一旦第一个函数完成后调用第二个函数，它是使用 Swift 的 SKAction 重复计数

Question

我有这个功能：

func midGroundlevelMovingSlow() {

    let moveLevelImage = SKAction.moveByX(-self.frame.size.width, y: 0, duration: gameSpeed)
    let replaceLevelImage = SKAction.moveByX(self.frame.size.width, y: 0, duration: 0)
    let movingAndReplacingLevelImage = SKAction.repeatAction(SKAction.sequence([moveLevelImage,replaceLevelImage]), count: 5)

    mgImage.runAction(movingAndReplacingLevelImage)
    mgImage2.runAction(movingAndReplacingLevelImage)
}

一旦完成了它的 5 循环，我希望调用这个函数：

func midGroundlevelMovingMedium() {

    let moveLevelImage = SKAction.moveByX(-self.frame.size.width, y: 0, duration: gameSpeed * 2)
    let replaceLevelImage = SKAction.moveByX(self.frame.size.width, y: 0, duration: 0)
    let movingAndReplacingLevelImage = SKAction.repeatAction(SKAction.sequence([moveLevelImage,replaceLevelImage]), count: 5)

    mgImage.runAction(movingAndReplacingLevelImage)
    mgImage2.runAction(movingAndReplacingLevelImage)
}

谁能帮我解决这个问题？

谢谢，

史蒂文

Answer 1

SKSpriteNode 有一个带有完成处理程序的运行操作函数。

sprite.runAction(action: SKAction) { () -> Void in
        // call next function
    }

您可以在该完成处理程序中运行您的下一个方法。

如果您想等待两个函数完成，您可以使用计数器并在调用下一个方法之前检查两者是否都已完成。

类似

func midGroundlevelMovingSlow(completion: (()->Void)?) {

    let moveLevelImage = SKAction.moveByX(-self.frame.size.width, y: 0, duration: gameSpeed)
    let replaceLevelImage = SKAction.moveByX(self.frame.size.width, y: 0, duration: 0)
    let movingAndReplacingLevelImage = SKAction.repeatAction(SKAction.sequence([moveLevelImage,replaceLevelImage]), count: 5)

    var callCount = 2

    mgImage.runAction(movingAndReplacingLevelImage){ () -> Void in
        if --callCount == 0 {
            completion?()
        }
    }

    mgImage2.runAction(movingAndReplacingLevelImage){ () -> Void in
        if --callCount == 0 {
            completion?()
        }
    }
}