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我在单独的文件中有 2 个类：让我们称它们为 NotificationManagement 和 NotificationReceiver

在我的 NotificationReceiver 类中，我有以下代码：

 class NotificationReceiver: UIViewController{
 //...   

 func showNotificationHandler(alert: UIAlertAction!) {

        var storyboard : UIStoryboard = UIStoryboard(name: Constants.storyboardName, bundle: nil)
        var vc : Article = storyboard.instantiateViewControllerWithIdentifier(Constants.articleVCTitle) as! Article
        self.presentViewController(vc, animated: false, completion: nil)


    }
    func dismissNotificationHandler(alert: UIAlertAction!) {
        Constants.articleAddress = ""
    }

    func showAlert(title:String, message:String) {

        let alert = UIAlertController(title: title,
            message: message, preferredStyle: .Alert)
        let okAction = UIAlertAction(title: Messages.showMsg, style: .Default , handler: showNotificationHandler)
        let dismissAction = UIAlertAction(title: Messages.discardMsg, style: .Cancel, handler: dismissNotificationHandler)
        alert.addAction(dismissAction)
        alert.addAction(okAction)
        self.presentViewController(alert, animated: false, completion: nil)
        }
  //...
}

现在我想把这个通知的处理放在 NotificationManagement 类而不是 NotificationReceiver 类中。但是，如果我确实把它放在那里，我需要在处理程序中使用类似的东西，以传递有关当前视图控制器的信息：

func showNotificationHandler(alert: UIAlertAction!, currentViewController: UIViewController) {

        var storyboard : UIStoryboard = UIStoryboard(name: Constants.storyboardName, bundle: nil)
        var vc : Article = storyboard.instantiateViewControllerWithIdentifier(Constants.articleVCTitle) as! Article
        currentViewController.presentViewController(vc, animated: false, completion: nil)  
    }

但是后来我无法编译，并且错误确实具有误导性，例如编译器找不到样式 .Default （这是因为我向 showNotificationHandler 添加了参数）。我想像这样使用 NotificationReceiver 中的函数：

let notificationManagement = NotificationManagement()
notificationManagement.showAlert("title",message: "message")
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