Flutter how to show Admob App open Ad only when variable divisble by 3-CodePudding

I want to show the Admob App an open Ad and for that I followed this link and is working fine. Now Ad is showing whenever I opened the app but I want to show ad only when a variable is divisible by 3. For like ad comes to the user 3 times and the ad dispose of and for the six-time show ad again site i followed

CodePudding user response：

You can wrap these in an if statement when the variable is divisible by 3.

    //Load AppOpen Ad
    appOpenAdManager.loadAd();

    //Show AppOpen Ad After 8 Seconds
    Future.delayed(const Duration(seconds: 8)).then((value) {
      //Here we will wait for 8 seconds to load our ad
      //After 8 seconds it will go to HomePage
      appOpenAdManager.showAdIfAvailable();
      Navigator.push(
        context,
        MaterialPageRoute(
          builder: (context) => const HomePage(),
        ),
      );
    });

CodePudding user response：

Use any local storage libraries / shared preferences

create variable as storing open openedCount=0;

read it from storage/preference

Now wrap your like

if(openedCount % 3  == 0)
{
openedCount =0;
//store it again
appOpenAdManager.loadAd();

    //Show AppOpen Ad After 8 Seconds
    Future.delayed(const Duration(seconds: 8)).then((value) {
      //Here we will wait for 8 seconds to load our ad
      //After 8 seconds it will go to HomePage
      appOpenAdManager.showAdIfAvailable();
      Navigator.push(
        context,
        MaterialPageRoute(
          builder: (context) => const HomePage(),
        ),
      );
    });
}
else{
// increment the value of openedCount in the storage 
Navigator.push(
        context,
        MaterialPageRoute(
          builder: (context) => const HomePage(),
        ),
      );
}