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  • 我有一个由 3 个不同的导航图使用的通用片段。这个片段有 3 个不同的 navGraphViewModels 但只使用其中一个，无论使用它们的片段 ID 调用 它，这不是一个好的设计。我需要使这个片段更通用
  • 我正在实现这 3 个 navGraphViewModels，其中 2 个是不必要的。（因为你不能在不实例化这些 navGraphViewModels 的情况下声明。）请任何优秀的架构师或程序设计师都可以为我找到解决方案。
@AndroidEntryPoint
    class CommonFragment: BaseFragment<CommonFragmentBinding>() {
        //todo should be implemented more generic
        private val bookSharedViewModel: BookSharedViewModel by navGraphViewModels(R.id.book)
        private val tripSharedViewModel: TripsSharedViewModel by navGraphViewModels(R.id.trips)
        private val homeSharedViewModel: HomeSharedViewModel by navGraphViewModels(R.id.home)
        private val viewModel: CommonFragmentViewModel by viewModels()
        private var label: String? = null
    
        override fun inflateBinding(
                inflater: LayoutInflater,
                container: ViewGroup?
        ): CommonFragmentBinding =
                CommonFragmentBinding.inflate(
                        inflater,
                        container,
                        false
                )
    
        @SuppressLint("RestrictedApi")
        override fun onCreate(savedInstanceState: Bundle?) {
            super.onCreate(savedInstanceState)
            label = findNavController().previousBackStackEntry?.destination?.displayName
            viewModel.editTripModel = getFromSharedModel()
        }
    
        private fun getFromSharedModel() = when (label) {
            "${BuildConfig.APPLICATION_ID}:id/bookFragment" -> bookSharedViewModel.editTripModel
            "${BuildConfig.APPLICATION_ID}:id/tripEditFragment" -> tripSharedViewModel.editTripModel
            "${BuildConfig.APPLICATION_ID}:id/homeEditFragment" -> homeSharedViewModel.editTripModel
            else -> null
        }
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