在HATEOAS_embedded PagedResource中命名Spring页面键

Question

我们目前正在最新的项目中实现一个API库。我们正在尝试使用Spring HATEOAS和HAL作为适当的库来生成json HAL responsed。

使用Spring Boot1.2.5。释放提供的Spring HATEOAS版本。目前我们正在将实际的JPA实体映射为资源。

即，我们的实体如下所示：

import lombok.Getter;
import lombok.Setter;

@Getter
@Setter
@Entity
@Table(name = "users")
public class User {

    @Id
    private UUID id;

    @Column(nullable = false, length = 70)
    private String firstName;

    @Column(nullable = false, length = 70)
    private String lastName;
}

我们的存储库：

public interface UserRepository extends PagingAndSortingRepository<User, UUID>, JpaSpecificationExecutor<User> {
    User findByUsername(String username);
}

我们的服务：

@Service
@Transactional
public class UserService implements UserDetailsService {

    private UserRepository userRepository;

    @Autowired
    public UserService(UserRepository userRepository) {
        this.userRepository = userRepository;
    }

    @Transactional
    public Page<User> findAll(Pageable pageable) {
        return this.userRepository.findAll(pageable);
    }
}

RestController：

@RestController
@RequestMapping(value = "/users", produces =MediaType.APPLICATION_JSON_VALUE) // this is done otherwise objectmapper returns xml for some reason
public UserRestController {

    public UserService userService;
    public EntityLinks entityLinks;

    @Autowired
    public UserRestController(UserService userService, EntityLinks entityLinks) {
        this.userService = userService;
        this.entityLinks = entityLinks;
    }

    @RequestMapping(method = RequestMethod.GET)
    public PagedResource<Resource<User>> getUsers(Pageable pageable, PagedResourcesAssembler<User> pagedResourcesAssembler) {
        Page<User> userPage = this.userService.findAll(pageable);

        PagedResources<Resource<User>> userPagedResources =
            pagedResourcesAssembler.toResource(
                    usersPage,
                    linkTo(methodOn(UserRestController.class).showUsers(pageable, pagedResourcesAssembler)).withSelfRel());

        return userPagedResources;

    }
}

假设我将默认页面大小设置为1，则页面将正确生成。设置上一页/下一页的链接。正在生成“Embeddeble”属性，但我们在第一页得到“userList”：[{user}]，第二页得到“user$$\u jvst163_9List”：[{user}]。

有没有一种方法可以真正命名页面列表属性（我一直在网上寻找一些东西，但没有用）。

例如：

{
  "_links": {
    "self": {
      "href": "http://localhost:8080/users{?page,size,sort}",
      "templated": true
    },
    "next": {
      "href": "http://localhost:8080/users?page=1&size=1&sort=firstname,lastname,asc"
    }
  },
  "_embedded": {
    "userList": [{
      "id": "2027bea9-cfdc-4724-b29c-39b3f64cbfd5",
      "firstname": "admin",
      "lastname": "asdf",
    }]
  },
  "page": {
    "size": 1,
    "totalElements": 2,
    "totalPages": 2,
    "number": 0
  }
}
{
  "_links": {
    "self": {
      "href": "http://localhost:8080/users{?page,size,sort}",
      "templated": true
    },
    "prev": {
      "href": "http://localhost:8080/users?page=0&size=1&sort=firstname,lastname,asc"
    }
  },
  "_embedded": {
    "user_$$_jvst163_9List": [{
      "id": "52c0c09e-c386-4aec-9723-f8beaf99adc5",
      "username": "admin",,
      "firstname": "firstname",
      "lastname": "lastname"
    }]
  },
  "page": {
    "size": 1,
    "totalElements": 2,
    "totalPages": 2,
    "number": 1
  }
}

Answer 1

据我所知，作者自己对这个问题的评论是，找到的解决方案是将@Relation（collectionRelation=“users”）添加到用户类中。

...
@Table(name = "users")
@Relation(collectionRelation = "users")
public class User {
...