON对象,如下所示:

{
  "amount": -4.81,
  "date": "2014-06-17T21:18:00.000+0000",
  "description": "Pizza",
  "_links": {
    "self": {
      "href": "http://localhost:8080/transactions/5"
    },
    "category": {
      "href": "http://localhost:8080/transactions/5/category"
    },
    "account": {
      "href": "http://localhost:8080/transactions/5/account"
    }
  }
}

但现在的目标是仅检索该URL下的最新事务,因为我不想序列化整个数据库表.所以我写了一个控制器:

@Controller
public class TransactionController {
    private final TransactionRepository transactionRepository;

    @Autowired
    public TransactionController(TransactionRepository transactionRepository) {
        this.transactionRepository = transactionRepository;
    }

    // return the 5 latest transactions
    @RequestMapping(value = "/transactions", method = RequestMethod.GET)
    public @ResponseBody List<Transaction> getLastTransactions() {
        return  transactionRepository.findAll(new PageRequest(0, 5, new Sort(new Sort.Order(Sort.Direction.DESC, "date")))).getContent(); …