ios - 为什么使用 Alamofire 请求失败？

Question

我想获取github/users的数据，我尝试了两种方法:

第一种方法成功:

AF.request("https://api.github.com/users").responseJSON { response in
    print("UOUOUOUOUOU", response.description)
}

但是第二次失败了，为什么？

Network.request(Request.users).responseDecodable { (response: AFDataResponse<UserList>) in
      switch response.result {
      case .success(let value):
        print("SUCCESS")
      case .failure(let error):
            print("FAIL")
      }
    }

我需要使用第二种方法。

这是我的类(class)网络

class Network {
  let session: Session
  let evaluators = ["api.github.com/users": PinnedCertificatesTrustEvaluator(certificates: [
        Certificates.github
        ])
  ]

  private init() {
    session = Session(serverTrustManager: ServerTrustManager(evaluators: evaluators)
    )
  }

  private static let shared = Network()
  static func request(_ convertible: URLRequestConvertible) -> DataRequest {
    return shared.session.request(convertible)
  }
}

我收到这个错误:

FAIL serverTrustEvaluationFailed(reason: Alamofire.AFError.ServerTrustFailureReason.noRequiredEvaluator(host: "api.github.com"))
2019-12-17 12:03:21.649222-0400 SSLOwner[5533:83111] Task <57DC5433-F132-4453-BB68-CCF90B5F6058>.<1> HTTP load failed, 0/0 bytes (error code: -999 [1:89])

编辑 1:这是我的 Struct UserList 和 User

import Foundation
struct UserList: Codable {
  let users: [User]

  enum CodingKeys: String, CodingKey {
    case users = "items"
  }
}


struct User: Codable {
  let displayName: String

}

与

session = Session()

我收到这个错误:

responseSerializationFailed(reason: Alamofire.AFError.ResponseSerializationFailureReason.decodingFailed(error: Swift.DecodingError.typeMismatch(Swift.Dictionary, Swift.DecodingError.Context(codingPath: [], debugDescription: "Expected to decode Dictionary but found an array instead.", underlyingError: nil))))

Answer 1

您的上一个错误明确说明了错误所在。 API 调用的响应是一个数组，而不是您假设的字典。将您的预期响应转换为用户列表，然后它应该可以工作:

Network.request(Request.users).responseDecodable { (response: AFDataResponse<[User]>) in
  switch response.result {
  case .success(let value):
    print("SUCCESS")
  case .failure(let error):
        print("FAIL")
  }
}