Question

我试图在SceneKit中获取触摸位置的条件。我正在触摸SpriteKit对象并编写该代码以获取SKnode对象：

class GameViewController: UIViewController , SCNSceneRendererDelegate {

var overlay: SKScene!

...

override public func touchesEnded(_ touches: Set<UITouch>, with event: UIEvent!) {

   for touch: AnyObject in touches {
        let location: CGPoint  = touch.location(in: overlay)
        let touchedNode: SKNode = overlay.nodes(at: location

        if(touchedNode.name == "WalkAnimationButton") {

            self.tap()
        }
        else if(touchedNode.name == "CameraButton") {

            sceneView.allowsCameraControl = true
        }
        else if(touchedNode.name == "WrenchButton") {

            sceneView.showsStatistics = true
        }
        else if(touchedNode.name == "CharacterButton") {
            if(currentCharacter < characterPaths.count-1) {
                currentCharacter = currentCharacter+1
            }
            else{
                currentCharacter = 0
            }
        }

    }

    self.setCharacterFromModelWithName(name: characterPaths[currentCharacter] as! NSString)
}

Xcode写了我之前从未遇到过的错误：
无法转换类型＆＃39; [SKNode]＆＃39;的值指定类型＆＃39; SKNode＆＃39;

Answer 1

要调用触摸的节点，我们应该使用该代码：

{{1}}

因为我们调用了SpriteKit对象的名称，它正在用作SKNode类的成员。

SpriteKit Swift3触及了SKNode

1 个答案: