CAShapeLayer 在动画 Swift 期间检测触摸答案

【问题标题】：CAShapeLayer detect touch during animation SwiftCAShapeLayer 在动画 Swift 期间检测触摸
【发布时间】：2016-05-08 17:14:28
【问题描述】：

我可以像这样检测到 CAShapeLayer 的触摸 (touchesEnded)：

let touchLocation : CGPoint = (touch as! UITouch).locationInView(self.view)

for shape in shapes{
    if CGPathContainsPoint(shape.path, nil, touchLocation, false){
        print("Layer touch")
    }
}

我可以像这样为 CAShapeLayer 的路径设置动画：

let newShapePath = UIBezierPath(arcCenter: toPoint, radius: 20, startAngle: CGFloat(0), endAngle: CGFloat(M_PI * 2), clockwise: true).CGPath

// animate the `path`
let animation = CABasicAnimation(keyPath: "path")
animation.toValue = newShapePath
animation.duration = CFTimeInterval(duration)

animation.timingFunction = CAMediaTimingFunction(name: kCAMediaTimingFunctionEaseIn)
animation.fillMode = kCAFillModeBoth
animation.removedOnCompletion = false

shape.addAnimation(animation, forKey: animation.keyPath)

但是当动画发生时，在 CAShapeLayer 上没有检测到触摸。是否可以在为路径设置动画时检测到 CAShapeLayer 上的触摸？

【问题讨论】：

标签： ios swift core-animation cashapelayer cabasicanimation

【解决方案1】：

您可以访问图层的presentationLayer 以执行此操作。这将为您提供动画时给定图层的“飞行中”值的粗略近似值。例如：

for shape in shapes {

    // gets the layer's presentation layer if it exists – else fallback on the model layer
    let presentationLayer = shape.presentationLayer() as? CAShapeLayer ?? shape

    if CGPathContainsPoint(presentationLayer.path, nil, touchLocation, false){
        print("Layer touch")
    }
}

另外，作为旁注，如果您不使用动画委托，通常认为使用removedOnCompletion = false 是不好的做法。您应该只更新图层的模型值以表示其新状态，而不是让动画逗留。您可以通过CATransaction 执行此操作，以确保不会生成隐式动画。例如：

let animation = CABasicAnimation(keyPath: "path")
animation.fromValue = shape.path
animation.toValue = newShapePath
animation.duration = CFTimeInterval(duration)

animation.timingFunction = CAMediaTimingFunction(name: kCAMediaTimingFunctionEaseIn)

shape.addAnimation(animation, forKey: animation.keyPath)

// update the layer's model values
CATransaction.begin()
CATransaction.setDisableActions(true)
shape.path = newShapePath
CATransaction.commit()

【讨论】：