ios - Swift 使协议扩展成为通知观察者

Question

让我们考虑以下代码：

protocol A {
    func doA()
}

extension A {
  func registerForNotification() {
      NSNotificationCenter.defaultCenter().addObserver(self, selector: Selector("keyboardDidShow:"), name: UIKeyboardDidShowNotification, object: nil)
  }

  func keyboardDidShow(notification: NSNotification) {

  }
}

现在看一个实现 A 的 UIViewController 子类：

class AController: UIViewController, A {
   override func viewDidLoad() {
      super.viewDidLoad()
      self.registerForNotification()
      triggerKeyboard()
   }

   func triggerKeyboard() {
      // Some code that make key board appear
   }

   func doA() {
   }
}

但令人惊讶的是，这会因错误而崩溃：

keyboardDidShow:]: 无法识别的选择器发送到实例 0x7fc97adc3c60

那么我应该在视图控制器本身中实现观察者吗？它不能留在扩展中吗？

以下事情已经尝试过。

使 A 成为类协议。将keyboardDidShow 添加到协议本身作为签名。

protocol A:class {
   func doA()
   func keyboardDidShow(notification: NSNotification)
}

Answer 1

我通过实现较新的- addObserverForName:object:queue:usingBlock:方法NSNotificationCenter并直接调用该方法解决了类似的问题。

extension A where Self: UIViewController  {
    func registerForNotification() {
        NSNotificationCenter.defaultCenter().addObserverForName(UIKeyboardDidShowNotification, object: nil, queue: nil) { [unowned self] notification in
            self.keyboardDidShow(notification)
        }
    }

    func keyboardDidShow(notification: NSNotification) {
        print("This will get called in protocol extension.")
    }
}

此示例将导致keyboardDidShow在协议扩展中被调用。

Answer 2

除了詹姆斯保兰托尼奥的回答。unregisterForNotification可以使用关联对象来实现方法。

var pointer: UInt8 = 0

extension NSObject {
    var userInfo: [String: Any] {
        get {
            if let userInfo = objc_getAssociatedObject(self, &pointer) as? [String: Any] {
                return userInfo
            }
            self.userInfo = [String: Any]()
            return self.userInfo
        }
        set(newValue) {
            objc_setAssociatedObject(self, &pointer, newValue, .OBJC_ASSOCIATION_RETAIN)
        }
    }
}

protocol A {}
extension A where Self: UIViewController {

    var defaults: NotificationCenter {
        get {
            return NotificationCenter.default
        }
    }

    func keyboardDidShow(notification: Notification) {
        // Keyboard did show
    }

    func registerForNotification() {
        userInfo["didShowObserver"] = defaults.addObserver(forName: .UIKeyboardDidShow, object: nil, queue: nil, using: keyboardDidShow)
    }

    func unregisterForNotification() {
        if let didShowObserver = userInfo["didShowObserver"] as? NSObjectProtocol {
            defaults.removeObserver(didShowObserver, name: .UIKeyboardDidShow, object: nil)
        }
    }
}

Answer 3

在 Swift 中使用选择器要求你的具体类必须继承自 NSObject。要在协议扩展中强制执行此操作，您应该使用where. 例如：

protocol A {
    func doA()
}

extension A where Self: NSObject {
  func registerForNotification() {
      NSNotificationCenter.defaultCenter().addObserver(self, selector: Selector("keyboardDidShow:"), name: UIKeyboardDidShowNotification, object: nil)
  }

  func keyboardDidShow(notification: NSNotification) {

  }
}

Answer 4

Answer 5

我使用NSObjectProtocol如下解决了它，

@objc protocol KeyboardNotificaitonDelegate: NSObjectProtocol {
func keyboardWillBeShown(notification: NSNotification)
func keyboardWillBeHidden(notification: NSNotification)
}

extension KeyboardNotificaitonDelegate {

func registerForKeyboardNotifications() {
    //Adding notifies on keyboard appearing
    NotificationCenter.default.addObserver(self, selector: #selector(keyboardWillBeShown(notification:)), name: NSNotification.Name.UIKeyboardWillShow, object: nil)
    NotificationCenter.default.addObserver(self, selector: #selector(keyboardWillBeHidden(notification:)), name: NSNotification.Name.UIKeyboardWillHide, object: nil)
}

func deregisterFromKeyboardNotifications() {
    //Removing notifies on keyboard appearing
    NotificationCenter.default.removeObserver(self, name: NSNotification.Name.UIKeyboardWillShow, object: nil)
    NotificationCenter.default.removeObserver(self, name: NSNotification.Name.UIKeyboardWillHide, object: nil)
}
}