Spring REST API 多个 RequestParams 与控制器实现

Question

我想知道在给出多个请求参数的 GET 请求的情况下执行控制器的正确方法。在我对 REST 的理解中，拥有一个带有用于过滤/排序的附加参数的端点比多个端点（每种情况一个）要好得多。我只是想知道这种端点的维护和可扩展性。请看下面的例子：

@RestController
@RequestMapping("/customers")
public class CustomerController {

    @Autowired
    private CustomerRepository customerRepo;

    @GetMapping
    public Page<Customer> findCustomersByFirstName(
                @RequestParam("firstName") String firstName,
                @RequestParam("lastName") String lastName,
                @RequestParam("status") Status status, Pageable pageable) {

        if (firstName != null) {
            if (lastName != null) {
                if (status != null) {
                    return customerRepo.findByFirstNameAndLastNameAndStatus(
                                                    firstName, lastName, status, pageable);
                } else {
                    return customerRepo.findByFirstNameAndLastName(
                                                    firstName, lastName, pageable);
                }
            } else {
                // other combinations omitted for sanity
            }
        } else {
            // other combinations omitted for sanity
        }
    }
}

这样的端点使用起来似乎很方便（参数的顺序无关紧要，所有这些都是可选的......），但维护这样的东西看起来就像地狱一样（组合的数量可以是巨大的）。

我的问题是 - 处理此类事情的最佳方法是什么？ “专业”的 API 是如何设计的？

Answer 1

作为除其他解决方案之外的替代解决方案，您可以在存储库中使用JpaSpecificationExecutor<T>，并根据您的参数创建一个规范对象并将其传递给findAll 方法。

因此，您的存储库应从JpaSpecificationExecutor<Customer> 接口扩展，如下所示：

@Repository
interface CustomerRepository extends JpaSpecificationExecutor<Customer> {

}

您的控制器应该获得所需的参数作为Map<String, String 以获得动态行为。

@RestController
@RequestMapping("/customers")
public class CustomerController {
    private final CustomerRepository repository;

    @Autowired
    public CustomerController(CustomerRepository repository) {
        this.repository = repository;
    }

    @GetMapping
    public Page<Customer> findAll(@RequestBody HashMap<String, String> filters, Pageable pageable) {
        return repository.findAll(QueryUtils.toSpecification(filters), pageable);
    }
}

而且，您应该定义一个方法来将提供的参数转换为Specification<Customer>：

class QueryUtils {
    public static Specification<Customer> toSpecification(Map<String, String> filters) {
        Specification<Customer> conditions = null;

        for (Map.Entry<String, String> entry : filters.entrySet()) {
            Specification<Customer> condition = Specification.where((root, query, cb) -> cb.equal(root.get(entry.getKey()), entry.getValue()));
            if (conditions == null) {
                conditions = condition;
            } else {
                conditions = conditions.and(condition);
            }
        }

        return conditions;
    }
}

另外，您可以使用Meta 模型进行更好的条件查询并将其与提供的解决方案相结合。

Answer 2

实际上，您自己回答了一半，查询参数用于过滤目的，正如您在代码中看到的那样，这将通过 GET 请求被允许。但是您关于验证的问题是一种权衡。

例如；如果你不想进行这种检查，你可以依赖@RequestParam中默认的强制required = true，并在响应中立即处理。

或者您也可以在 @Valid 的支持下使用 @RequestBody 以获得更清晰的错误信息；例如

@PostMapping(value = "/order")
public ResponseEntity<?> submitRequest(@RequestBody @Valid OrderRequest requestBody, 
            Errors errors) throws Exception {

        if (errors.hasErrors())
            throw new BusinessException("ERR-0000", "", HttpStatus.NOT_ACCEPTABLE);

        return new ResponseEntity<>(sendOrder(requestBody), HttpStatus.CREATED);
}

// Your Pojo
public class OrderRequest {
    @NotNull(message = "channel is required")
    private String channel;

    @NotNull(message = "Party ID is required")
    private long partyId;
}

更多信息请查看@Valid usage in Spring

这种方式将您的验证机制从控制器层解耦到业务层。这反过来会节省大量样板代码，但正如您注意到的那样，改为 POST。

所以一般来说，您的问题没有直接的答案，而简短的答案取决于您，所以选择对您来说容易且功能强大且维护较少的最佳实践

Answer 3

最好有一个带有此类验证的 POST 请求，而不是 GET 请求。您可以对控制器使用以下方法。

@PostMapping(value = "/findCustomer",produces = MediaType.APPLICATION_JSON_VALUE)
public ResponseEntity<?> findCustomersByFirstName(@Valid @RequestBody Customer customer){
   return customerRepo.findByFirstNameAndLastNameAndStatus(customer.getFirstName, customer.getLastName(), customer.getStatus(), pageable);

}

按如下方式使用 DTO。

public class Customer {

private String firstName;
private String lastName;
private String status;

public String getFirstName() {
    return firstName;
}

public void setFirstName(String firstName) {
    this.firstName= firstName;
}

public String getLastName() {
    return lastName;
}

public void setLastName(String lastName) {
    this.lastName= lastName;
}

public String getStatus() {
    return status;
}

public void setStatus(String status) {
    this.status= status;
}

public LivenessInputModel(String firstName, String lastName, String status) {
    this.firstName= firstName;
    this.lastName= lastName;
    this.status= status;
}

public LivenessInputModel() {

}

}

并添加控制器级别的异常建议以返回错误响应。

@ControllerAdvice
public class ControllerExceptionAdvice {

private static final String EXCEPTION_TRACE = "Exception Trace:";

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

public ControllerExceptionAdvice() {
    super();
}

@ExceptionHandler({ BaseException.class })
public ResponseEntity<String> handleResourceException(BaseException e, HttpServletRequest request,
                                                      HttpServletResponse response) {

    log.error(EXCEPTION_TRACE, e);

    HttpHeaders responseHeaders = new HttpHeaders();

    responseHeaders.setContentType(MediaType.APPLICATION_JSON);

    BaseExceptionResponse exceptionDto = new BaseExceptionResponse(e);

    return new ResponseEntity<>(exceptionDto.toString(), responseHeaders, e.getHttpStatus());
}


@ExceptionHandler({ Exception.class })
public ResponseEntity<String> handleException(Exception e, HttpServletRequest request,
                                              HttpServletResponse response) {

    log.error(EXCEPTION_TRACE, e);

    HttpHeaders responseHeaders = new HttpHeaders();

    responseHeaders.setContentType(MediaType.APPLICATION_JSON);

    HttpStatus httpStatus = HttpStatus.INTERNAL_SERVER_ERROR;

    BaseExceptionResponse exceptionDto = new BaseExceptionResponse(httpStatus.value(),
            ExceptionMessages.INTERNAL_DEFAULT);

    return new ResponseEntity<>(exceptionDto.toString(), responseHeaders, httpStatus);
}


@ExceptionHandler({ MethodArgumentNotValidException.class })
public ResponseEntity<String> handleValidationException(MethodArgumentNotValidException e,
                                                        HttpServletRequest request, HttpServletResponse response) {

    log.error(EXCEPTION_TRACE, e);

    HttpHeaders responseHeaders = new HttpHeaders();

    responseHeaders.setContentType(MediaType.APPLICATION_JSON);

    ValidationException validationEx = new ValidationException(e);
    BaseExceptionResponse exceptionDto = new BaseExceptionResponse(validationEx);

    return new ResponseEntity<>(exceptionDto.toString(), responseHeaders, validationEx.getHttpStatus());
}


@ExceptionHandler({ HttpMediaTypeNotSupportedException.class, InvalidMimeTypeException.class,
        InvalidMediaTypeException.class, HttpMessageNotReadableException.class })
public ResponseEntity<String> handleMediaTypeNotSupportException(Exception e, HttpServletRequest request,
                                                                 HttpServletResponse response) {

    log.error(EXCEPTION_TRACE, e);

    HttpHeaders responseHeaders = new HttpHeaders();

    responseHeaders.setContentType(MediaType.APPLICATION_JSON);

    HttpStatus httpStatus = HttpStatus.BAD_REQUEST;

    BaseExceptionResponse exceptionDto = new BaseExceptionResponse(httpStatus.value(),
            ExceptionMessages.BAD_REQUEST_DEFAULT);

    return new ResponseEntity<>(exceptionDto.toString(), responseHeaders, httpStatus);
}


@ExceptionHandler({ HttpRequestMethodNotSupportedException.class })
public ResponseEntity<String> handleMethodNotSupportException(Exception e, HttpServletRequest request,
                                                              HttpServletResponse response) {

    log.error(EXCEPTION_TRACE, e);

    HttpHeaders responseHeaders = new HttpHeaders();

    responseHeaders.setContentType(MediaType.APPLICATION_JSON);

    HttpStatus httpStatus = HttpStatus.METHOD_NOT_ALLOWED;

    BaseExceptionResponse exceptionDto = new BaseExceptionResponse(httpStatus.value(),
            ExceptionMessages.METHOD_NOT_ALLOWED);

    return new ResponseEntity<>(exceptionDto.toString(), responseHeaders, httpStatus);
}

@ExceptionHandler({ MissingServletRequestParameterException.class })
public ResponseEntity<String> handleMissingServletRequestParameterException(Exception e, HttpServletRequest request,
                                                                            HttpServletResponse response) {

    log.error(EXCEPTION_TRACE, e);

    HttpHeaders responseHeaders = new HttpHeaders();

    responseHeaders.setContentType(MediaType.APPLICATION_JSON);

    HttpStatus httpStatus = HttpStatus.BAD_REQUEST;

    BaseExceptionResponse exceptionDto = new BaseExceptionResponse(httpStatus.value(),
            ExceptionMessages.BAD_REQUEST_DEFAULT);

    return new ResponseEntity<>(exceptionDto.toString(), responseHeaders, httpStatus);
}

}

Answer 4

美好的一天。我不能称自己为专业人士，但这里有一些技巧可以让这个控制器看起来更好。

• 使用 DTO 而不是使用一组参数

public class CustomerDTO {

    private String firstName;
    private String lastName;
    private String status;

}

使用这个类，您的方法的签名将如下所示：

@GetMapping
public Page<Customer> findCustomersByFirstName(CustomerDTO customerDTO, Pageable pageable) {
    ...
}

• 如果需要，请使用验证

例如，您可以将其中一些字段设为必填：

public class CustomerDTO {

    @NotNull(message = "First name is required")
    private String firstName;
    private String lastName;
    private String status;

}

不要忘记在控制器中的 DTO 参数之前添加 @Valid 注解。

• 在 if-else 中使用规范而不是这个块

这里有一个很好的指南 - REST Query Language with Spring Data JPA Specifications

• 使用服务层，不需要从控制器调用repository

@GetMapping
public Page<Customer> findCustomersByFirstName(@Valid CustomerDTO customerDTO, BindingResult bindingResult, Pageable pageable) {
    if (bindingResult.hasErrors()) {
        // error handling
    }
    return customerService.findAllBySpecification(new CustomerSpecification(customerDTO));
}

您的控制器不应包含有关处理实体或某些业务内容的任何逻辑。它只是关于处理请求/错误、重定向、视图等......

Answer 5

处理此类问题的最佳方法是什么？

处理它的最佳方法是使用已有的工具。当您使用 Spring Boot 时，我假设 Spring Data JPA 然后为 Spring Data JPA 启用 QueryDsl 支持和 Web 支持扩展。

然后你的控制器就变成了：

@GetMapping
public Page<Customer> searchCustomers( 
        @QuerydslPredicate(root = Customer.class) Predicate predicate, Pageable pageable) {
   return customerRepo.findBy(predicate, pageable);
}

您的存储库被简单地扩展为支持 QueryDsl：

public interface CustomerRepository extends PagingAndSortingRepository<Customer, Long>, 
            QueryDslPredicateExecutor<Customer>{

}

您现在可以通过任何参数组合进行查询，而无需编写任何进一步的代码。

https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#core.web.type-safe https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#core.extensions.querydsl