React Navigation：如何使用子堆栈中的特定选项卡返回根选项卡导航器

Question

我在 React Native 中使用 react-navigation v2，每当想要返回根导航器中的 特定 选项卡时都会卡住。

我有以下路由堆栈：

const HomeStack = createStackNavigator(
  {
    Home: Home,
    CreateInvoice: CreateInvoiceScreen,
    InvoiceSummary: InvoiceSummaryScreen,
    PinEntry: PinEntryScreen
  },
  {
    navigationOptions: {
      header: null
    }
  }
);

const CustomersStack = createStackNavigator(
  {
    Customers: CustomersScreen,
    Details: CustomerDetailsScreen
  },
  {
    navigationOptions: {
      header: null
    }
  }
);

const Tab = createBottomTabNavigator(
  {
    Home: HomeStack,
    Transactions: TransactionsTab,
    Customers: CustomersStack,
    Settings: SettingsTab
  }
);

const Routers = createStackNavigator({
  splash: {
    screen: SplashScreen,
    navigationOptions: {...navigationOptions}
  },
  login: {
    screen: LoginScreen,
    navigationOptions: {...navigationOptions}
  },
  home: {
    screen: HomeScreen,
    navigationOptions: {...navigationOptions}
  }
});

我现在在 PinEntry 屏幕中，我想返回 TabNavigator 中的 Transactions 选项卡。我可以使用以下脚本返回 Home 选项卡：

const resetAction = StackActions.reset({
  index: 0,
  actions: [NavigationActions.navigate('Home')]
});
this.props.navigation.dispatch(resetAction);

但我的目标是回到特定的标签位置，在本例中是 'Transactions' 标签。

我已经用谷歌搜索了很多次，但没有任何解决方案，我们将不胜感激。

Answer 1

我找到了解决办法：

const resetAction = StackActions.reset({
  index: 0,
  key: null,
  actions: [NavigationActions.navigate({ routeName: 'Home' })]
});
const goToTransaction = NavigationActions.navigate({
  routeName: 'Transactions'
});
this.props.navigation.dispatch(resetAction);
this.props.navigation.dispatch(goToTransaction);