Flutter - Always execute a function when the page appears

Question

How could I make the name() function run whenever the Page1 page appeared?

In the code below before going to Page2 I execute the dispose() Already inside Page2 if I click the back button or the physical button of Android the function name() is not executed, but if I click the 'go to Page1' button the function name() is executed.

Could you help me to always execute the name() function when Page1 appears?

import 'package:flutter/material.dart';

void main() {
  runApp(new MyApp());
}

class MyApp extends StatelessWidget {  
  @override
  Widget build(BuildContext context) {
    return new MaterialApp(
      home: new MyHomePage(),
      routes: <String, WidgetBuilder> {
        '/page2': (BuildContext context) => new Page2(),
      },
    );
  }
}

class MyHomePage extends StatefulWidget {
  @override
  _MyHomePageState createState() => new _MyHomePageState();
}

class _MyHomePageState extends State<MyHomePage> {
  String nameScreen;

  String name() {
    return 'foo1';
  }

  @override
  void initState() {
    super.initState();
    this.nameScreen = name();

  }

  @override
  void dispose() {
    this.nameScreen = '';
    super.dispose();
  }

  @override
  Widget build(BuildContext context) {
    return new Scaffold(
      appBar: new AppBar(
        title: new Text('Page 1'),
        backgroundColor: new Color(0xFF26C6DA),
      ),
      body: new Center(
        child: new Column(
          mainAxisAlignment: MainAxisAlignment.center,
          children: <Widget>[
            new RaisedButton(
              child: const Text('go to Page2'),
              onPressed: () async {
                dispose();
                bool isLoggedIn = await Navigator.of(context).pushNamed('/page2');
                if (isLoggedIn) {
                  setState((){
                    this.nameScreen = name();
                  });
                }
              },
            ),            
            new Text(
              '$nameScreen',              
            ),
          ],
        ),
      ),
    );
  }
}

class Page2 extends StatelessWidget{
  @override
  Widget build(BuildContext context) {
    return new Scaffold( 
      appBar: new AppBar(
        title: new Text('Page 2'),
        backgroundColor: new Color(0xFFE57373)
      ),
      body: new Center(
        child: new Column(
          mainAxisAlignment: MainAxisAlignment.center,
          children: <Widget>[
            new RaisedButton(
              child: const Text('go back to Page1'),
              onPressed: () {
                Navigator.pop(context, true);
              }
            ),
          ],
        ),
      ),
    );
  }
}

Answer 1

There is no need to call dispose at all when you are willing to pop and change State later, since dispose will remove the current object from the tree, which does not translate to the logic you are trying to develop.

You can indeed override the BackButton and pass the same call of Navigator.pop(context, result) to it. Check the following example I have tweaked your code a little bit to show you the difference between each State of your nameScreen field. I hope this helps you.

class MyHomePage extends StatefulWidget {
  @override
  _MyHomePageState createState() => _MyHomePageState();
}

class _MyHomePageState extends State<MyHomePage> {
  String nameScreen = "";

  String name() {
    return 'foo1';
  }

  @override
  void initState() {
    super.initState();
    this.nameScreen = "From initState";

  }

@override
void dipose(){
    super.dispose();
}

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: const Text('Page 1'),
        backgroundColor: Color(0xFF26C6DA),
      ),
      body: Center(
        child: Column(
          mainAxisAlignment: MainAxisAlignment.center,
          children: <Widget>[
            RaisedButton(
              child: const Text('go to Page2'),
              onPressed: () async {
                //dispose(); ///No need for dispose
                String result = await Navigator.of(context).pushNamed('/page2');

                  setState((){
                    this.nameScreen = result;
                  });

              },
            ),
            Text(
              '$nameScreen',
            ),
          ],
        ),
      ),
    );
  }
}

class Page2 extends StatelessWidget{
  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
          leading: IconButton(icon: Icon(Icons.arrow_back), onPressed: ()async{
            Navigator.pop(context,"From BackButton");
          }),
          title: const Text('Page 2'),
          backgroundColor: Color(0xFFE57373)
      ),
      body: Center(
        child: Column(
          mainAxisAlignment: MainAxisAlignment.center,
          children: <Widget>[
            RaisedButton(
                child: const Text('go back to Page1'),
                onPressed: () {
                  Navigator.pop(context, "From RaisedButton");
                }
            ),
          ],
        ),
      ),
    );
  }

Answer 2

One way of doing this is to use the .whenComplete() method on the Navigator widget.

Suppose you are going to the second page from the first page. Here you have to pass the functionThatSetsTheState as a pointer to the navigation part of your code.

The function looks like this and should be in a Stateful Widget.

void functionThatSetsTheState(){
 setState(() {});
}

Your navigation code for OnPressed, OnTap, OnLongPress, etc.

Navigator.of(context).push(MaterialPageRoute(builder: (BuildContext context) => SecondPage())).whenComplete(() => {functionThatSetsTheState()});

Answer 3

You can use RouteObserves if you want to execute some function whenever your page appears, you will have to implement RouteAware on the page where you want to run execute the function whenever the screens appears, you're gonna have to do something like this on ur Page1

final RouteObserver<PageRoute> routeObserver = RouteObserver<PageRoute>(); // add this on your main class
void main() {
  runApp(MaterialApp(
    home: Container(),
    navigatorObservers: [routeObserver], // add observer here;
  ));
} 





// your page where func should run whenever this page appears
class MyHomePage extends StatefulWidget with RouteAware {
  @override
  _MyHomePageState createState() => _MyHomePageState();
}

class _MyHomePageState extends State<MyHomePage> {
  String nameScreen = "";

  String name() {
    return 'foo1';
  }

  @override
  void initState() {
    super.initState();
    this.nameScreen = "From initState";

  }

 @override
  void didChangeDependencies() {
    super.didChangeDependencies();
    routeObserver.subscribe(this, ModalRoute.of(context));
  }

  @override
  void dispose() {
    routeObserver.unsubscribe(this);
    super.dispose();
  }

// implementing RouteAware method
void didPush() {
// Route was pushed onto navigator and is now topmost route.
    name(); // your func goes here
  }


  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: const Text('Page 1'),
        backgroundColor: Color(0xFF26C6DA),
      ),
      body: Center(
        child: Column(
          mainAxisAlignment: MainAxisAlignment.center,
          children: <Widget>[
            RaisedButton(
              child: const Text('go to Page2'),
              onPressed: () async {
                //dispose(); ///No need for dispose
                String result = await Navigator.of(context).pushNamed('/page2');

                  setState((){
                    this.nameScreen = result;
                  });

              },
            ),
            Text(
              '$nameScreen',
            ),
          ],
        ),
      ),
    );
  }
}

you can head over to this link for more explanation https://api.flutter.dev/flutter/widgets/RouteObserver-class.html

Answer 4

You can override the back button on the second screen. And instead of system closing, do

WillPopScope(
    onWillPop: () {
       print('back pressed');
       Navigator.pop(context, "From BackButton");
       return true;
      },
    child: Scaffold(...)