How to select conditional createStackNavigator screen? - react-native

Question

I'm trying to select conditional screen based on Redux data. I have two screens in a screen stack, and this screen stack is also a part of TabNavigator (Logged out screen should be shown with Tabs too). If user is logged in, I need to show CheckScreen, otherwise CheckScreenLoggedOut. I tried setting initialRouteName conditionally but it shows CheckScreenLoggedOut only after logging in.
How can I implement this logic?
Here is my createStackNavigator part:
const CheckScreenStack = createStackNavigator(
{
CheckSendLoggedOut: {
screen: IntroScreen,
navigationOptions: ({navigation}) => ({
headerLeft: (
<View>
<TestModeIcon/>
</View>
),
}),
},
CheckScreen: {
screen: CheckScreen,
navigationOptions: ({navigation}) => ({
headerRight: (
<View>
<TouchableOpacity title='' onPress={() => { navigation.navigate('NotificationsScreen'); }}>
<NotificationTabBarIcon/>
</TouchableOpacity>
</View>
)
}),
},
},
{
initialRouteName: (typeof store.getState().userData !== "undefined") ? 'CheckScreen' : 'CheckSendLoggedOut'
}
);

Answer 1

I have already achieved this by accessing the redux state with connect, you can do something like this:
const CheckScreenStack = ({ userData }) => {
const Stack = createStackNavigator(
{
...
},
{
initialRouteName: (typeof userData !== "undefined") ? 'CheckScreen' : 'CheckSendLoggedOut'
}
);
return Stack;
}
const mapStateToProps = ({ userData }) => {
return { userData };
};
export default connect(
mapStateToProps,
{},
)(CheckScreenStack);