如何缩短此代码以防止代码重复？

Question

我有这个问题很久了。我正在尝试可视化 3 门问题，只是为了好玩和练习 Swift。所以我有：

3 个门，因此有 3 个不同的 IBActions & 所有门的 3 种功能。这些功能完全相同，只是每个代码中的门数不同。所以我想知道，我可以缩短这段代码吗？：

func openSecondChoice(whatDoorIsClickedOn: Int)
    {
        if whatDoorIsClickedOn == 1
        {
            if whatDoorIsClickedOn == doorWithNumber
            {
                UIButtonDoor1.setBackgroundImage( UIImage (named: "doorWithMoney"), for: UIControlState.normal)
            }
            else
            {
                UIButtonDoor1.setBackgroundImage( UIImage (named: "doorWithGoat"), for: UIControlState.normal)
            }
        }
        if whatDoorIsClickedOn == 2
        {
            if whatDoorIsClickedOn == doorWithNumber
            {
                UIButtonDoor2.setBackgroundImage( UIImage (named: "doorWithMoney"), for: UIControlState.normal)
            }
            else
            {
                UIButtonDoor2.setBackgroundImage( UIImage (named: "doorWithGoat"), for: UIControlState.normal)
            }
        }
        if whatDoorIsClickedOn == 3
        {
            if whatDoorIsClickedOn == doorWithNumber
            {
                UIButtonDoor3.setBackgroundImage( UIImage (named: "doorWithMoney"), for: UIControlState.normal)
            }
            else
            {
                UIButtonDoor3.setBackgroundImage( UIImage (named: "doorWithGoat"), for: UIControlState.normal)
            }
        }
    }

哎呀！这段代码太丑了！例如，如果用户按下 door1，我将调用函数“openSecondChoise(whatDoorIsClickedOn: 1)”。有没有办法缩短这个？谢谢！我这里不使用类，我应该使用它们吗？

Answer 1

通常，当您开始使用1、2、3 等为变量名称添加后缀时，是时候改用数组了。这就是数组的用途。

使用包含您的UIButtonDoor1...UIButtonDoor3 的数组uiButtonDoors，您的函数可能如下所示：

func openSecondChoice(whatDoorIsClickedOn: Int) {
    let imageName = whatDoorIsClickedOn == doorWithNumber ? "doorWithMoney" : "doorWithGoat"
    uiButtonDoors[whatDoorIsClickedOn - 1].setBackgroundImage(UIImage(named: imageName), for: UIControlState.normal)
}

Answer 2

func openSecondChoice(whatDoorIsClickedOn: Int) {
  let imageName = whatDoorIsClickedOn == doorWithNumber ? "doorWithMoney" : "doorWithGoat"
  let image = UIImage(named: imageName)

  let button: UIButton

  switch whatDoorIsClickedOn {
  case 1:
    button = UIButtonDoor1
  case 2:
    button = UIButtonDoor2
  case 3:
    button = UIButtonDoor3
  default:
    fatalError("Cannot be. Switch must be exhaustive, that's why we need to use 'default' for a switch on Int.")
  }

  button.setBackgroundImage(image, for: .normal)
}

对于更短的版本，请查看@tuple_cat 的答案。

Answer 3

另一种方法，为了好玩。

import UIKit

class DoorGame {
    func setupButtons() {
        let buttons = [UIButton(), UIButton(), UIButton()]

        for (index, button) in buttons.enumerated() {
            button.addTarget(self, action: #selector(buttonTapped(_:)), for: .touchUpInside)
            button.setTitle("\(index)", for: .normal)
        }

        let winningIndex = Int(arc4random_uniform(UInt32(buttons.count)))

        buttons[winningIndex].tag = 1
    }

    @objc func buttonTapped(_ sender: UIButton) {
        let imageName = sender.tag == 1 ? "doorWithMoney" : "doorWithGoat"
        let image = UIImage(named: imageName)

        sender.setBackgroundImage(image, for: .normal)
    }
}