swift - 无法识别的选择器发送到实例，通知中心获取键盘大小

Question

如果我尝试点击文本字段，则会收到错误，与尝试获取移动 iOS 设备上键盘大小的这几行代码有关。通知中心的代码行位于重写的 ViewDidAppear 内。

 NotificationCenter.default.addObserver(self, selector: Selector(("keyboardWillShow:")), name: UIResponder.keyboardDidShowNotification, object: nil)

 NotificationCenter.default.addObserver(self, selector: Selector(("keyboardWillHide:")), name: UIResponder.keyboardDidHideNotification, object: nil)

func keyboardWillShow(notification: NSNotification) {
    if let userInfo = notification.userInfo {
        if let keyboardSize = (userInfo[UIResponder.keyboardFrameBeginUserInfoKey] as? NSValue)?.cgRectValue {
            self.bottomConstraint.constant = keyboardSize.height
        }
    }
}

func keyboardWillHide(notification: NSNotification) {
    self.bottomConstraint.constant = 0

}

Answer 1

使用类型安全语法

#selector(keyboardWillShow)

和

@objc func keyboardWillShow(_ notification: Notification) { ...

<小时/>

但是我强烈建议使用基于现代闭包的语法

NotificationCenter.default.addObserver(forName: UIResponder.keyboardWillShowNotification, object: nil, queue: .main) { [weak self] notification in
    if let userInfo = notification.userInfo,
       let keyboardSize = (userInfo[UIResponder.keyboardFrameBeginUserInfoKey] as? NSValue)?.cgRectValue {
            self?.bottomConstraint.constant = keyboardSize.height
    }
}
NotificationCenter.default.addObserver(forName: UIResponder.keyboardWillHideNotification, object: nil, queue: .main) { [weak self] _ in
    self?.bottomConstraint.constant = 0
}