与提供商 flutter 的身份验证 Firebase

Question

我尝试在此处使用 Porvider 实现此身份验证 Firebase 问题是我想保存变量 loggedIn 的 state 即使我关闭并 reponning 应用程序所以我使用此代码但不工作

    class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return Consumer<AuthenticationService>(builder: (_, auth, __) {
      if (auth.getCurrentUser() != null)
        return HomeView();
      else
        return TermsView();
    });
  }
}

这里方法获取当前用户

Future getCurrentUser() async {
FirebaseUser _user ;
_user = await FirebaseAuth.instance.currentUser();

return _user ;
}

有没有办法检查Consumer中的登录？

Answer 1

您需要使用FirebaseAuth.currentUser来检查用户是否已登录，我在Consumer中使用FutureBuilder来完成这项工作。

void main() {
  WidgetsFlutterBinding.ensureInitialized();

  runApp(
    ChangeNotifierProvider<Auth>(
      create: (_) => Auth(),
      child: MaterialApp(home: MyApp()),
    ),
  );
}

class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return Consumer<Auth>(
      builder: (_, auth, __) {
        return FutureBuilder<FirebaseUser>(
          future: auth.currentUser,
          builder: (context, snapshot) {
            if (snapshot.hasData) return HomeScreen();
            return LoginScreen();
          },
        );
      },
    );
  }
}

class HomeScreen extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return Scaffold(
      backgroundColor: Color(0xFF4C3E13),
      appBar: AppBar(title: Text('Home Screen')),
      floatingActionButton: FloatingActionButton.extended(
        label: Text('Sign out'),
        onPressed: () async {
          final auth = Provider.of<Auth>(context, listen: false);
          await auth.signOut();
        },
      ),
    );
  }
}

class LoginScreen extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(title: Text('Login Screen')),
      body: Center(
        child: RaisedButton(
          onPressed: () async {
            final auth = Provider.of<Auth>(context, listen: false);
            await auth.signIn(email: 'user@user.com', password: 'user1234');
          },
          child: Text('Sign In'),
        ),
      ),
    );
  }
}

class Auth with ChangeNotifier {
  final _auth = FirebaseAuth.instance;

  Future<void> signOut() async {
    await _auth.signOut();
    notifyListeners();
  }

  Future<void> signIn({@required String email, @required String password}) async {
    await _auth.signInWithEmailAndPassword(email: email, password: password);
    notifyListeners();
  }

  Future<FirebaseUser> get currentUser async {
    return await _auth.currentUser();
  }
}

Answer 2

我尝试在 Firebase 与 Porvider 在这里实现此身份验证问题是我想保存变量 loggedIn 的 state 即使我关闭并响应应用程序所以我使用了此代码但不工作

    class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return Consumer<AuthenticationService>(builder: (_, auth, __) {
      if (auth.getCurrentUser() != null)
        return HomeView();
      else
        return TermsView();
    });
  }
}

这里方法来获取当前用户

Future getCurrentUser() async {
FirebaseUser _user ;
_user = await FirebaseAuth.instance.currentUser();

return _user ;
}

有没有办法在Consumer内部检查登录？