I am trying to create a custom search for my users repository. I have a custom restcontroller for it
#BasePathAwareController
#RequestMapping("/users")
#MultipartConfig(fileSizeThreshold = 20971520)
public class UserController implements ResourceProcessor<Resource<User>>,{
#Autowired
UserRepository userReposiotry;
#Autowired
private EntityLinks entityLinks;
#RequestMapping(value = "/search/getAvatar", method = RequestMethod.GET, produces = "image/jpg")
public ResponseEntity<InputStreamResource> downloadImage(#RequestParam("username") String username)
throws IOException {
ClassPathResource file = new ClassPathResource("uploads/" + username+ "/avatar.jpg");
return ResponseEntity
.ok()
.contentLength(file.contentLength())
.contentType(
MediaType.parseMediaType("application/octet-stream"))
.body(new InputStreamResource(file.getInputStream()));
}
#Override
public Resource<User> process(Resource<User> resource) {
LinkBuilder lb = entityLinks.linkFor(User.class);
resource.add(new Link(lb.toString()));
**// How can I add the search/getAvatar to the user search resource?**
return resource;
}
}
The first issue is that I get a 404 when trying to call /users/search/getAvatar?username=Tailor
The second is that how can I add this to the users search links?
Thank you
To add a search link, you need to extend RepositorySearchesResource as illustrated here:
Spring Data Rest Add custom endpoint to specific reposiotry
Spring Data Rest - Add link to search endpoint
As pointed out in the comments, be sure to check the domain type so as to add search link only for relevant repository.
Related
I have created a custom controller which needs to convert entities to resources. I have annotated my repositories with #RepositoryRestResource annotation. I want to know if there is a way I can invoke the default functionality of spring Data REST from my custom controller which serializes the entities to resources with links to other entities embedded in them.
I don't want to return entities from my handler method but Resources.
Thanks.
Very simple, using objects Resource or Resources. For example - in this controller we add custom method which return list of all user roles which are enums:
#RepositoryRestController
#RequestMapping("/users/roles")
public class RoleController {
#GetMapping
public ResponseEntity<?> getAllRoles() {
List<Resource<User.Role>> content = new ArrayList<>();
content.addAll(Arrays.asList(
new Resource<>(User.Role.ROLE1),
new Resource<>(User.Role.ROLE2)));
return ResponseEntity.ok(new Resources<>(content));
}
}
To add links to resource you have to use object RepositoryEntityLinks, for example:
#RequiredArgsConstructor
#RepositoryRestController
#RequestMapping("/products")
public class ProductController {
#NonNull private final ProductRepo repo;
#NonNull private final RepositoryEntityLinks links;
#GetMapping("/{id}/dto")
public ResponseEntity<?> getDto(#PathVariable("id") Integer productId) {
ProductProjection dto = repo.getDto(productId);
return ResponseEntity.ok(toResource(dto));
}
private ResourceSupport toResource(ProductProjection projection) {
ProductDto dto = new ProductDto(projection.getProduct(), projection.getName());
Link productLink = links.linkForSingleResource(projection.getProduct()).withRel("product");
Link selfLink = links.linkForSingleResource(projection.getProduct()).slash("/dto").withSelfRel();
return new Resource<>(dto, productLink, selfLink);
}
}
For more example see my 'how-to' and sample project.
I'm using Spring boot 1.5.3, Spring Data REST, Spring HATEOAS. Spring Data REST is amazing and does a perfect job, but sometimes it's needed custom business logic and therfore I need to create a custom controller.
I'm going to use a #RepositoryRestController to benefit of Spring Data REST functionality (http://docs.spring.io/spring-data/rest/docs/current/reference/html/#customizing-sdr.overriding-sdr-response-handlers).
Because Spring Data REST use HATEOAS by default, I'm using that. I need a controller like this:
#RepositoryRestController
#RequestMapping(path = "/api/v1/workSessions")
public class WorkSessionController {
#Autowired
private EntityLinks entityLinks;
#Autowired
private WorkSessionRepository workSessionRepository;
#Autowired
private UserRepository userRepository;
#PreAuthorize("isAuthenticated()")
#RequestMapping(method = RequestMethod.POST, path = "/start")
public ResponseEntity<?> start(#RequestBody(required = true) CheckPoint checkPoint) {
Authentication auth = SecurityContextHolder.getContext().getAuthentication();
if (checkPoint == null) {
throw new RuntimeException("Checkpoint cannot be empty");
}
if (workSessionRepository.findByAgentUsernameAndEndDateIsNull(auth.getName()).size() > 0) {
// TODO return exception
throw new RuntimeException("Exist a open work session for the user {0}");
}
// ...otherwise it's opened a new work session
WorkSession workSession = new WorkSession();
workSession.setAgent(userRepository.findByUsername(auth.getName()));
workSession.setCheckPoint(checkPoint);
workSession = workSessionRepository.save(workSession);
Resource<WorkSession> resource = new Resource<>(workSession);
resource.add(entityLinks.linkFor(WorkSession.class).slash(workSession.getId()).withSelfRel());
return ResponseEntity.ok(resource);
}
}
Because the parameter CheckPoint must be an existant resource, I want the client send the link of the resourse (like you can do in Spring Data REST POST methods).
Unfortunately when I try to do that, server side I receive an empty CheckPoint object.
I already read Resolving entity URI in custom controller (Spring HATEOAS) and converting URI to entity with custom controller in spring data rest? and expecially Accepting a Spring Data REST URI in custom controller.
I'm wondering if there is a best practice to do this avoiding to expose id to the client, followind so the HATEOAS principles.
Try to change you controller like this:
#RepositoryRestController
#RequestMapping(path = "/checkPoints")
public class CheckPointController {
//...
#Transactional
#PostMapping("/{id}/start")
public ResponseEntity<?> startWorkSession(#PathVariable("id") CheckPoint checkPoint) {
//...
}
}
That will mean: "For CheckPoint with given ID start new WorkSession".
Your POST request will be like: localhost:8080/api/v1/checkPoints/1/start.
I have a class:
public class user{
private String id;
private MultiPartFile file;
**Getters And Setters**
}
And in the Controller:
#PostMapping(value="/upload)
public void upload(User user){
}
In the front end I post data with form-data.I can get the user object.
But when I add #RequestBody and #RequestParam,it can't works.
in my opinion，#RequestParam is used to binding parameter to simple class . when I use #RequestBody ,spring will find HttpMessageConverter to convert http request body to class.But I'm not sure about that.Does anyone can explain to me?
So, I believe we are talking about org.springframework.web.multipart.MultipartFile, which is to be used together with #RequestParam variable. The mechanism is somewhat special in this case.
I had a similar problem, and what I ended up using was org.springframework.web.multipart.commons.CommonsMultipartResolver. From frontend I've constructed multipart request with two parts, in your scenario it could be user (containing just JSON data) and file (containing the file itself), e.g.:
#PostMapping(value="/upload")
public void upload(#RequestParam("user") User user, #RequestParam("file") MultipartFile file){
...
}
But then, you need to configure custom serialization of the User part, which can be done using org.springframework.web.multipart.commons.CommonsMultipartResolver. You can configure it using bean config like this:
#Configuration
public class MappingConfig {
#Order(Integer.MIN_VALUE)
#Bean(name = "multipartResolver")
public CommonsMultipartResolver multipartResolver() {
return new CommonsMultipartResolver();
}
#Bean
public Converter<String, User> stringToUser() {
return new Converter<String, User>() {
#Override
public User convert(String jsonString) {
return new Gson().fromJson(jsonString, User.class);
}
};
}
...
}
Also, as you can see I am using Gson manually, I couldn't find a better way how to do it. Also, it doesn't play with Java 8 lambdas, so it cannot be shortened (because of explicit types are needed for it to work).
I hope that this will at least points you to a right path.
I want to customize my spring-data-rest search method path by passing parameter as a path variable like follows
http://localhost:8080/orders/search/customers/{customerId}
findByCustomer(#PathVariable("customerId") Integer customer);
The search resource listh the links as follows
http://localhost:8080/orders/search/customers/%7BcustomerId%7D
How to expose search url with path params?
You can use custom handler similar to this:
#RepositoryRestController
public class OrderController {
#Autowired
OrderRepository orderRepository;
#GetMapping("/orders/search/customers/{id}")
public #ResponseBody ResponseEntity<?> getByCustomers(#PathVariable Integer customer) {
Order order = orderRepository.findOne(id);
if(order == null) return ResponseEntity.status(HttpStatus.NOT_FOUND).body(null);
Resource<Order> resource = new Resource<Order>(order);
return ResponseEntity.ok(resource);
}
}
More about this can be found here.
Use HttpServletRequest to get the request url:
findByCustomer(#PathVariable("customerId") Integer customer, HttpServletRequest request){
String request = request.getRequestURL().toString(); // StringBuffer, so use append if you want to...
[...]
}
also you can use request.getQueryString() to get the query part after ?.
In our Spring-Data-Rest Project we have a custom (fuzzy) search on a /buergers/search/findBuergerFuzzy?searchString="..." endpoint.
Is it possible to add a link for it on the /buergers/search endpoint (Without overriding the automatically exposed Repository findBy Methods)?
The Controller exposing the search:
#BasePathAwareController
#RequestMapping("/buergers/search/")
public class BuergerSearchController {
#Autowired
QueryService service;
#RequestMapping(value = "/findBuergerFuzzy", method = RequestMethod.GET)
public
#ResponseBody
ResponseEntity<?> findBuergerFuzzy(PersistentEntityResourceAssembler assembler, #Param("searchString") String searchString) {
if (searchString.length() < 3)
throw new IllegalArgumentException("Search String must be at least 3 chars long.");
List<Buerger> list = service.query(searchString, Buerger.class, new String[]{"vorname", "nachname", "geburtsdatum", "augenfarbe"});
final List<PersistentEntityResource> collect = list.stream().map(assembler::toResource).collect(Collectors.toList());
return new ResponseEntity<Object>(new Resources<>(collect), HttpStatus.OK);
}
}
UPDATE: This is an outdated workaround answer. Upgrade to Spring HATEOAS 1.0.
Old Workaround:
Digging the spring-data-rest source i found the RepositorySearchesResource which seems to solve the problem.
#Component
public class SearchResourcesProcessor implements ResourceProcessor<RepositorySearchesResource> {
#Override
public RepositorySearchesResource process(RepositorySearchesResource repositorySearchesResource) {
final String search = repositorySearchesResource.getId().getHref();
final Link findFullTextFuzzy = new Link(search + "/findFullTextFuzzy{?q}").withRel("findFullTextFuzzy");
repositorySearchesResource.add(findFullTextFuzzy);
return repositorySearchesResource;
}
}
Because we generate this code by templates, this is sufficient and fullfills our needs. Make sure to check the comments for the right and safe way.
Version
migrate-to-1.0.changes
ResourceSupport is now RepresentationModel
Resource is now EntityModel
Resources is now CollectionModel
PagedResources is now PagedModel
Code
The code for new version:
import org.springframework.data.rest.webmvc.RepositorySearchesResource;
import org.springframework.hateoas.Link;
import org.springframework.hateoas.server.RepresentationModelProcessor;
import org.springframework.stereotype.Component;
#Component
public class RepositorySearchesProcessor implements RepresentationModelProcessor<RepositorySearchesResource> {
#Override
public RepositorySearchesResource process(RepositorySearchesResource model) {
System.out.println(model.getDomainType());
model.add(Link.of(model.getRequiredLink("self").getHref() + "/findFullTextFuzzy{?q}").withRel("findFullTextFuzzy"));
return model;
}
}
How
About how to find what resource or model you use, after setting breakpoints in each method of RepresentationModel, you maybe find something useful :