如何获取 CAKeyframeAnimation 对象的当前屏幕 xy 位置？答案

【问题标题】：How to get current screen xy position of a CAKeyframeAnimation object?如何获取 CAKeyframeAnimation 对象的当前屏幕 xy 位置？
【发布时间】：2019-02-06 19:12:56
【问题描述】：

在 iOS 12 beta 12 下的 Swift 4.2 中，我想获取我沿 CAKeyframeAnimation 中的路径设置动画的点的当前位置。

动画的代码是这样的：

    @IBOutlet weak var dot: UIImageView!

    func animateDot() {        
        // Oval path.
        let ovalWidth = UIScreen.main.bounds.width * 0.7
        let ovalHeight = UIScreen.main.bounds.height * 0.75
        let ovalPath = UIBezierPath(ovalIn: CGRect(x: -ovalWidth / 2 , y: -ovalHeight / 2, width: ovalWidth, height: ovalHeight))
        let orbit = CAKeyframeAnimation(keyPath: "position")
        orbit.path = ovalPath.cgPath
        orbit.duration = CFTimeInterval(2.5)
        orbit.isAdditive = true
        orbit.repeatCount = 6
        orbit.calculationMode = kCAAnimationPaced
        orbit.rotationMode = kCAAnimationRotateAuto

        self.dot.layer.add(orbit, forKey: "orbit")
    }

我的问题是：一旦启动并运行，我如何在 Swift 中查询动画以获取点中心的当前屏幕 xy？

【问题讨论】：

标签： cakeyframeanimation ios12 swift4.2

【解决方案1】：

我发现可以通过查询点关联的active表示层来获取当前点所在位置的边框：

        if (self.dot.layer.presentation() != nil)
        {
            let dotLocation = self.dot.layer.presentation()?.frame
            os_log("self.dot.layer.presentation()?.frame.minX %12f", type: .debug, dotLocation!.minX)
            os_log("self.dot.layer.presentation()?.frame.maxX %12f", type: .debug, dotLocation!.maxX)
            os_log("self.dot.layer.presentation()?.frame.minY %12f", type: .debug, dotLocation!.minY)
            os_log("self.dot.layer.presentation()?.frame.maxY %12f", type: .debug, dotLocation!.maxY)

            var dotScreenLocation: CGPoint = CGPoint(x: 0, y: 0)
            dotScreenLocation.x = dotLocation!.minX + ((dotLocation!.maxX - dotLocation!.minX) / 2)
            dotScreenLocation.y = dotLocation!.minY + ((dotLocation!.maxY - dotLocation!.minY) / 2)
            os_log("self.dot.layer.presentation()?.frame center x,y %12f,%12f", type: .debug, dotScreenLocation.x, dotScreenLocation.y)
}

【讨论】：