Is there any way to pop back to previous route by a condition in Flutter?

Question

I'm currently doing a purchase feature on my App. Whenever free users click on a premium feature, it'll pop to the previous screen and users can now using the premium feature.

This is the current code I have:

Check on premium feature:

if (!PremiumUtils.checkLifetimePremium(user?.premiumStatus) &&
    !PremiumUtils.checkSubscriptionPremium(user?.premiumExpireDate)) {
  Navigator.push(context, _createUpgradeRoute());
}

If the user is a free user, display the Upgrade UI. And I used the PremiumUtils to check whether the purchase is done or not, to navigate between routes:

part of pages;

class Upgrade extends StatelessWidget {
  const Upgrade({Key? key}) : super(key: key);

  @override
  Widget build(BuildContext context) {
    UserModel? user = context.watch<AppViewModel>().userModel;

    return ChangeNotifierProvider(
        create: (context) => UpgradeViewModel(),
        child: Builder(
          builder: (context) {
            return Consumer<UpgradeViewModel>(
              builder: (context, provider, child) {
                if (PremiumUtils.checkLifetimePremium(user?.premiumStatus) ||
                    PremiumUtils.checkSubscriptionPremium(user?.premiumExpireDate)) {
                     Navigator.pop(context);
                }
                return the UI;
            }
          }
        ));
  }
}

This obviously causing the setState() or markNeedsBuild() called during build exception. And I can't think of any way around.

Please give me some idea about this problem!

Answer 1

This answer solved it.

Wrap it in Future.microtask. This will schedule it to happen on the next async task cycle (i.e. after build is complete).

Future.microtask(() => Navigator.push(
  context, 
  MaterialPageRoute(builder: (context) => LoginView())
))