Question

我正在尝试将UIGestureRecognizer集成到我的快速游乐场中，这样无论我在操场上点什么，我的精灵就会到达那一点。我已经尝试了一切！观看无数的youtube视频，然后进行堆栈溢出以获得答案，但每个人都开始创建一个名为视图的类，这是一个UIView，在他们的代码中他们一直指的是“自我”。我做了一个名为“view”的变量，它是一个SKView。我尝试获取部分代码，将其放入我的代码中并进行更改，但它只是不起作用。这是我到目前为止的工作。

view.addGestureRecognizer(UIGestureRecognizer(target: view, action:  #selector(handleTap(sender:))))

func handleTap(sender: UITapGestureRecognizer) {
        player.position = sender.location(in: view)
}

我的游乐场一直告诉我，我正在使用未解析的标识符'handleTap（sender :)'

Answer 1

UIGestureRecognizer示例，在Playground中具有不同的手势识别器状态

<强> swift3 +

导入UIKit 导入PlaygroundSupport

类ViewController：UIViewController {

var yellowView: UIView!
var redView: UIView!
var yellowViewOrigin: CGPoint!

override func loadView() {

    // UI

    let view = UIView()
    view.backgroundColor = .white

    yellowView = UIView()
    yellowView.backgroundColor = .yellow
    view.addSubview(yellowView)

    redView = UIView()
    redView.backgroundColor = .red
    view.addSubview(redView)

    // Layout
    redView.translatesAutoresizingMaskIntoConstraints = false
    yellowView.translatesAutoresizingMaskIntoConstraints = false
    NSLayoutConstraint.activate([
        yellowView.topAnchor.constraint(equalTo: view.topAnchor, constant: 20.0),
        yellowView.leadingAnchor.constraint(equalTo: view.leadingAnchor, constant: 20.0),
        yellowView.heightAnchor.constraint(equalToConstant: 80.0),
        yellowView.widthAnchor.constraint(equalToConstant: 80.0),

        redView.bottomAnchor.constraint(equalTo: view.bottomAnchor, constant: -20.0),
        redView.trailingAnchor.constraint(equalTo: view.trailingAnchor, constant: -20.0),
        redView.heightAnchor.constraint(equalToConstant: 80.0),
        redView.widthAnchor.constraint(equalToConstant: 80.0)

        ])

    self.view = view
}

override func viewDidLoad() {
    super.viewDidLoad()
    let pan = UIPanGestureRecognizer(target: self, action: #selector(self.handlePanGesture(_:)))
    yellowView.addGestureRecognizer(pan)
    yellowViewOrigin = yellowView.frame.origin
    view.bringSubview(toFront: yellowView)
}

@objc func handlePanGesture(_ sender: UIPanGestureRecognizer) {
    let targetView = sender.view!
    let translation = sender.translation(in: view)

    switch sender.state {
    case .began,.changed:
        targetView.center = CGPoint(x: targetView.center.x + translation.x
            ,y: targetView.center.y + translation.y)
        sender.setTranslation(CGPoint.zero, in: view)
    case .ended:
        if targetView.frame.intersects(redView.frame){
            UIView.animate(withDuration: 0.3) {
                targetView.alpha = 0.0
            }

        }
        else{
        UIView.animate(withDuration: 0.3) {
            targetView.frame.origin = self.yellowViewOrigin
        }
        }
        break
    default:
        break
    }

}

}

PlaygroundPage.current.liveView = ViewController（）

Answer 2

我认为您需要在函数声明前添加@objc标记，以便handleTap对#selector可见。

@objc func handleTap(sender: UITapGestureRecognizer) {
    player.position = sender.location(in: view) 
}

Answer 3

必须从func handleTap类型调用

view。

请尝试将其添加到SKView扩展程序中：

    extension SKView {
        func handleTap(sender: UITapGestureRecognizer) {
                player.position = sender.location(in: view)
        }
    }

Swift Playgrounds中的UIGestureRecognizer

3 个答案: