无法在 Swift 游乐场中向 SKScene 添加滑动识别器？答案

【问题标题】：Unable to add swipe recognizer to SKScene in Swift playground?无法在 Swift 游乐场中向 SKScene 添加滑动识别器？
【发布时间】：2017-03-18 23:14:41
【问题描述】：

好的，我对 Swift Playground 的工作方式还有些陌生，但我正在尝试将 Swift 3 中的滑动手势识别器添加到我的 Swift Playground 中。关注这个http://www.spritekitlessons.com/gesture-recognizer-with-sprite-kit-and-swift/我现在有：

func swipedRight(sender:UISwipeGestureRecognizer){
    print("swiped right")
}

func swipedLeft(sender:UISwipeGestureRecognizer){
    print("swiped left")
}

func swipedUp(sender:UISwipeGestureRecognizer){
    print("swiped up")
}

func swipedDown(sender:UISwipeGestureRecognizer){
    print("swiped down")
}

let degree = CGFloat(M_PI_2) / 90


class GameScene: SKScene {
    var selectedNode: SKNode?
    var shakeAction: SKAction?

    override func didMove(to view: SKView) {

        /* Setup your scene here */

        let swipeRight:UISwipeGestureRecognizer = UISwipeGestureRecognizer(target: self, action: Selector(("swipedRight:")))
        swipeRight.direction = .right
        view.addGestureRecognizer(swipeRight)


        let swipeLeft:UISwipeGestureRecognizer = UISwipeGestureRecognizer(target: self, action: Selector(("swipedLeft:")))
        swipeLeft.direction = .left
        view.addGestureRecognizer(swipeLeft)


        let swipeUp:UISwipeGestureRecognizer = UISwipeGestureRecognizer(target: self, action: Selector(("swipedUp:")))
        swipeUp.direction = .up
        view.addGestureRecognizer(swipeUp)


        let swipeDown:UISwipeGestureRecognizer = UISwipeGestureRecognizer(target: self, action: Selector(("swipedDown:")))
        swipeDown.direction = .down
        view.addGestureRecognizer(swipeDown)
    }

let frame = CGRect(x: 0, y: 0, width: 1000, height: 600) //view size
let view = SKView(frame: frame)

let scene = GameScene(size: frame.size)
view.presentScene(scene)
PlaygroundPage.current.liveView = view

这可以编译，但是当我滑动时，我得到一个无法识别的选择器错误，即使我确实在选择器中包含了函数：

我也尝试将函数放在类中。如何向 Swift Playground SKScene 添加滑动识别器？

【问题讨论】：

标签： swift swift3 swift-playground skscene

【解决方案1】：

您将选择器作为字符串传递，如错误日志中所述，它们肯定有问题

尝试使用新的选择器语法 - #selector(methodName)。

例子：

class ViewController: UIViewController {

    override func viewDidLoad() {
        super.viewDidLoad()

        /* Swift 3 */

        let swipe = UISwipeGestureRecognizer(target: self, action:#selector(handleSwipe))
        view.addGestureRecognizer(swipe)

    }

    func handleSwipe() {
        print("Swiped!")
    }
}

不推荐使用字符串作为选择器。如果methodName() 方法不存在，则使用新的选择器语法，您将收到编译错误——您的应用不会因为“无法识别的选择器”而崩溃。

【讨论】：