swift - 使用未解析的标识符 'result' swift 3

Question

我正在为我的应用程序编写登录页面，但收到此错误社交媒体应用程序，xcode 8.3.2，swift 3我已经尝试过文件检查器中的目标成员身份，但没有任何改变我还删除了测试单元(UITest 和 Test)并更新它们，它也不起作用。在第 41 行，我收到此错误“使用未解析的标识符‘结果’”

下图解释了代码 Picture

import UIKit

class LoginViewController : UIViewController
{
    @IBOutlet weak var txt_username: UITextField!
    @IBOutlet weak var txt_password: UITextField!
    override func viewDidLoad() {
        super.viewDidLoad()
    }
    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
    }
    @IBAction func btn_log_in_click(_ sender: Any){
        let server=MultipartUtility (Url:"http://x.x.x.x/appname-api/user/login/")
//I.m hiding the real ip and details for posting this
            server.AddFormField("username", value: txt_username.text)
            server.AddFormField("password", value: txt_password.text)


let task = URLSession.shared.dataTask(with: server.execute())
{Data,URLResponse,error in
    if error != nil{
    print(error as Any)
        return
    }
    do{
        let json = try JSONSerialization.jsonObject(with: Data!, options: .allowFragments)
        if let json_result = json as? [String: Any]{
            let result = json_result ["result"] as? String
            if result == "0"
            {
                DispatchQueue.main.async {
                    let alert = UIAlertController(title:"Incorrect Username",message : "The username you entered doesn't appear to belong to an account. Please check your username and try again", preferredStyle : .alert)
                    let alert_action = UIAlertAction(title: "Try Again", style: .default, handler: nil)
                    alert.addAction(alert_action)
                    self.present(alert, animated: true, completion: nil)
                }
            }
        }
        else{

            DispatchQueue.main.async {
            UserDefaults.standard.set(result!, forKey: "user_id")
//" use of unresolved identifier 'result' "
            let current_view=UIApplication.shared.windows[0] as UIWindow
            let new_view=(self.storyboard? .instantiateViewController(withIdentifier: "tab_bar"))! as UIViewController
            UIView.transition(from: (current_view.rootViewController? .view)!, to:new_view.view , duration: 0.65, options: .transitionFlipFromRight, completion: {(action) in current_view.rootViewController=new_view
            })
            }
        }
    }
    catch{
    }
}
        task.resume()

      }
}

Answer 1

if let json_result = json as? [String: Any]
                {
                    let result = json_result ["result"] as? String
                    if result == "0"
                    {
                        DispatchQueue.main.async {
                            let alert = UIAlertController(title:"Incorrect Username",message : "The username you entered doesn't appear to belong to an account. Please check your username and try again", preferredStyle : .alert)
                            let alert_action = UIAlertAction(title: "Try Again", style: .default, handler: nil)
                            alert.addAction(alert_action)
                            self.present(alert, animated: true, completion: nil)
                        }
                    }
                    else
                    {
                        DispatchQueue.main.async {
                            UserDefaults.standard.set(result!, forKey: "user_id")
                            //" use of unresolved identifier 'result' "
                            let current_view=UIApplication.shared.windows[0] as UIWindow
                            let new_view=(self.storyboard? .instantiateViewController(withIdentifier: "tab_bar"))! as UIViewController
                            UIView.transition(from: (current_view.rootViewController? .view)!, to:new_view.view , duration: 0.65, options: .transitionFlipFromRight, completion: {(action) in current_view.rootViewController=new_view
                            })
                        }

                    }
                }
                else{
                // Error in jsonSerialization
                }