php - swift 5错误: “Expected to decode Dictionary but found an array instead.”

Question

我是Swift编程的新手，我正在研究在网上找到的登录和注册表xcode示例。
它说，它适用于各种后端。所以我将其更改为与我的login.php文件一起使用。但我只是到目前为止...

func handleResponse(for request: URLRequest,
                    completion: @escaping (Result<[User], Error>) -> Void) {

    let session = URLSession.shared

    let task = session.dataTask(with: request) { (data, response, error) in

        DispatchQueue.main.async {

            guard let unwrappedResponse = response as? HTTPURLResponse else {
                completion(.failure(NetworkingError.badResponse))
                return
            }

            print(unwrappedResponse.statusCode)

            switch unwrappedResponse.statusCode {

            case 200 ..< 300:
                print("success")

            default:
                print("failure")
            }

            if let unwrappedError = error {
                completion(.failure(unwrappedError))
                return
            }

            if let unwrappedData = data {

                do {
                    let json = try JSONSerialization.jsonObject(with: unwrappedData, options: [])
                    print(json)
                    if let users = try? JSONDecoder().decode([User].self, from: unwrappedData) {
                        completion(.success(users))
                    } else {
                        let errorResponse = try JSONDecoder().decode(ErrorResponse.self, from: unwrappedData)
                        completion(.failure(errorResponse))
                    }

                } catch {
                    completion(.failure(error))
                }
            }
        }
    }

    task.resume()
}

我有这个请求功能。

func request(endpoint: String,
             loginObject: Login,
             completion: @escaping (Result<User, Error>) -> Void) {

    guard let url = URL(string: baseUrl + endpoint) else {
        completion(.failure(NetworkingError.badUrl))
        return
    }

    var request = URLRequest(url: url)

    do {
        let loginData = try JSONEncoder().encode(loginObject)
        request.httpBody = loginData
        print(loginObject)
    } catch {
        completion(.failure(NetworkingError.badEncoding))
    }

    request.httpMethod = "POST"
    request.addValue("application/json", forHTTPHeaderField: "Content-Type")

    handleResponse(for: request, completion: completion)
}

然后得到这个错误

typeMismatch(Swift.Dictionary<Swift.String, Any>, Swift.DecodingError.Context(codingPath: [], debugDescription: "Expected to decode Dictionary<String, Any> but found an array instead.", underlyingError: nil))

所以我在网上搜索，主要是在这里，出了什么问题。我测试了代码，找出了错误的出处，我发现了，我应该将代码更改为...

        if let unwrappedData = data {

            do {
                let json = try JSONSerialization.jsonObject(with: unwrappedData, options: [])
                print(json)
                if let users = try? JSONDecoder().decode([User].self, from: unwrappedData) {
                    completion(.success(users))
                } else {
                    let errorResponse = try JSONDecoder().decode(ErrorResponse.self, from: unwrappedData)
                    completion(.failure(errorResponse))
                }

            } catch {
                completion(.failure(error))
            }
        }

所以我认为这是解决我的问题的正确方法。但不幸的是，更改后我又遇到了另一个错误...

Member 'success' in 'Result<User, Error>' produces result of type 'Result<Success, Failure>', but context expects 'Result<User, Error>'

而且我什至无法构建和运行代码。有谁能够帮助我？
如有必要，我将我的login.php从数组更改为字典。

这是完工关闭吗？

enum MyResult<T, E: Error> {

    case success(T)
    case failure(E) }



func handleResponse(for request: URLRequest,
                        completion: @escaping (Result<User, Error>) -> Void) {
        ... }


enum NetworkingError: Error {
    case badUrl
    case badResponse
    case badEncoding }

ErrorResponse.swift

import Foundation

struct ErrorResponse: Decodable, LocalizedError {
    let reason: String

    var errorDescription: String? { return reason } }

Login.swift

import Foundation

struct Login: Encodable {
    let username: String
    let password: String }

User.swift

import Foundation

struct User: Decodable {
    let user_id: Int
    let username: String
    let password: String
    let firstname: String
    let surname: String
    let activated: Int
    let reg_time: String
}

打印(JSON)...

({
        activated = 1;
        firstname = Thomas;
        password = Maggie;
        "reg_time" = "0000-00-00 00:00:00";
        surname = Ghost;
        "user_id" = 2;
        username = "testuser";
    })

现在我几乎要重新营业了。我发现我在mysql中的user_id是int(3)，但是swift 5并没有将它当作它。当我取出let user_id:int时，我终于摆脱了最后一条错误消息。但是现在我可以通过单击任何按钮以及任何用户名和密码登录，无论它是对还是错。

Answer 1

我相信问题在于完成结帐的签名。看起来您正在使用Swift.Result，但改变了通用的Success和Failure类型。您可以将封闭函数发布到要通过完成封闭的位置吗？

OK，尝试将handleResponse更改为:

unc handleResponse(for request: URLRequest,
                    completion: @escaping (Result<[User], Error>) -> Void) {
    ... }

并确保ErrorResponse实现 Error