无法从 API 获取访问令牌

Question

我是 Swift 新手，刚刚遇到一个问题。欢迎任何帮助和建议，我也见过 Alomofire，但由于一些错误我无法设置 Alamofire，而且我也需要 Alamofire 的帮助:|

request.addValue("application/json", forHTTPHeaderField:   
      "Accept")
    request.addValue("application/x-www-form-urlencoded", 
      forHTTPHeaderField:   "Content-Type")
    let postString = ["grant_type"  : "password" ,
                      "username"    : EntPhoneNumber.text! ,
                      "password"    : EntPassword.text! ,] as [String : 
 Any]
    do {
        
        
        request.httpBody = try JSONSerialization.data(withJSONObject: 
 postString)
    }catch let error {
        print(error.localizedDescription)
        DisplayMessage(UserMessage: "Something went wrong , please try 
 again!")
        return
    }
    
    let task = URLSession.shared.dataTask(with: request)
    {
        (data : Data? , response : URLResponse? , error : Error?) in
        
        self.removeActivtyIndicator(activityIndicator: 
 MyActivityIndicator)
        
        
        
        if error != nil
        {
            self.DisplayMessage(UserMessage: "1Could not successfully 
 perform this request , please try again later.")
            print("error = \(String(describing : error))")
            return
        }
        
        
        
        // let's convert response sent from a server side code to a 
 NSDictionary object:
        
        do { let json = try JSONSerialization.jsonObject(with: data!, 
 options: .mutableContainers) as? NSDictionary
            
            print(json!)
            if let parseJSON = json
            {
                let userID = parseJSON["grant_type"] as? String
                print("access_token : \(String(describing: userID))");
                
                if userID == nil
                {
                    //display an alert dialog with a friendly error 
message
                    self.DisplayMessage(UserMessage: "2Could not 
successfully perform this request , please try later.")
                    return
                }
                else
                {
                    self.DisplayMessage(UserMessage: "3successfully 
loged in.")
                }
            }
            

...

如您所见，这是我的代码，我正在向 API 发布请求以获取

access_token

但我收到错误：

{错误=“无效请求”; "error_description" = "缺少必需的 'grant_type' 参数。"; }

这些是我应该发布到 API 的参数，在 postman 中它可以正常工作，但我的代码在编译器中根本不起作用。

Answer 1

你把事情搞得太复杂了

let postString = ["grant_type"  : "password" ,
                      "username"    : EntPhoneNumber.text! ,
                      "password"    : EntPassword.text! ,] as [String :
                        Any]

    Alamofire.request(URL, method: .post, parameters: postString, encoding: URLEncoding.methodDependent, headers: nil).responseJSON { response in

        if let json = response.result.value {
            print("Response: ",json)
        }

    }

试试这个。

这就是您所需要的。如果您有任何疑问，请在这里询问。

Answer 2

点击此链接使用 CocoaPods 安装 Alamofire

https://www.raywenderlich.com/156971/cocoapods-tutorial-swift-getting-started在此处输入链接描述

我已经为 API 请求创建了这个自定义方法。

func request(_ method: HTTPMethod
    , _ URLString: String
    , parameters: [String : AnyObject]? = [:]
    , headers: [String : String]? = [:]
    , completion:@escaping (Any?) -> Void
    , failure: @escaping (Error?) -> Void) {

    Alamofire.request(URLString, method: method, parameters: parameters, encoding: JSONEncoding.default, headers: headers)
        .responseJSON { response in

            switch response.result {
            case .success:
                completion(response.result.value!)
            case .failure(let error):
                failure(error)
            }
    }
}

Answer 3

发送请求时需要对查询中的登录名和密码参数进行编码。让我们创建一个单独的

Network

类，它具有

loginRequest(username:password)

函数，该函数使用用户名和密码参数向服务器发出 API 请求。登录方法有一个

closure

，它返回 userId 字符串或错误。

class Networking {

    enum NetworkError: Error {
        case parsingData
        case jsonParsing
        case emptyData
        case apiError(error: Error)
    }

    func parseJSON(from data: Data) throws -> Any {
        do {
            let json = try JSONSerialization.jsonObject(with: data, options: .allowFragments)
            return json
        } catch {
            return error
        }
    }

    func loginRequest(username: String, password: String, completion: @escaping (String?, NetworkError?) -> ()) {

        var request = URLRequest(url: URL(string: "api.nahadeh.com/connect/token")!)
        request.addValue("application/json", forHTTPHeaderField: "Accept")
        request.addValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")

        let queryParams = "grant_type=password&username=\(username)&password=\(password)"
        request.httpBody = queryParams.data(using: .utf8, allowLossyConversion: false)

        URLSession.shared.dataTask(with: request) { (data, response, error) in

            guard error != nil else {
                completion(nil, NetworkError.apiError(error: error!))
                return
            }

            guard let data = data else {
                return completion(nil, NetworkError.emptyData)
            }

            do {
                let json = try self.parseJSON(from: data) as! [String: Any]
                guard let accessToken = json["access_token"] as? String else {
                    completion(nil, NetworkError.jsonParsing)
                    return
                }
                completion(accessToken, nil)
            } catch {
                completion(nil, NetworkError.jsonParsing)
            }
        }.resume()
    }
}

现在，您可以从 UIViewController 子类创建 Networking 类的实例，并使用用户名和密码参数调用登录方法

class LoginViewController: UIViewController {

    let networking = Networking()

    func login(username: String, password: String) {

        networking.loginRequest(username: username, password: password) { (token, error) in

            if let token = token {
                print(token)
                self.displayMessage("Successfully loged in. Go to home screen")
                return
            }

            if let nError = error {
                switch nError {
                case .jsonParsing:
                    self.displayMessage("jsonParsingError")
                case .emptyData:
                    self.displayMessage("emptyData")
                case .apiError(let error):
                    print(error.localizedDescription)
                    self.displayMessage("Could not successfully perform this request , please try later.")
                case .parsingData:
                    self.displayMessage("parsingData")
                }
            }
        }
    }

    func displayMessage(_ message: String, completion: (() -> Void)? = nil) {
        let alert = UIAlertController(title: "Message", message: message, preferredStyle: .alert)
        let okAction = UIAlertAction(title: "OK", style: .default, handler: nil)
        alert.addAction(okAction)
        present(alert, animated: true, completion: completion)
    }
}

使用 Alamofire 发出 API 请求非常容易，只需传递参数，它就会为你完成工作

  func loginRequest(username: String, password: String, completion: @escaping (String?, Error?) -> ()) {

        let params: [String: String] = [
            "grant_id": "password",
            "username": username,
            "password" : password
        ]

        Alamofire.request("api.nahadeh.com/connect/token", method: .post, parameters: params, encoding: .httpBody, headers: nil).validate().response { (response) in
            switch response.result {
            case .success(let value):
                let accessToken = value["access_token"]
                completion(accessToken, nil)
            case .failure(let error):
                completion(nil, error)
            }
        }
    }