应用程序崩溃：异常 - 发送到实例的无法识别的选择器

Question

我有一个 UITableviewController，它具有以下逻辑，一旦键盘像这样切换，就可以向上/向下滑动整个视图：

class ChatDetailViewController: UIViewController, UITableViewDelegate, UITableViewDataSource {
 // variables...

 override func viewDidLoad() {
        super.viewDidLoad()
        // do stuff...
        NotificationCenter.default.addObserver(self, selector: Selector(("keyboardWillShow:")), name: UIResponder.keyboardWillShowNotification, object: nil)
        NotificationCenter.default.addObserver(self, selector: Selector(("keyboardWillHide:")), name: UIResponder.keyboardWillHideNotification, object: nil)
       // do other stuff...
}
...
func keyboardWillShow(notification: NSNotification) {
        if let keyboardSize = (notification.userInfo?[UIResponder.keyboardFrameBeginUserInfoKey] as? NSValue)?.cgRectValue {
            self.view.frame.origin.y -= keyboardSize.height
        }
    }

    func keyboardWillHide(notification: NSNotification) {
        self.view.frame.origin.y = 0
    }
...
}

切换键盘然后应用程序崩溃，出现以下异常：ChatDetailViewController keyboardWillShow:]：无法识别的选择器发送到实例 0x7f82fc41fdf0

错误消息乍一看似乎很清楚，但我仍然无法弄清楚我的选择器出了什么问题。没有代码警告，没有拼写错误，...

我哪里做错了？

Answer 1

在Swift 4中我以前用过这样的

override func viewDidLoad() {
    super.viewDidLoad()   
    NotificationCenter.default.addObserver(self, selector: #selector(self.keyboardWillShow), name: NSNotification.Name.UIKeyboardWillShow, object: nil)
    NotificationCenter.default.addObserver(self, selector: #selector(self.keyboardWillHide), name: NSNotification.Name.UIKeyboardWillHide, object: nil)
}

@objc func keyboardWillShow(notification: NSNotification) {
     print("keyboardWillShow")
}

@objc func keyboardWillHide(notification: NSNotification){
     print("keyboardWillHide")
}

在 Swift 5 中，他们将访问键盘通知的方式重命名为 UIResponder:

override func viewDidLoad() {
    super.viewDidLoad()   
    NotificationCenter.default.addObserver(self, selector: #selector(self.keyboardWillShow), name: UIResponder.keyboardWillShowNotification, object: nil)
    NotificationCenter.default.addObserver(self, selector: #selector(self.keyboardWillHide), name: UIResponder.keyboardWillHideNotification, object: nil)
}

Answer 2

您忘记添加@objc 部分：

NotificationCenter.default.addObserver(self,
                                               selector: #selector(keyboardWillShow),
                                               name: UIResponder.keyboardWillShowNotification,
                                               object: nil)
        NotificationCenter.default.addObserver(self,
                                               selector: #selector(keyboardWillHide),
                                               name: UIResponder.keyboardWillHideNotification,
                                               object: nil)
@objc func keyboardWillShow(_ sender: Notification) {}
@objc func keyboardWillHide(_ sender: Notification) {}

Answer 3

三期：

选择器创建错误（使用#selector）
动作签名错误（缺少下划线）
动作必须标记为@objc

并且 Swift 中的类型是 Notification 没有 NS 前缀

override func viewDidLoad() {
    super.viewDidLoad()
    // do stuff...
    NotificationCenter.default.addObserver(self, selector: #selector(keyboardWillShow), name: UIResponder.keyboardWillShowNotification, object: nil)
    NotificationCenter.default.addObserver(self, selector: #selector(keyboardWillHide), name: UIResponder.keyboardWillHideNotification, object: nil)
    // do other stuff...
}

    ...

@objc func keyboardWillShow(_ notification: Notification) {
    if let keyboardSize = (notification.userInfo?[UIResponder.keyboardFrameBeginUserInfoKey] as? NSValue)?.cgRectValue {
        self.view.frame.origin.y -= keyboardSize.height
    }
}

@objc func keyboardWillHide(_ notification: Notification) {
    self.view.frame.origin.y = 0
}

...

应用程序崩溃：异常 - 发送到实例的无法识别的选择器

App crashing: Exception - unrecognised selector sent to instance

nsnotificationcenter

swift