我为Fragment创建了这个扩展函数:

fun Fragment.safeNavigate(
    @IdRes actionId: Int,
    @Nullable args: Bundle? = null,
    @Nullable navOptions: NavOptions? = null,
    @Nullable navigatorExtras: Navigator.Extras? = null
) {
    NavHostFragment.findNavController(this).apply {
        if (currentDestination?.label == this@safeNavigate::class.java.simpleName) {
            navigate(actionId, args, navOptions, navigatorExtras)
        }
    }
}

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)

今天

def navigationVersion = "2.2.1"

这个问题仍然存在。我在Kotlin上的方法是:

// To avoid "java.lang.IllegalArgumentException: navigation destination is unknown to this NavController", se more https://stackoverflow.com/q/51060762/6352712
fun NavController.navigateSafe(
    @IdRes destinationId: Int,
    navDirection: NavDirections,
    callBeforeNavigate: () -> Unit
) {
    if (currentDestination?.id == destinationId) {
        callBeforeNavigate()
        navigate(navDirection)
    }
}

fun NavController.navigateSafe(@IdRes destinationId: Int, navDirection: NavDirections) {
    if (currentDestination?.id == destinationId) {
        navigate(navDirection)
    }
}

将我的答案优雅地扔到处理两种情况(双击，同时点击两个按钮)的环中，但尽量不掩盖真正的错误。

我们可以使用navigateSafe()函数来检查我们试图导航到的目的地从当前目的地是否是无效的，但从前一个目的地是否是有效的。如果是这种情况，代码假设用户双击或同时点击两个按钮。

然而，这个解决方案并不完美，因为它可能会掩盖一些小众情况下的实际问题，即我们试图导航到恰好是父端的目的地。但据推测，这种情况不太可能发生。

代码:

fun NavController.navigateSafe(directions: NavDirections) {
    val navigateWillError = currentDestination?.getAction(directions.actionId) == null

    if (navigateWillError) {
        if (previousBackStackEntry?.destination?.getAction(directions.actionId) != null) {
            // This is probably some user tapping two different buttons or one button twice quickly
            // Ignore...
            return
        }

        // This seems like a programming error. Proceed and let navigate throw.
    }

    navigate(directions)
}

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

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