Question

我有一个微服务，它通过多次调用另一个微服务来实现某种优化功能（第二个微服务计算所谓的目标函数值，第一个micriservice更改此tagrget函数的参数）

这导致有必要在Rest Controller层中编写一些逻辑。为了清楚起见，下面将显示一些简化的代码

@RestController
public class OptimizerController {
   private OptimizationService service;
   private RestTemplate restTemplate;

   @GetMapping("/run_opt")
   public DailyOptResponse doOpt(){
      Data iniData = service.prepareData(null);
      Result r = restTemplate.postForObject(http://calc-service/plain_calc", iniData, Result.class);

      double dt = service.assessResult(r);

      while(dt > 0.1){
          Data newData = service.preapreData(r);
          r = restTemplate.postForObject(http://calc-service/plain_calc", newData , Result.class);
          dt = service.assessResult(r);
      }

      return service.prepareResponce(r);
   }

正如我在示例中看到的那样，所有人都在努力使rest控制器尽可能简单，并将所有逻辑移至服务层。但是，如果我必须从服务层调用其他一些微服务怎么办？我应该在服务层中保留数据形式的逻辑并将其返回到控制器层，还是在服务层中使用RestTemplate对象？

感谢您的帮助

Answer 1

很简单。

整个逻辑都在服务层（包括其他服务）中。

简单的例子：

控制器：

@RestController
@RequestMapping("/api/users")
public class UserController {

    private final UserManager userManager;

    @Autowired
    public UserController(UserManager userManager) {
        super();
        this.userManager = userManager;
    }

    @GetMapping()
    public List<UserResource> getUsers() {
        return userManager.getUsers();
    }

    @GetMapping("/{userId}")
    public UserResource getUser(@PathVariable Integer userId) {
        return userManager.getUser(userId);
    }

    @PutMapping
    public void updateUser(@RequestBody UserResource resource) {
        userManager.updateUser(resource);
    }
}

服务：

@Service
public class UserManager {

    private static final Logger log = LoggerFactory.getLogger(UserManager.class);

    private final UserRepository userRepository;
    private final UserResourceAssembler userResourceAssembler;
    private final PictureManager pictureManager;

    @Autowired
    public UserManager(
            UserRepository userRepository,
            UserResourceAssembler userResourceAssembler,
            PictureManager pictureManager
    ) {

        super();
        this.userRepository = userRepository;
        this.userResourceAssembler = userResourceAssembler;
        this.pictureManager= pictureManager;
    }


    public UserResource getUser(Integer userId) {
        User user = userRepository.findById(userId).orElseThrow(() -> new NotFoundException("User with ID " + userId + " not found!"));
        return userResourceAssembler.toResource(user);
    }

    public List<UserResource> getUsers() {
        return userResourceAssembler.toResources(userRepository.findAll());
    }

    public void updateUser(UserResource resource) {
        User user = userRepository.findById(resource.getId()).orElseThrow(() -> new NotFoundException("User with ID " + resource.getId() + " not found!"));
        PictureResource pictureResource = pictureManager.savePicture(user);
        user = userResourceAssembler.fromResource(user, resource);
        user = userRepository.save(user);

        log.debug("User {} updated.", user);
    }
}

服务2：

@Service
public class PictureManager {

    private static final Logger log = LoggerFactory.getLogger(PictureManager.class);
    private final RestTemplate restTemplate;

    @Autowired
    public PictureManager(RestTemplate restTemplate) {
        super();
        this.restTemplate = restTemplate;
    }


    public PictureResource savePicture(User user) {
        //do some logic with user
        ResponseEntity<PictureResource> response = restTemplate.exchange(
            "url",
            HttpMethod.POST,
            requestEntity,
            PictureResource.class);

        return response.getBody();
    }
}

存储库：

public interface UserRepository extends JpaRepository<User, Integer> {

    User findByUsername(String username);

}

具有硬逻辑的“微服务”的架构模式（春季启动）

1 个答案: