如何在两个视图不正确时自动翻转它们

Posted 2023-04-18

【中文标题】如何在两个视图不正确时自动翻转它们

【英文标题】：How to flip two views when their incorrect automatically

【发布时间】：2021-09-15 18:28:08

【问题描述】：

我正在寻找创建一个记忆游戏，我现在正在研究这个功能。我已经大致了解了项目的逻辑，但我想补充一点细节。我想补充的细节之一是，当两张选择的卡片不正确时，它们会在 1 秒后自行转动。我尝试使用 Handler().postDelay 但它似乎没有用。

这里是代码

class GamePlay1Fragment : Fragment() 

//    lateinit var front_anim: AnimatorSet
//    lateinit var back_anim: AnimatorSet

    private lateinit var pieces: List<ImageView>
    private lateinit var gameCards: List<GameCard>
    private var indexOfSelectedPiece: Int? = null


    override fun onCreateView(
        inflater: LayoutInflater,
        container: ViewGroup?,
        savedInstanceState: Bundle?
    ): View? 
        val binding: Gameplay1FragmentBinding = DataBindingUtil.inflate(
            inflater,
            R.layout.gameplay1_fragment,
            container, false)


        binding.backButtonView.setOnClickListener  v: View ->
            v.findNavController().navigate(GamePlay1FragmentDirections.actionGamePlay1FragmentToLobbyFragment())
        


        val images = mutableListOf(memorybatcardfront, memorycatcardfront, memorycowcardfront,
            memorydragonfront, memorygarbagemancardfront, memoryghostdogcardfront)

        images.addAll(images)

        images.shuffle()

        pieces = listOf(binding.card1back, binding.card2back, binding.card3back, binding.card4back, binding.card5back,binding.card6back, binding.card7back,
                        binding.card8back, binding.card9back, binding.card10back, binding.card11back, binding.card12back)

        gameCards = pieces.indices.map  index ->
            GameCard(images[index])
        


            pieces.forEachIndexed  index, piece ->
                piece.setOnClickListener 

                    updatingModels(index)

                    updatingViews()
                
            



        return binding.root
    

    private fun updatingViews() 
        gameCards.forEachIndexed  index, gameCard ->
          val piece = pieces[index]
            piece.setImageResource(if (gameCard.isFacedUp) gameCard.id else allcardbacks)
        
    

    private fun updatingModels(position: Int) 
        val gameCard = gameCards[position]

        if (gameCard.isFacedUp) return


        if (indexOfSelectedPiece == null) 
            restoreGameCards()
            indexOfSelectedPiece = position

        
        else 
            checkingForMatch(indexOfSelectedPiece!!, position)
            indexOfSelectedPiece = null
        
        gameCard.isFacedUp = !gameCard.isFacedUp
    

    private fun restoreGameCards() 
        val handler = Handler()
        for (gameCard in gameCards) 
            if (!gameCard.isMatched) 
                handler.postDelayed(runnable, 1000)
                gameCard.isFacedUp = false

            

        
    

    private fun checkingForMatch(position1: Int, position2: Int) 
        if (gameCards[position1].id == gameCards[position2].id) 
          gameCards[position1].isMatched = true
            gameCards[position2].isMatched = true
        
    

    private val runnable = Runnable()
        kotlin.run 
        
    

如果有其他方法可以做到这一点，我会全力以赴。欣赏它。

【参考方案1】：

问题是您在调用处理程序时同步翻转卡片（即它们同时执行），但您需要在处理程序延迟结束时翻转卡片。

要解决此问题，您需要在Handler 的Runnable 内传输卡片翻转：

private fun restoreGameCards() 
    val handler = Handler(Looper.getMainLooper())
    for (gameCard in gameCards) 
        if (!gameCard.isMatched) 
            handler.postDelayed(
                gameCard.isFacedUp = false // Executes after 1000 millisec
            , 1000)
        
    

【讨论】：

它确实有效，但不是我想要的。延迟确实会发生，但现在您可以在所有被点击的卡片面朝下之前点击多张卡片。

所以，你需要绑定用户点击多张卡片后的延迟..以及用户完成的时候；我猜你需要翻转这些卡片？

对。翻转第二张牌后，如果没有匹配，则在两张牌翻回之前有 1 秒的延迟。