在调用navigate之前检查currentDestination可能会有帮助。

例如，如果您在导航图fragmentA和fragmentB上有两个片段目的地，并且从fragmentA到fragmentB只有一个动作。当你已经在fragmentB上时，调用navigate(R.id.action_fragmentA_to_fragmentB)将导致IllegalArgumentException。因此，在导航之前，你应该总是检查currentDestination。

if (navController.currentDestination?.id == R.id.fragmentA) {
    navController.navigate(R.id.action_fragmentA_to_fragmentB)
}

I caught this exception after some renames of classes. For example: I had classes called FragmentA with @+is/fragment_a in navigation graph and FragmentB with @+id/fragment_b. Then I deleted FragmentA and renamed FragmentB to FragmentA. So after that node of FragmentA still stayed in navigation graph, and android:name of FragmentB's node was renamed path.to.FragmentA. I had two nodes with the same android:name and different android:id, and the action I needed were defined on node of removed class.

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

用try-catch(简单的方法)包装你的导航调用，或者确保在短时间内只有一个导航调用。这个问题可能不会消失。复制更大的代码片段在你的应用程序和尝试。

你好。基于上面的一些有用的回答，我想分享我的解决方案，可以扩展。

下面是导致我的应用程序崩溃的代码:

@Override
public void onListItemClicked(ListItem item) {
    Bundle bundle = new Bundle();
    bundle.putParcelable(SomeFragment.LIST_KEY, item);
    Navigation.findNavController(recyclerView).navigate(R.id.action_listFragment_to_listItemInfoFragment, bundle);
}

一个很容易重现这个错误的方法是用多个手指在项目列表上点击，点击每个项目就会在导航到新屏幕上解决(基本上和人们注意到的一样——在很短的时间内点击两次或两次以上)。我注意到:

第一次导航调用总是正常工作; 第二个和所有其他的导航方法调用在IllegalArgumentException中解析。

在我看来，这种情况可能会经常出现。因为重复代码是一种糟糕的做法，有一点影响总是好的，我想到了下一个解决方案:

public class NavigationHandler {

public static void navigate(View view, @IdRes int destination) {
    navigate(view, destination, /* args */null);
}

/**
 * Performs a navigation to given destination using {@link androidx.navigation.NavController}
 * found via {@param view}. Catches {@link IllegalArgumentException} that may occur due to
 * multiple invocations of {@link androidx.navigation.NavController#navigate} in short period of time.
 * The navigation must work as intended.
 *
 * @param view        the view to search from
 * @param destination destination id
 * @param args        arguments to pass to the destination
 */
public static void navigate(View view, @IdRes int destination, @Nullable Bundle args) {
    try {
        Navigation.findNavController(view).navigate(destination, args);
    } catch (IllegalArgumentException e) {
        Log.e(NavigationHandler.class.getSimpleName(), "Multiple navigation attempts handled.");
    }
}

}

因此上面的代码只改变了一行:

Navigation.findNavController(recyclerView).navigate(R.id.action_listFragment_to_listItemInfoFragment, bundle);

:

NavigationHandler.navigate(recyclerView, R.id.action_listFragment_to_listItemInfoFragment, bundle);

它甚至变得更短了一点。代码在发生崩溃的确切位置进行了测试。没有经历过，并将使用相同的解决方案为其他导航，以避免同样的错误进一步。

任何想法都欢迎!

到底是什么导致了崩溃

记住，当我们使用方法navigation . findnavcontroller时，我们使用相同的导航图、导航控制器和后堆栈。

We always get the same controller and graph here. When navigate(R.id.my_next_destination) is called graph and back-stack changes almost instantly while UI is not updated yet. Just not fast enough, but that is ok. After back-stack has changed the navigation system receives the second navigate(R.id.my_next_destination) call. Since back-stack has changed we now operate relative to the top fragment in the stack. The top fragment is the fragment you navigate to by using R.id.my_next_destination, but it does not contain next any further destinations with ID R.id.my_next_destination. Thus you get IllegalArgumentException because of the ID that the fragment knows nothing about.

这个确切的错误可以在NavController.java方法findDestination中找到。

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。

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)