如何编写在spring-data-hatoas中公开资源或资源列表的分页控制器
问题内容:
使用spring-data,我想为我的Person
实体编写两个方法。
Person.java:
public class Person {
@Id
String id;
String name;
Integer age;
// getters/setters omitted for clarity
}
我也写了PersonResouce
:
public class PersonResource extends Resource<Person> {
public PersonResource(Person content, Link... links) {
super(content, links);
}
}
我还添加了一个PersonResourceAssembler
:
public class PersonResourceAssembler extends ResourceAssemblerSupport<Person, PersonResource> {
public PersonResourceAssembler() {
super(PersonController.class, PersonResource.class);
}
public PersonResource createResource(Person person) {
PersonResource personResource = new PersonResource(person);
Link link = linkTo(PersonController.class).slash(person.getId()).withSelfRel();
personResource.add(link);
return personResource;
}
@Override
public PersonResource toResource(Person person) {
PersonResource resource = createResource(person);
return resource;
}
}
这是我的PersonController
:
@RestController
@RequestMapping("persons")
public class PersonController {
@Autowired
private PersonService personService;
@GetMapping
public HttpEntity<List<PersonResource>> showAll(@PageableDefault(size = 20) Pageable pageable, PersonDTO condition) {
Page<Person> page = personService.findAll(pageable, condition);
Iterable<Person> personList = page.getContent();
PersonResourceAssembler assembler = new PersonResourceAssembler();
List<PersonResource> resources = assembler.toResources(personList);
return new HttpEntity<>(resources);
}
@RequestMapping(name = "{id}", produces= MediaType.APPLICATION_JSON_VALUE)
public HttpEntity<PersonResource> showOne(@PathVariable("id") Long id, PersonDTO condition) {
condition.setId(id);
Person person = personService.get(id);
PersonResourceAssembler assembler = new PersonResourceAssembler();
PersonResource resource = assembler.toResource(person);
return new HttpEntity<>(resource);
}
}
这是列表的响应:
[
{
"createdById": 1,
"createdDate": "2017-09-21T10:21:05.741Z",
"deleted": false,
"email": null,
"firstName": "User49",
"id": 52,
"lastModifiedById": null,
"lastName": "robot",
"links": [
{
"href": "http://localhost:8080/users/52",
"rel": "self"
}
],
"middleName": null,
"mobile": "010101010001149",
"roleList": [
{
"createdById": null,
"createdDate": "2017-09-21T10:21:05.580Z",
"deleted": false,
"id": 2,
"lastModifiedById": null,
"name": "USER",
"userList": null,
"version": 0
}
],
"username": "user49",
"version": 0
}
]
这是对一种资源的响应:
{
"_links": {
"self": {
"href": "http://localhost:8080/users/52"
}
},
"createdById": 1,
"createdDate": "2017-09-21T10:21:05.741Z",
"deleted": false,
"email": null,
"firstName": "User49",
"id": 52,
"lastModifiedById": null,
"lastName": "robot",
"middleName": null,
"mobile": "010101010001149",
"roleList": [
{
"createdById": null,
"createdDate": "2017-09-21T10:21:05.580Z",
"deleted": false,
"lastModifiedById": null,
"name": "USER",
"userList": null
}
],
"username": "user49"
}
我看了看文档,似乎可以使用
PagedResources 创建分页。
我还希望我的响应看起来像RepositoryRestController
响应,这意味着列表响应:
- 放在
entities
键“ _embedded”下 - 放在
links
键“ _links”下 - 放在
page
键“页面”下
我尝试了一下,PagedResources
但是它的工作原理似乎与之不同Resource
,不能仅仅代替它。
我想看到使用PagedResource的控制器/汇编器。
解
我的解决方案是这样做
@GetMapping
public ResponseEntity<?> findAll(PagedResourcesAssembler<Person> pageAssembler, @PageableDefault(size = 20) Pageable pageable, UserDTO condition) {
Page<User> userList = userService.findAll(pageable, condition);
PagedResources<?> resources = pageAssembler.toResource(userList, new UserResourceAssembler());
return ResponseEntity.ok(resources);
}
问题答案:
尝试使用PagedResourcesAssembler构建页面资源:
@RestController
@RequestMapping("persons")
public class PersonController {
@Autowired private PersonService personService;
@Autowired private PagedResourcesAssembler<Person> assembler;
@Autowired private EntityLinks links;
@GetMapping("/paged")
public ResponseEntity<?> getPaged(Pageable pageable) {
Page<Person> personsPage = personService.getPaged(pageable);
Link pageSelfLink = links.linkFor(Person.class).slash("/paged").withSelfRel();
PagedResources<?> resources = assembler.toResource(personPage, this::toResource, pageSelfLink);
return ResponseEntity.ok(resources);
}
private ResourceSupport toResource(Person person) {
Link pesonLink = links.linkForSingleResource(person).withRel("person");
Link selfLink = links.linkForSingleResource(person).withSelfRel();
return new Resource<>(person, personLink, selfLink);
}
}