Question

我需要运行一个未知量（来自字典数组）的SKAction序列，这些序列是动态创建的，并且在循环中使用延迟随着我循环而增加而偏移。循环的打印输出应按以下顺序出现

show line 5
hiding everything
show line 6
hiding everything
show line 2
hiding everything

获取此功能我使用此代码

func display(winningLines: [Int : AnyObject]) {

    var delay = 0.0
    var actions = [SKAction]()

    let hideAction = SKAction.run {
        print("hiding everything")
    }

    for winningLine in winningLines {

        let displayAction = SKAction.run {
            print("line \(winningLine.key)")
        }

        let wait = SKAction.wait(forDuration: delay)            
        var action: SKAction!

        //repeatForever = true
        if !repeatForever {
             action = SKAction.sequence([wait, displayAction, SKAction.wait(forDuration: 1.0), hideAction])
        }
        else {
            let seq = SKAction.sequence([wait, displayAction, SKAction.wait(forDuration: 1.0), hideAction])
            action = SKAction.repeatForever(seq)
        }
        self.run(action)

        delay += 1.0
    }
}

如果序列只需要发生一次而不是按预期输出，那么有时我需要序列才能永远重复（直到用户停止操作）。永远重复该操作不起作用，因为它重复每个单独的序列并扰乱顺序。第一个序列每1秒重复一次，第二个序列每2秒重复一次......等等。

有关如何让它永远正常循环的任何建议吗？

line 5
hiding everything
line 5
line 6
hiding everything
line 5
hiding everything
line 2
hiding everything
line 5
line 6
hiding everything
hiding everything
line 5
hiding everything
hiding everything
line 5
line 6
line 2
hiding everything
line 5
hiding everything
hiding everything

我还尝试将操作附加到一组SKActions并在最后循环它们但它不起作用

var actions = [SKAction]()

...

actions.append(action)

for action in actions {
    self.run(action)
}

Answer 1

如果您唯一需要关注的是未知数量的序列，您可以执行以下操作：

创建所有必需的序列
将它们放在另一个序列中
永远循环

因此，例如，此代码将创建四个框，并将按顺序应用淡入/淡出（代码中的displayAction / hideAction）到每个框，并将永远重复：

import SpriteKit

class GameScene: SKScene {

var boxes:[SKSpriteNode] = []

override func didMove(to view: SKView) {

    let distance:CGFloat = 10

    for i in 0...3 {

        let box = SKSpriteNode(color: .purple, size: CGSize(width: 50, height: 50))
        box.name = String(i) //name == index
        box.position.x = CGFloat(i) * (box.size.width + distance)
        box.alpha = 0.0
        addChild(box)
        boxes.append(box)

    }
}

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

    var actions:[SKAction] = []

    let action = SKAction.sequence([

        SKAction.fadeIn(withDuration: TimeInterval(1)),
        SKAction.fadeOut(withDuration: 0),
    ])

    for index in 0...boxes.count {

        let sequence = SKAction.sequence([
            SKAction.run(action, onChildWithName: String(index)),
            SKAction.wait(forDuration: 1)
        ])

        actions.append(sequence)

    }

    self.run(SKAction.repeatForever(SKAction.sequence(actions)))
}
}

如何循环动态延迟的SKActions序列

1 个答案: