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) {

  }
}

现在看一下实现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？

已经尝试过。

制作A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

我通过实施- addObserverForName:object:queue:usingBlock:的较新NSNotificationCenter方法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.")
    }
}

此示例将导致在协议扩展中调用keyboardDidShow。

Answer 2

除了James Paolantonio的回答。可以使用关联对象实现unregisterForNotification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

为避免崩溃，请在使用该协议的Swift类中实现observer方法。

实现必须在Swift类本身，而不仅仅是协议扩展，因为选择器总是引用Objective-C方法，并且协议扩展中的函数不能用作Objective-C选择器。如果Swift类继承自Objective-C类，那么来自Swift类的方法可用作Objective-C选择器

“If your Swift class inherits from an Objective-C class, all of the methods and properties in the class are available as Objective-C selectors.”

此外，在Xcode 7.1中，self在AnyObject调用中将其指定为观察者时，必须向下转换为addObserver。

protocol A {
    func doA()
}

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

    func keyboardDidShow(notification: NSNotification) {
        print("will not appear")
    }
}

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

    func triggerKeyboard(){
        // Some code that makes the keyboard appear
    }

    func doA(){
    }

    func keyboardDidShow(notification: NSNotification) {
        print("got the notification in the class")
    }
}

Answer 4

在Swift中使用选择器要求您的具体类必须从NSObject继承。要在协议扩展中强制执行此操作，您应使用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 5

我使用NSObjectProtocol解决了它，如下所示

@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)
}
}

Swift使协议扩展成为通知观察者

5 个答案: