永远移动精灵

Question

当我按下屏幕上的按钮时，我试图将精灵向右移动。但是，当我尝试这样做时，我只有将精灵移动到某个点的解决方案。所以...我希望精灵永远向右移动，或者直到我做其他事情。

这是在 Xcode 中使用 SpriteKit 中的 Swift。

override func touchesBegan(touches: Set<UITouch>, withEvent event: UIEvent?) {

    for touch : AnyObject in touches{

        let pointInTouch = touch.locationInNode(self)
        let tappedNode = nodeAtPoint(pointInTouch)
        let tappeNodeName = tappedNode.name

        if tappeNodeName == "Btn"{

            player.physicsBody?.velocity = CGVectorMake(0, 0)

            let action = SKAction.applyImpulse(CGVectorMake(400, 0), duration: 1)
            player.runAction(SKAction.repeatActionForever(action))

            print("Touched!")                
        }
    }
}

Answer 1

您可以简单地将节点向右移动，直到它确实存在于场景中

class GameScene: SKScene {

    private var player: SKSpriteNode!

    override func touchesBegan(touches: Set<UITouch>, withEvent event: UIEvent?) {

        guard let touch = touches.first else { return }
        let pointInTouch = touch.locationInNode(self)
        let tappedNode = nodeAtPoint(pointInTouch)
        if tappedNode.name == "Btn"{
            let moveToRight = SKAction.moveToX(self.frame.width + player.frame.width, duration: 5)
            player.runAction(moveToRight)
        }

    }
}

更新：恒速

如果您希望玩家以恒定速度移动，您可以使用此代码。

如您所见，我正在使用space / speed 计算持续时间。您只需要为您的场景找到最佳的速度常数值。

class GameScene: SKScene {

    private var player: SKSpriteNode!

    override func touchesBegan(touches: Set<UITouch>, withEvent event: UIEvent?) {

        guard let touch = touches.first else { return }
        let pointInTouch = touch.locationInNode(self)
        let tappedNode = nodeAtPoint(pointInTouch)


        let deltaX = self.frame.width + player.frame.width - player.position.x
        let speed: CGFloat = 10 // <-- change this to find the best value for you
        let duration: NSTimeInterval = NSTimeInterval(deltaX / speed)

        if tappedNode.name == "Btn"{
            let moveToRight = SKAction.moveByX(deltaX, y:0, duration: duration)
            player.runAction(moveToRight)
        }

    }
}