Flutter：只有在登录成功后按下返回按钮时才会出现主屏幕

Question

提供者代码

  Future<void> login(String email, String password) async {
  }
    try {
      final response = await http.post(
        url,
        body: json.encode(
          {
            'email': email,
            'password': password,
            'returnSecureToken': true,
          },
        ),
      );
      final responseData = json.decode(response.body);

      if (responseData['error'] != null) {
        throw HttpException(responseData['error']['message']);
      }

      _token = responseData['idToken'];
      _userId = responseData['localId'];
      _expiryDate = DateTime.now().add(
        Duration(
          seconds: int.parse(
            responseData['expiresIn'],
          ),
        ),
      );
      notifyListeners();
    } catch (error) {
      throw error;
    }

主要的

class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return MultiProvider(
      providers: [
        ChangeNotifierProvider(
          create: (ctx) => Auth(),
        ),
      ],
      child: Consumer<Auth>(
        builder: (ctx, auth, _) => MaterialApp(
          title: 'Medicare',
          theme: ThemeData(
            primaryColor: kPrimaryColor,
            scaffoldBackgroundColor: Colors.white,
            appBarTheme: AppBarTheme(backgroundColor: Colors.white),
          ),
          home: auth.isAuth ? HomeScreen() : WelcomePage(),
          routes: {
            AuthScreen.routeName: (ctx) => AuthScreen(),
          },
        ),
      ),
    );
  }
}

欢迎屏幕

认证屏幕

主屏幕

当用户成功登录时，我对我的逻辑中缺少的内容感到困惑，除非我按下模拟器/真实设备的后退按钮，否则不会出现主屏幕。 我正在使用 state 管理和 firebase 后端的提供程序。

编辑：这是我的 function，用于带有错误处理程序的登录/注册用户。

    try {
      if (_authMode == AuthMode.Login) {
        // Login User
        await Provider.of<Auth>(context, listen: false).login(
          _authData['email'],
          _authData['password'],
        );
      } else {
        // Sign Up User
        await Provider.of<Auth>(context, listen: false).signup(
          _authData['email'],
          _authData['password'],
        );
      }
    } on HttpException catch (error) {
      var errorMessage = 'Authentication failed';
      if (error.toString().contains('EMAIL_EXISTS')) {
        errorMessage = 'This email address is already in use.';
      } else if (error.toString().contains('INVALID_EMAIL')) {
        errorMessage = 'This is not a valid email address';
      } else if (error.toString().contains('WEAK_PASSWORD')) {
        errorMessage = 'This password is too weak';
      } else if (error.toString().contains('EMAIL_NOT_FOUND')) {
        errorMessage = 'Could not find a user with that email';
      } else if (error.toString().contains('INVALID_PASSWORD')) {
        errorMessage = 'Invalid Password.';
      }
      _showErrorDialog(errorMessage);
    } catch (error) {
      const errorMessage =
          'Coult not authenticate you. Please try again later.';
      _showErrorDialog(errorMessage);
    }

我怀疑由于我所做的 httpexception，我无法推送一个上下文。

Answer 1

我没有深入了解 go 但我在大括号级别的代码中发现了错误。

在提供程序代码中 function 未来登录，以空括号结束！

检查您的代码

Answer 2

Future<void> login(String email, String password) async {

     // empty
     // empty
      // empty
      // empty

  }
    try {
      final response = await http.post(
        url,
        body: json.encode(
          {
            'email': email,
            'password': password,
            'returnSecureToken': true,
          },
        ),
      );
      final responseData = json.decode(response.body);

      if (responseData['error'] != null) {
        throw HttpException(responseData['error']['message']);
      }

      _token = responseData['idToken'];
      _userId = responseData['localId'];
      _expiryDate = DateTime.now().add(
        Duration(
          seconds: int.parse(
            responseData['expiresIn'],
          ),
        ),
      );
      notifyListeners();
    } catch (error) {
      throw error;
    }