Question

我有这个问题很长一段时间。我试图想象3门问题，只是为了乐趣和练习Swift。所以我有：

3门，因此3个不同的IBAction＆amp; 所有门的3个功能。这些功能都完全相同，但每个代码中只有门的数量不同。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)
            }
        }
    }

育！这段代码太丑了！例如，如果用户按下door1，我正在调用函数“openSecondChoise（whatDoorIsClickedOn：1）”。有没有办法缩短这个？谢谢！我不在这里使用课程，我应该使用它们吗？

Answer 1

通常，当您使用onCreate，Fragment，@Override public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) { View rootView = inflater.inflate(R.layout.fragment_detail, container, false); Bundle bundle = getArguments(); // .. Other initializations updateViews(bundle); return rootView; }等开始为变量名称添加后缀时，是时候使用数组了。这就是数组的用途。

对于包含1 ... 2的数组3，您的函数可能如下所示：

uiButtonDoors

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)
}

对于较短的版本，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)
    }
}

如何缩短此代码以防止重复代码？

3 个答案: