How to display popup after navigating back to previous screen in flutter?

Question

I have a scenario where i have one popup in screen A .It will be triggerd from initstate() . User can navigate to screen B after clicking on button inside the popup . How can i show the popup if user come back to screen 1 by clicking on arrow back button ?

To get a better answer, post some code. How are you currently opening the screen? — MSpeed, Sep 23 '21 at 13:56
Please provide enough code so others can better understand or reproduce the problem. — Community, Oct 01 '21 at 18:00

score 1 · Answer 1 · answered Sep 23 '21 at 14:00

You can achieve with the help of .then() : https://api.flutter.dev/flutter/dart-async/Future/then.html

                         Navigator.of(context)
                             .push(CupertinoPageRoute<Screen1>(
                              builder: (context) => Screen2(
                                ))).then((value) {

                             //ShowPopUpMenu() <-- Your PopUpMenu.

                          });

EyaS · Answer 2 · 2022-02-24T10:28:46.410

Add a boolean parameter to your screen if you want the popup:

class PopScreen extends StatefulWidget {
  PopScreen (this.displayPopup);
  final bool displayPopup;
  @override
  _PopScreenState createState() => _PopScreenState();
}

class _PopScreenState extends State<PopScreen> {
  @override
  void initState() {
    super.initState();
    if(widget.displayPopup){
      // display your popup here
    }
  }
}

call to the navigator:

Navigator.pushNamed(context, '/yourscreen', arguments: true);

How to display popup after navigating back to previous screen in flutter?

2 Answers2