在其状态类中访问有状态窗口小部件的功能?扑 [英] Accessing a function of stateful widget in its state class? flutter

问题描述

我试图在onPressed方法退出按钮后调用_signOut函数.但是它没有(识别函数或让我调用它)，但是我可以调用call widget.Onsignedout，它的父级回调，一切都按预期工作.除了我在auth.signout上注销用户，然后回叫只是为了更新表单.如何从状态类访问_signOut()?谢谢

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),
          ),
        ),
      ),
    );
  }
}

推荐答案

您需要将方法公开.现在，由于名称前有下划线(_)，该方法是私有的，您无法访问它.

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.

只需将名称从_signOut更改为signOut，然后您就可以通过widget.signOut()调用它.

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

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