如何访问 Siri Remote play/pause 按钮和覆盖菜单按钮

Question

如何使用 Siri 遥控器访问 play/pause 按钮并覆盖菜单按钮？我目前正在使用它，但它对我不起作用。当我使用这段代码时，我的程序崩溃了，但只有当我调用它时才会崩溃四个示例按下暂停按钮编码器目前位于 didMoveToView 下面，紧挨着 touchesBegan

let tapGesture = UITapGestureRecognizer(target: self, action: "handleTap:")
tapGesture.allowedPressTypes = [NSNumber(integer: UIPressType.Menu.rawValue)]
self.view.addGestureRecognizer(tapGesture)

Answer 1

您的问题是您正在调用一个名为 handleTap: 的函数来接收一个参数，但您没有一个名为 handleTap: 的函数。这就是 action 在这一行中所代表的：

let tapGesture = UITapGestureRecognizer(target: self, action: Selector("handleTap:"))

将您的 func tapped() 更改为：

func handleTap(sender: UITapGestureRecognizer) {
    if sender.state == UIGestureRecognizerState.Ended {
        print("Menu button released")
    }
}

Answer 2

我通过将 tapRecognizer 选择器移动到我之前设置的触摸处理函数中解决了我的问题，因此代码现在如下所示：

private func handleTouches(touches: Set<UITouch>) {
    for touch in touches {
        let touchLocation = touch.locationInNode(self)
        lastTouch = touchLocation

        let tapRecognizer = UITapGestureRecognizer(target: self, action: Selector("handleTap:"))
        tapRecognizer.allowedPressTypes = [NSNumber(integer: UIPressType.PlayPause.rawValue)];
        self.view!.addGestureRecognizer(tapRecognizer)
    }
}

func handleTap(sender: UITapGestureRecognizer) {
    if sender.state == UIGestureRecognizerState.Ended {
        print("Menu button released")
    }     
 }

Answer 3

我对 Swift 4 使用以下内容（根据问题：@selector() in Swift?）

let tapGesture = UITapGestureRecognizer(target: self, action: #selector(handleTap(sender:)))
tapGesture.allowedPressTypes = [NSNumber(integer: UIPressType.Menu.rawValue)]
self.view.addGestureRecognizer(tapGesture)

@objc func handleTap(sender: UITapGestureRecognizer) {
    if sender.state == UIGestureRecognizerState.Ended {
        print("Menu button released")
    }
}

如何访问 Siri Remote play/pause 按钮和覆盖菜单按钮

How to access Siri Remote play/pause button and override menu button

swift

tvos

uitapgesturerecognizer

siri-remote