[英]How to get a response in HAL-Format with Spring-Hateoas
Basically i have the same issue like the member who posted this question 基本上我有像发布此问题的成员一样的问题
When i request a single user in my Application, i get the response in the HAL-format, like i wish 当我在我的应用程序中请求单个用户时,我得到了HAL格式的响应,就像我希望的那样
http://localhost:8080/api/v1/users/25 with GET : http:// localhost:8080 / api / v1 / users / 25 with GET :
{
"userId": "25",
"firstname": "Beytullah",
"lastname": "Güneyli",
"username": "gueneylb",
"_links": {
"self": {
"href": "http://localhost:8080/api/v1/users/25"
},
"roles": [
{
"href": "http://localhost:8080/api/v1/roles/33"
},
{
"href": "http://localhost:8080/api/v1/roles/34"
}
]
}
}
But, when i request all users , i get the response in non-HAL-Format, like this: 但是,当我请求所有用户时,我得到非HAL格式的响应,如下所示:
http://localhost:8080/api/v1/users with GET : http:// localhost:8080 / api / v1 / GET 用户 :
[...
{
"userId": "25",
"firstname": "Beytullah",
"lastname": "Güneyli",
"username": "gueneylb",
"links": [
{
"rel": "self",
"href": "http://localhost:8080/api/v1/users/25"
},
{
"rel": "roles",
"href": "http://localhost:8080/api/v1/roles/33"
},
{
"rel": "roles",
"href": "http://localhost:8080/api/v1/roles/34"
}
]
},
...]
Here are my methods: 这是我的方法:
@RequestMapping(value = "users", method = RequestMethod.GET, produces = MediaTypes.HAL_JSON_VALUE)
public List<UserResource> list() throws MyException, NotFoundException {
List<User> userList= userRepository.findAll();
List<UserResource> resources = new ArrayList<UserResource>();
for (User user : userList) {
resources.add(getUserResource(user));
}
if(userList == null)
throw new MyException("List is empty");
else
return resources;
}
@RequestMapping(value = "users/{id}", method = RequestMethod.GET)
public UserResource get(@PathVariable Long id) throws NotFoundException {
User findOne = userRepository.findOne(id);
if (findOne == null){
log.error("Unexpected error, User with ID " + id + " not found");
throw new NotFoundException("User with ID " + id + " not found");
}
return getUserResource(findOne);
}
private UserResource getUserResource(User user) throws NotFoundException {
resource.add(linkTo(UserController.class).slash("users").slash(user.getId()).withSelfRel());
for(Role role : user.getRoles()){
resource.add(linkTo(RoleController.class).slash("roles").slash(role.getId()).withRel("roles"));
}
return resource;
}
You can see that both methods invoke the getUserResource(User user)
method. 您可以看到两个方法都调用getUserResource(User user)
方法。
But when i get all users in my database, the format of the _links
is not like i want. 但是当我在我的数据库中获得所有用户时, _links
的格式不是我想要的。 I think it must be something about the List
of resources i return. 我认为它必须是我返回的资源List
。 Maybe because of that it has no HAL-Format. 也许是因为它没有HAL格式。 I also tried a Set
instead of List
but it gave me the same response 我也尝试过Set
而不是List
但是它给了我相同的响应
You should return Resources<UserResource>
in your list()
method. 您应该在list()
方法中返回Resources<UserResource>
。
return new Resources(resources);
You could also add a self-link to the resources itself to point to the list resource. 您还可以向资源本身添加自我链接以指向列表资源。
Furthermore, I would suggest using a RessourceAssembler
to create the resource instance - see http://docs.spring.io/spring-hateoas/docs/0.23.0.RELEASE/reference/html/#fundamentals.resource-assembler . 此外,我建议使用RessourceAssembler
来创建资源实例 - 请参阅http://docs.spring.io/spring-hateoas/docs/0.23.0.RELEASE/reference/html/#fundamentals.resource-assembler 。
Additionally, you could add paging to your list resource. 此外,您可以向列表资源添加分页。 For this you need to: 为此,您需要:
findAll
method in your repository that returns a Page<User>
. 在存储库中使用findAll
方法返回Page<User>
。 PagedResourcesAssembler<User>
in your controller 在控制器中自动装配PagedResourcesAssembler<User>
PagedResources<UserResource>
in your list method 在list方法中返回PagedResources<UserResource>
PagedResourcesAssembler
to convert the Page
into PagedResources
使用PagedResourcesAssembler
将Page
转换为PagedResources
This would result in something like this: 这将导致类似这样的事情:
private final PagedResourcesAssembler<User> pagedResourcesAssembler;
@RequestMapping(value = "users", method = RequestMethod.GET)
public ResponseEntity<PagedResources<UserResource>> list(Pageable pageable) {
final Page<User> page = repository.findAll(pageable);
final Link link = ControllerLinkBuilder.linkTo(ControllerLinkBuilder.methodOn(this.getClass()).list(null)).withSelfRel();
PagedResources<UserResource> resources = page.getContent().isEmpty() ?
(PagedResources<UserResource>) pagedResourcesAssembler.toEmptyResource(page, ShippingZoneResource.class, link)
: pagedResourcesAssembler.toResource(page, resourceAssembler, link);
return ResponseEntity.ok(resources);
}
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