类型“Notification.Name”（又名“NSNotification.Name”）没有成员“UIResponder”

Question

我有这个代码：

        // MARK: - Listen for Keyboard Events and move the screen up upon keyboard is loaded
    NotificationCenter.default.addObserver(self, selector: #selector(keyboardWillChange), name: UIResponder.keyboardWillShowNotification, object: nil)
    NotificationCenter.default.addObserver(self, selector: #selector(keyboardWillChange), name: UIResponder.keyboardWillHideNotification, object: nil)
    NotificationCenter.default.addObserver(self, selector: #selector(keyboardWillChange), name: UIResponder.keyboardWillChangeFrameNotification, object: nil)
}

deinit {
    NotificationCenter.default.removeObserver(self, name: UIResponder.keyboardWillShowNotification, object: nil)
    NotificationCenter.default.removeObserver(self, name: UIResponder.keyboardWillHideNotification, object: nil)
    NotificationCenter.default.removeObserver(self, name: UIResponder.keyboardWillChangeFrameNotification, object: nil)
}

@objc func keyboardWillChange(notification: Notification) {

    guard let keyboardRect = (notification.userInfo?[UIResponder.keyboardFrameEndUserInfoKey] as? NSValue)?.cgRectValue else {
        return
    }
    if notification.name == Notification.Name.UIResponder.keyboardWillShowNotification ||
        notification.name == Notification.Name.UIKeyboardWillChangeFrame {

                view.frame.origin.y = -keyboardRect.height
    }else {
        view.frame.origin.y = 0
    }


}

但是 Xcode 11 不断告诉错误：

类型“Notification.Name”（又名“NSNotification.Name”）在此行中没有成员“UIResponder” ：

if notification.name == Notification.Name.UIResponder.keyboardWillShowNotification ||

有人知道如何解决吗？

谢谢你的帮助！

Answer 1

利用

if notification.name == UIResponder.keyboardWillShowNotification ||
    notification.name == UIResponder.keyboardWillChangeFrameNotification {

代替

if notification.name == Notification.Name.UIResponder.keyboardWillShowNotification ||
    notification.name == Notification.Name.UIKeyboardWillChangeFrame {

如错误中所见，它没有诸如 Notification.Name UIResponder 之类的调用。