我正在尝试将我的AF请求转换为路由器结构,以获得更干净的项目。我收到了一个错误:
协议类型'Any‘的值不能符合'Encodable';只有struct/enum/class类型才能符合协议。
请帮我修一下密码。谢谢!
我的URL将有一个用户名的占位符,密码将在body中发送。响应将是Bool (success), username and bearer token。
下面是我的AF请求
let username = usernameTextField.text
let password = passwordTextField.text
let loginParams = ["password":"\(password)"]
AF.request("https://example.com/users/\(username)/login",
method: .post,
parameters: loginParams,
encoder: JSONParameterEncoder.default,
headers: nil, interceptor: nil).response { response in
switch response.result {
case .success:
if let data = response.data {
do {
let userLogin = try JSONDecoder().decode(UsersLogin.self, from: data)
if userLogin.success == true {
defaults.set(username, forKey: "username")
defaults.set(password, forKey: "password")
defaults.set(userLogin.token, forKey: "token")
print("Successfully get token.")
} else {
//show alert
print("Failed to get token with incorrect login info.")
}
} catch {
print("Error: \(error)")
}
}
case .failure(let error):
//show alert
print("Failed to get token.")
print(error.errorDescription as Any)
}
}到目前为止,对于转换为AF路由器结构,我有以下内容:
import Foundation
import Alamofire
enum Router: URLRequestConvertible {
case login(username: String, password: String)
var method: HTTPMethod {
switch self {
case .login:
return .post
}
}
var path: String {
switch self {
case .login(let username):
return "/users/\(username)/login"
}
}
var parameters: Parameters? {
switch self {
case .login(let password):
return ["password": password]
}
}
// MARK: - URLRequestConvertible
func asURLRequest() throws -> URLRequest {
let url = try Constants.ProductionServer.baseURL.asURL()
var request = URLRequest(url: url.appendingPathComponent(path))
// HTTP Method
request.httpMethod = method.rawValue
// Common Headers
request.setValue(ContentType.json.rawValue, forHTTPHeaderField: HTTPHeaderField.acceptType.rawValue)
request.setValue(ContentType.json.rawValue, forHTTPHeaderField: HTTPHeaderField.contentType.rawValue)
// Parameters
switch self {
case .login(let password):
request = try JSONParameterEncoder().encode(parameters, into: request) //where I got the error
}
return request
}
}
class APIClient {
static func login(password: String, username: String, completion: @escaping (Result<UsersLogin, AFError>) -> Void) {
AF.request(Router.login(username: username, password: password)).responseDecodable { (response: DataResponse<UsersLogin, AFError>) in
completion(response.result)
}
}
}LoginViewController类(我在这里替换了AF.request代码)
APIClient.login(password: password, username: username) { result in
switch result {
case .success(let user):
print(user)
case .failure(let error):
print(error.localizedDescription)
}可编编码UsersLogin模型
struct UsersLogin: Codable {
let success: Bool
let username: String
let token: String?
enum CodingKeys: String, CodingKey {
case success = "success"
case username = "username"
case token = "token"
}
}发布于 2020-05-27 20:27:35
花了我一段时间但终于修好了。我也清理了代码。
enum Router: URLRequestConvertible {
case login([String: String], String)
var baseURL: URL {
return URL(string: "https://example.com")!
}
var method: HTTPMethod {
switch self {
case .login:
return .post
}
}
var path: String {
switch self {
case .login(_, let username):
return "/users/\(username)/login"
}
}
func asURLRequest() throws -> URLRequest {
print(path)
let urlString = baseURL.appendingPathComponent(path).absoluteString.removingPercentEncoding!
let url = URL(string: urlString)
var request = URLRequest(url: url!)
request.method = method
switch self {
case let .login(parameters, _):
request = try JSONParameterEncoder().encode(parameters, into: request)
}
return request
}
}用法
let username = usernameTextField.text
AF.request(Router.login(["password": password], username)).responseDecodable(of: UsersLogin.self) { (response) in
if let userLogin = response.value {
switch userLogin.success {
case true:
print("Successfully get token.")
case false:
print("Failed to get token with incorrect login info.")
}
} else {
print("Failed to get token.")
}
}发布于 2021-11-12 11:56:31
我用这种方式解决了一个类似的问题。我创建了一个协议Routable
enum EncodeMode {
case encoding(parameterEncoding: ParameterEncoding, parameters: Parameters?)
case encoder(parameterEncoder: ParameterEncoder, parameter: Encodable)
}
protocol Routeable: URLRequestConvertible {
var baseURL: URL { get }
var path: String { get }
var method: HTTPMethod { get }
var encodeMode: EncodeMode { get }
}
extension Routeable {
// MARK: - URLRequestConvertible
func asURLRequest() throws -> URLRequest {
let url = baseURL.appendingPathComponent(path)
var urlRequest: URLRequest
switch encodeMode {
case .encoding(let parameterEncoding, let parameters):
urlRequest = try parameterEncoding.encode(URLRequest(url: url), with: parameters)
case .encoder(let parameterEncoder, let parameter):
urlRequest = URLRequest(url: url)
urlRequest = try parameterEncoder.encode(AnyEncodable(parameter), into: urlRequest)
}
urlRequest.method = method
return urlRequest
}
}我的路由器看起来就像这个
enum WifiInterfacesRouter: Routeable {
case listActive(installationId: Int16?)
case insert(interface: WifiInterface)
var encodeMode: EncodeMode {
switch self {
case .listActive(let installationId):
guard let installationId = installationId else {
return .encoding(parameterEncoding: URLEncoding.default, parameters: nil)
}
return .encoding(parameterEncoding: URLEncoding.default, parameters: ["idInstallation": installationId])
case .insert(let interface):
return .encoder(parameterEncoder: JSONParameterEncoder.default, parameter: interface)
}
}
var baseURL: URL {
return URL(string: "www.example.com/wifiInterfaces")!
}
var method: HTTPMethod {
switch self {
case .listActive: return .get
case .insert: return .post
}
}
var path: String {
switch self {
case .listActive: return "listActive"
case .insert: return "manage"
}
}
}解决构建错误
协议'Encodable‘作为一种类型不能符合协议本身
我使用了有用的AnyCodable库。Codable的类型擦除实现。
发布于 2020-05-18 17:36:07
您不能使用带有Parameters类型的Encodable字典,因为[String: Encodable]字典不是Encodable,就像错误说的那样。我建议将asURLRequest流程的这一步移到一个单独的函数中,例如:
func encodeParameters(into request: inout URLRequest) {
switch self {
case let .login(parameters):
request = try JSONParameterEncoder().encode(parameters, into: request)
}
}不幸的是,对于有许多路由的路由器来说,这不是很好,所以我通常将我的路由分解成小枚举,并将参数移动到与路由器结合生成URLRequest的不同类型中。
https://stackoverflow.com/questions/61828715
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