Flutter - 如何将用户数据传递给所有视图

2 回答

• 0

  我建议进一步调查继承的小部件;下面的代码显示了如何使用它们异步更新数据：

  import 'dart:convert';
  
  import 'package:flutter/material.dart';
  import 'package:http/http.dart' as http;
  
  void main() {
    runApp(new MaterialApp(
        title: 'Inherited Widgets Demo',
        theme: new ThemeData(
          primarySwatch: Colors.blue,
        ),
        home: new Scaffold(
            appBar: new AppBar(
              title: new Text('Inherited Widget Example'),
            ),
            body: new NamePage())));
  }
  
  // Inherited widget for managing a name
  class NameInheritedWidget extends InheritedWidget {
    const NameInheritedWidget({
      Key key,
      this.name,
      Widget child}) : super(key: key, child: child);
  
    final String name;
  
    @override
    bool updateShouldNotify(NameInheritedWidget old) {
      print('In updateShouldNotify');
      return name != old.name;
    }
  
    static NameInheritedWidget of(BuildContext context) {
      // You could also just directly return the name here
      // as there's only one field
      return context.inheritFromWidgetOfExactType(NameInheritedWidget);
    }
  }
  
  // Stateful widget for managing name data
  class NamePage extends StatefulWidget {
    @override
    _NamePageState createState() => new _NamePageState();
  }
  
  // State for managing fetching name data over HTTP
  class _NamePageState extends State<NamePage> {
    String name = 'Placeholder';
  
    // Fetch a name asynchonously over HTTP
    _get() async {
      var res = await http.get('https://jsonplaceholder.typicode.com/users');
      var name = JSON.decode(res.body)[0]['name'];
      setState(() => this.name = name); 
    }
  
    @override
    void initState() {
      super.initState();
      _get();
    }
  
    @override
    Widget build(BuildContext context) {
      return new NameInheritedWidget(
        name: name,
        child: const IntermediateWidget()
      );
    }
  }
  
  // Intermediate widget to show how inherited widgets
  // can propagate changes down the widget tree
  class IntermediateWidget extends StatelessWidget {
    // Using a const constructor makes the widget cacheable
    const IntermediateWidget();
  
    @override
    Widget build(BuildContext context) {
      return new Center(
        child: new Padding(
          padding: new EdgeInsets.all(10.0),
          child: const NameWidget()));
    }
  }
  
  class NameWidget extends StatelessWidget {
    const NameWidget();
  
    @override
    Widget build(BuildContext context) {
      final inheritedWidget = NameInheritedWidget.of(context);
      return new Text(
        inheritedWidget.name,
        style: Theme.of(context).textTheme.display1,
      );
    }
  }
  

  回复于 2024-04-27T04:24:41+08:00
• 15

  因为这个问题，我遇到了另一个问题you can check it out here所以我提出的解决方案有点凌乱，我创建了一个单独的Instance dart页面并将其导入到每个页面 .

  GoogleSignInAccount Guser = googleSignIn.currentUser;
   FirebaseUser Fuser;
  

  我在登录时将用户存储在那里并检查每个StateWidget是否为null

  Future<Null> _ensureLoggedIn() async {
  
  if (Guser == null) Guser = await googleSignIn.signInSilently();
  if (Fuser == null) {
    await googleSignIn.signIn();
    analytics.logLogin();
  }
  if (await auth.currentUser() == null) {
    GoogleSignInAuthentication credentials =
    await googleSignIn.currentUser.authentication;
    await auth.signInWithGoogle(
      idToken: credentials.idToken,
      accessToken: credentials.accessToken,
    );
  }
  

  这是我的旧代码，我在我当前的应用程序上清理了它，但我现在没有那个代码 . 只需检查空用户并再次登录

  我也为大多数Firebase实例做了这个，因为我的应用程序上有超过3个页面而且Inherited Widgets工作太多了

  回复于 2024-04-27T04:24:41+08:00