Swift:简化菜单按钮逻辑 [英] Swift: Simplify menu button logic

问题描述

我在 Swift 中有一个游戏，其中我的每个关卡都是从基本 GameScene 类继承的单独类(对于我正在做的事情来说，这种方式要容易得多，不要评判我).我还有一个菜单，每个级别都有一个按钮.这是按钮加载关卡的方式:

I have a game in Swift where each of my levels is a separate class inherited from the base GameScene class (it's a lot easier this way for what I'm doing, don't judge me). I also have a menu which has a button for each level. This is how the buttons loads the level:

override func touchesBegan(_ touches: Set<UITouch>, with event: UIEvent?) {
  if let t = touches.first {
    let node = atPoint(t.location(in: self))
    if let name = node.name {
      let newScene: GameScene!
      switch Int(name)! {
      case 1:
        newScene = Level1(size: frame.size)
      case 2:
        newScene = Level2(size: frame.size)
      case 3:
        newScene = Level3(size: frame.size)
      case 4:
        newScene = Level4(size: frame.size)
      case 5:
        newScene = Level5(size: frame.size)
      case 6:
        newScene = Level6(size: frame.size)
      case 7:
        newScene = Level7(size: frame.size)
      default:
        newScene = Level1(size: frame.size)
      }
      view?.presentScene(newScene, transition: .crossFade(withDuration: 0.5))
    }
  }
}

对我来说，这个开关看起来非常丑陋和毫无意义，但我想不出避免它的方法.我希望这里有人能帮我解决这个问题，我只是想不出更好的选择.

To me, this switch looks incredibly ugly and pointless, but I can't think of a way to avoid it. I was hoping someone here could help me out with this, I just can't think of a better alternative.

推荐答案

您可以先将关卡的类名构造为字符串 "Level\(name)" ，然后再获取您需要的实际类通过将其名称传递给函数:

You could construct your Level's class name as a string "Level\(name)" first and then get the actual class you need by passing it's name to the function:

func classFromString(_ className: String) -> AnyClass! {
    let namespace = Bundle.main.infoDictionary!["CFBundleExecutable"] as! String
    let cls: AnyClass = NSClassFromString("\(namespace).\(className)")!
    return cls
}

使用示例:

let className = "Level1"
let levelInstance = (classFromString(className) as! GameScene).init(size: frame.size)

...当然最好的建议是避免这种架构

... but of course the best advice would be to avoid such kind of architecture

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