Accessing a function of stateful widget in its state class? flutter

Question

I am trying to call _signOut function after the onPressed method for logout button. However it doesnt (recognize the function or let me call it) I can however make a call widget.Onsignedout, a callback to it parent and everything works as intended. except i sign out the user at auth.signout and call back is just to update the form. how can I access the _signOut() from state class? thank you

import 'package:flutter/material.dart';
import 'package:login_demo/auth.dart';
import 'package:login_demo/root_page.dart';

class HomePage extends StatefulWidget {
  HomePage({this.auth, this.onSignedOut});
  final BaseAuth auth;
  //To call a function of a parent, you can use the callback pattern
  final VoidCallback onSignedOut;

  void _signOut() async {
    try {
      Text('Signing Out here');
      await auth.signOut();
      onSignedOut;
    } catch (e) {
      print(e);
    }
  }

  @override
  _HomePageState createState() => _HomePageState();
}

class _HomePageState extends State<HomePage> {
  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text('Homepage'),
        actions: <Widget>[
          FlatButton(
            child: Text(
              'Logout',
              style: TextStyle(fontSize: 17.0, color: Colors.white)
            ),
            onPressed: widget.onSignedOut,
          ),
        ],
      ),
      body: Container(
        child: Center(
          child: Text(
            'Welcome',
            style: TextStyle(fontSize: 32.0),
          ),
        ),
      ),
    );
  }
}

score 5 · Answer 1 · answered Aug 07 '18 at 07:22

5

You need to make your method public. Right now because of the underscore (_) before the name, the method is private and you cannot access it.

Simply change the name from _signOut to signOut and then you should be able to call it by widget.signOut().

answered Aug 07 '18 at 07:22

Marcin Szałek

4,609
5
30
43

making the method public didn't seem to work, I get the error. "Expression here is a type of 'void' and therefore cannot be used. If I add "auth.signOut();" to the callback onSignedout method. it seems to work. – Tegster007 Aug 07 '18 at 20:14
If you are doing `onPressed: widget.signOut` try changing it to `onPressed: () => widget.signOut()` and tell me if it works :) Also what you can do is remove `void` from `signOut()` method since it will return a `Future` – Marcin Szałek Aug 08 '18 at 08:00
1

I actually ended up moving the function down stateful state section and passing the two parameters through the stateful constructor. I read somewhere it was good practice. – Tegster007 Aug 20 '18 at 16:46

score 0 · Answer 2 · answered Dec 22 '21 at 18:10

0

First, change your _signOut to signOut so it can be accessed by another file. OR more simple solution, move your function inside the state of your homepagestate statefulWidget.

answered Dec 22 '21 at 18:10

Enrico Antonio

1

Do not copy-paste someone answer – Rishabh Deep Singh Dec 22 '21 at 20:02
Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Dec 22 '21 at 20:03

Accessing a function of stateful widget in its state class? flutter

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