I am calling the 2.3.1 Navigation and the same error occurs when the application configuration changes. When the cause of the problem was found through Debug, the GaphId in NavHostFragment did not take effect as the ID currently set by calling navController.setGraph(). The GraphId of NavHostFragment can only be obtained from the <androidx.fragment.app.FragmentContainerView/> tag. At this time, this problem will occur if there are multiple GraphIds dynamically set in your code. When the interface is restored, the Destination cannot be found in the cached GraphId. You can solve this problem by manually specifying the value of mGraphId in NavHostFragment through reflection when switching Graph.

navController.setGraph(R.navigation.home_book_navigation);
try {
    Field graphIdField = hostFragment.getClass().getDeclaredField("mGraphId");
    graphIdField.setAccessible(true);
    graphIdField.set(navHostFragment, R.navigation.home_book_navigation);
} catch (NoSuchFieldException | IllegalAccessException e) {
    e.printStackTrace();
}

在思考了Ian Lake在推特上的建议后，我想出了以下方法。将NavControllerWrapper定义如下:

class NavControllerWrapper constructor(
  private val navController: NavController
) {

  fun navigate(
    @IdRes from: Int,
    @IdRes to: Int
  ) = navigate(
    from = from,
    to = to,
    bundle = null
  )

  fun navigate(
    @IdRes from: Int,
    @IdRes to: Int,
    bundle: Bundle?
  ) = navigate(
    from = from,
    to = to,
    bundle = bundle,
    navOptions = null,
    navigatorExtras = null
  )

  fun navigate(
    @IdRes from: Int,
    @IdRes to: Int,
    bundle: Bundle?,
    navOptions: NavOptions?,
    navigatorExtras: Navigator.Extras?
  ) {
    if (navController.currentDestination?.id == from) {
      navController.navigate(
        to,
        bundle,
        navOptions,
        navigatorExtras
      )
    }
  }

  fun navigate(
    @IdRes from: Int,
    directions: NavDirections
  ) {
    if (navController.currentDestination?.id == from) {
      navController.navigate(directions)
    }
  }

  fun navigateUp() = navController.navigateUp()

  fun popBackStack() = navController.popBackStack()
}

然后在导航代码中:

val navController = navControllerProvider.getNavController()
navController.navigate(from = R.id.main, to = R.id.action_to_detail)

我在我的项目中也有同样的问题，首先我试图在视图上触发导航动作的点击，但经过一些实验后，我发现在真正缓慢的设备上，debounce应该是一个非常高的值，导致应用程序对快速设备的用户感到缓慢。

所以我为NavController提出了以下扩展，我认为它符合原始的API，并且易于使用:

fun NavController.safeNavigate(directions: NavDirections) {
    try {
        currentDestination?.getAction(directions.actionId) ?: return
        navigate(directions.actionId, directions.arguments, null)
    } catch (e : Exception) {
        logError("Navigation error", e)
    }
}

fun NavController.safeNavigate(directions: NavDirections, navOptions: NavOptions?) {
    try {
        currentDestination?.getAction(directions.actionId) ?: return
        navigate(directions.actionId, directions.arguments, navOptions)
    } catch (e : Exception) {
        logError("Navigation error", e)
    }
}

fun NavController.safeNavigate(directions: NavDirections, navigatorExtras: Navigator.Extras) {
    try {
        currentDestination?.getAction(directions.actionId) ?: return
        navigate(directions.actionId, directions.arguments, null, navigatorExtras)
    } catch (e : Exception) {
        logError("Navigation error", e)
    }
}

请注意，我正在使用SafeArgs和NavDirections。这些函数检查操作从当前目的地是否有效，并且仅在操作不为空时进行导航。如果Navigation库每次都返回正确的操作，那么try catch部分就不需要了，但我希望消除所有可能的崩溃。

我通过检查当前目标中是否存在下一个操作来解决这个问题

public static void launchFragment(BaseFragment fragment, int action) {
    if (fragment != null && NavHostFragment.findNavController(fragment).getCurrentDestination().getAction(action) != null) {       
        NavHostFragment.findNavController(fragment).navigate(action);
    }
}

public static void launchFragment(BaseFragment fragment, NavDirections directions) {
    if (fragment != null && NavHostFragment.findNavController(fragment).getCurrentDestination().getAction(directions.getActionId()) != null) {       
        NavHostFragment.findNavController(fragment).navigate(directions);
    }
}

这解决了一个问题，如果用户快速点击2个不同的按钮

看来你在完成任务。应用程序可能有一次性设置或一系列登录屏幕。这些有条件的屏幕不应该被认为是应用程序的起始目的地。

https://developer.android.com/topic/libraries/architecture/navigation/navigation-conditional

当我按了两次后退键时，我想到了这个问题。首先，我拦截KeyListener并覆盖KeyEvent.KEYCODE_BACK。我在名为OnResume的函数中添加了下面的代码，然后解决了这个问题/问题。

  override fun onResume() {
        super.onResume()
        view?.isFocusableInTouchMode = true
        view?.requestFocus()
        view?.setOnKeyListener { v, keyCode, event ->
            if (event.action == KeyEvent.ACTION_DOWN && keyCode == KeyEvent.KEYCODE_BACK) {
                activity!!.finish()
                true
            }
            false
        }
    }

当我第二次遇到这种情况时，它的状态与第一次相同，我发现我可能会使用add函数。让我们来分析一下这些情况。

首先，FragmentA导航到FragmentB，然后FragmentB导航到FragmentA，然后按下返回按钮…崩溃出现了。 其次，FragmentA导航到FragmentB，然后FragmentB导航到FragmentC, FragmentC导航到FragmentA，然后按下返回按钮…崩溃出现了。

所以我认为当按下返回按钮时，FragmentA会回到FragmentB或FragmentC，这会导致登录混乱。最后，我发现名为popBackStack的函数可以用于返回，而不是导航。

  NavHostFragment.findNavController(this@TeacherCloudResourcesFragment).
                        .popBackStack(
                            R.id.teacher_prepare_lesson_main_fragment,false
                        )

到目前为止，问题已经真正解决了。