Question

我有一个注册viewController（VC）。一旦用户提交了他的注册详细信息，我的远程服务器就会响应“registration_success”。收到此消息后，我能够提供一个新VC，请求用户检查其电子邮件以验证其电子邮件地址。我能够呈现新的视图控制器，但无法将用户的电子邮件地址传输到新的VC。我的代码如下：

     guard
          let registeredEmail = myArray["email"] as? String else{
          print("cannot find email address")
          return
          }

          let verifyEmailVC = self.storyboard?.instantiateViewController(withIdentifier: "emailConfirmVC") as! VerifyEmailViewController
          print("registeredEmail: \(registeredEmail)")
          verifyEmailVC.emailToVerify.text! = registeredEmail
          self.present(verifyEmailVC, animated: true, completion: nil)

类VerifyEmailViewController：

    class VerifyEmailViewController: UIViewController {

    @IBOutlet var emailToVerify: UILabel!

    override func viewDidLoad() {
        super.viewDidLoad()

    }

}

控制台：

 jsonObject: {
email = "tran123@gmail.com";
firstname = Wang;
lastname = tran;
message = "registration_success";
status = 200;
userId = 53;
}
message: registration_success
registeredEmail: tran123@gmail.com

我收到以下错误：

线程1：致命错误：在展开时出乎意料地发现nil 可选值

排成一行：verifyEmailVC.emailToVerify.text! = registeredEmail

Answer 1

我修改了您的代码，但很少修正，请尝试以下代码

let verifyEmailVC = self.storyboard?.instantiateViewController(withIdentifier: "emailConfirmVC") as! VerifyEmailViewController
      print("registeredEmail: \(registeredEmail)")
      verifyEmailVC.emailToVerifyStr = registeredEmail
      self.present(verifyEmailVC, animated: true, completion: nil)

在 VerifyEmailViewController

中

    class VerifyEmailViewController: UIViewController {

    var emailToVerifyStr = ""
    @IBOutlet var emailToVerify: UILabel!

    override func viewDidLoad() {
        super.viewDidLoad()

    }

    override func viewWillAppear(_ animated:Bool){
       super.viewWillAppear(true)
       self.emailToVerify.text = emailToVerifyStr
    }
  }

希望能帮助你：）

Answer 2

VerifyViewController中的标签为零，您正在尝试将值设置为该标签。尝试将字符串传递给控制器，如下所示

let verifyEmailVC = self.storyboard?.instantiateViewController(withIdentifier: "emailConfirmVC") as! VerifyEmailViewController
print("registeredEmail: \(registeredEmail)")
verifyEmailVC.email = registeredEmail
self.present(verifyEmailVC, animated: true, completion: nil)


class VerifyEmailViewController: UIViewController {

@IBOutlet var emailToVerify: UILabel!
var email: String?

override func viewDidLoad() {
    super.viewDidLoad()
    emailToVerify.text = email
}

Answer 3

首先，我想推荐使用像ViewController这样的高级对象使用的JSON对象。

您应该考虑使用“ Store ”对象接收JSON，创建用户对象。

然后，您可以直接创建一个init功能传递您的用户对象。

像这样：

- (instancetype)initWithUser(User *)user {
    UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"Main" bundle:[NSBundle mainBundle]];

   self = [storyboard instantiateViewControllerWithIdentifier:@"emailConfirmVC"];

   if(self) {
      //default initialization
   }

   return  self;
 }

在ViewController之间传输数据而不使用Segues - Swift

3 个答案: