Swift 特定的通知观察者初始化

Question

在尝试使用新的 Swift NotificationCenter 时，我尝试将观察者对象创建为属性(反对将观察者分配给自身的经典 Obj-C 模式):

private let keyboardWillShowObserver = {
    return NotificationCenter.default.addObserver(forName: .UIKeyboardWillShow, object: nil, queue: nil, using: self.keyboardWillShow(_:))
}()

private func keyboardWillShow(_ notification: Notification) {
    bottomVerticalSpaceConstraint.constant = 400
}

问题是我收到以下错误消息，即使我有同一个类的成员函数部分:

Value of type '(NSObject) -> () -> MyAwesomeViewController' has no member 'keyboardWillShow'

Answer 1

HomeViewController 是您的 View Controller

  private let keyboardWillShowObserver = {
        return NotificationCenter.default.addObserver(forName: .UIKeyboardWillShow, object: nil, queue: nil, using: { (Notification) in
            HomeViewController.keyboardWillShow (Notification)
        })

    }()

     static func keyboardWillShow(_ notification: Notification) {
       // bottomVerticalSpaceConstraint.constant = 400
    }