JSON 数据未在 swift3 中使用 Alamofire 进行解析

Question

我必须使用 alamofire 解析 json在使用 JSON 会话之前，它从 JSON 获取数据工作正常。现在我尝试使用 Alamofire 解析 JSON。这是使用 JSON 解析 JSON 的代码，这段代码工作正常

func auth(_ email:String,password:String) {
     var request = URLRequest(url:AppConstants.apiURLWithPathComponents("usersignin"))
        let session = URLSession.shared
        request.httpMethod = "POST"
        let bodyData = "email=(email)&passCode=(password)&deviceType=iOS&deviceId=(deviceToken)"
        request.httpBody = bodyData.data(using: String.Encoding.utf8);
        let task = session.dataTask(with: request, completionHandler: { (data, response, error) in
            do {
                if data != nil {
                    if let jsonData = try JSONSerialization.jsonObject(with: data!, options: JSONSerialization.ReadingOptions.mutableContainers) as? NSDictionary {
                        errorCode = String(describing: jsonData["errorCode"]!)
                        msg = jsonData["msg"] as! String
                        print(errorCode)
                        print(jsonData)
                        if(errorCode == "1"){
                            DispatchQueue.main.async(execute: {
                                self.activityIndicator.stopAnimating()
                            })
                        } else {
                            self.name = jsonData.value(forKey: "name") as! String
                            if let kidsURLDetails = jsonData["kidsURLDetails"] as? NSArray {
                                for i in 0 ..< kidsURLDetails.count {
                                    if kidsURLDetails[i] is NSDictionary {
                                        let url = kidsURLDetails[i] as? NSDictionary
                                        self.urls.append((url?["url"]) as! String)
                                    }
                                }
                            }
                            self.serverURL = self.urls.joined(separator: ",")
                            print("ServerURL (self.serverURL)")
                            let prefs:UserDefaults = UserDefaults.standard
                            prefs.setValue(self.name, forKey: "NAME")
                            DispatchQueue.main.async(execute: {
                                UIApplication.shared.endIgnoringInteractionEvents()
                                let controllerId = "NavID"
                                let storyboard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
                                let initViewController: UIViewController = storyboard.instantiateViewController(withIdentifier: controllerId) as UIViewController
                                self.present(initViewController, animated: true, completion: nil)
                            })
                        }
                    }
                } else {
                }
            } catch let err as NSError {
                print("JSON Error (err)")
            }
        })
        task.resume()
    }

在上面的代码中，我使用了带有传递参数的 post 方法，当我尝试在 almofire 中传递参数的 post 方法时，出现错误"调用中额外的参数'方法'"，用户名和密码来自文本字段，因此输入电子邮件和密码后，我已经使用 post 方法传递了参数。

这是我将在alamofire json解析中实现的代码

 var request = URLRequest(url:AppConstants.apiURLWithPathComponents("usersignin"))

        let bodyData = "email=(username)&passCode=(passcode)&deviceType=iOS&deviceId=123456"
        let deviceId = "1234"
        let params: [String: Any] = ["email": username, "passCode": passwordstring, "deviceType": "IOS","deviceId":deviceId]
        Alamofire.request(request, method: .post, parameters: params, encoding: JSONEncoding.default, headers: nil)
            .responseJSON { response in
                print(response.result.value as Any)  }

如果我能尝试这段代码工作

Alamofire.request("http://www.kids.com/rk/api/usersignin?email=[e-mail]&passCode=123456&deviceType=&deviceId=", method: .post, parameters: nil, encoding: JSONEncoding.default, headers: nil)
            .responseJSON { response in
                print(response.result.value as Any)          }

如何使用 Alamofire 解析传递参数的 JSON 发布方法。 我犯错的地方请帮助我

Answer 1

你在调用中传递了错误的 1 个参数类型，它应该是 URLConvertible(字符串或 URL(而不是 URLRequest。 尝试下面的代码。

let params: [String: Any] = ["email": username, "passCode": passwordstring, "deviceType": "IOS","deviceId":deviceId]
let url = URL(string: "http://www.kids.com/rk/api/usersignin")!
Alamofire.request(url, method: .post, parameters: params, encoding: JSONEncoding.default, headers: nil)
    .responseJSON { response in
}

____________编辑___________

这里的标头是后请求标头(如果有(或 nil

let params: [String: Any] = ["email": username, "passCode": passwordstring, "deviceType": "IOS","deviceId":deviceId]
let urlString = "http://www.kids.com/rk/api/usersignin"
guard let url = URL(string: urlString), var request = try? URLRequest(url: url, method: .post, headers: header) else{
    //
    return
}
request.httpBody = params.map{ "($0)=($1)" }.joined(separator: "&").data(using: .utf8)
Alamofire.request(request).responseJSON { response in
}

Answer 2

如果您需要使用 post 请求在 url 中发送参数，您应该使用URLEncoding.default编码而不是JSONEncoding.default 。当您需要将 JSON 数据作为内容类型为 application/json 的正文发送时，将使用 JSONEncoding 。

更改您的代码，例如：

Alamofire.request(url, method: .post, parameters: params, encoding: URLEncoding.default, headers: nil).responseJSON { response in
    print(response.result.value as Any)  
}

或者您可以删除编码参数，因为URLEncoding是 Alamofire 的默认编码。

Answer 3

 // URL
    let urlString = “URL_Here”
    var params = [String: Any]()
    //Contruct your params
    params = ["email": username, "passCode": passwordstring, "deviceType": "IOS","deviceId":deviceId]
    // Request
    Alamofire.request(urlString, method: .post, parameters: params, encoding: JSONEncoding.default, headers: nil)
        .validate(statusCode: 200..<300)
        .responseJSON { response in
            if (response.result.error == nil) {
              let value = response.result.value
                print(value)
            }
            else {
                let errorString = response.result.error
                print(errorString)
            }
    }