BlocProvider.of() 使用不包含 CLASS 类型的 Bloc 的上下文调用

Question

在 flutter 中，我刚刚学习如何在应用程序上使用 Bloc，我想尝试使用此功能实现简单的登录。在实现 bloc 的 class 之后在视图

上使用它

当我尝试将此代码用作

时出现错误

BlocProvider.of<LoginListingBloc>(context).dispatch(LoginEvent(loginInfoModel: testLogin));

里面RaisedButton

错误：

BlocProvider.of() called with a context that does not contain a Bloc of type LoginListingBloc.

我的看法：

class _HomePageState extends State<HomePage> {
  LoginListingBloc _loginListingBloc;

  @override
  void initState() {
    super.initState();
    _loginListingBloc =
        LoginListingBloc(loginRepository: widget.loginRepository);
  }

  ...
  @override
  Widget build(BuildContext context) {
    return BlocProvider(
      bloc: _loginListingBloc,
      child: Scaffold(
        appBar: AppBar(
            elevation: 5.0, title: Text('Sample Code', style: appBarTextStyle)),
        body: Center(
          child: RaisedButton(
              child: Text(
                'click here',
                style: defaultButtonStyle,
              ),
              onPressed: () {
                BlocProvider.of<LoginListingBloc>(context).dispatch(LoginEvent(loginInfoModel: testLogin));
              }),
        ),
      ),
    );
  }
}

LoginListingBloc class:

class LoginListingBloc extends Bloc<LoginListingEvent, LoginListingStates> {
  final LoginRepository loginRepository;

  LoginListingBloc({this.loginRepository});

  @override
  LoginListingStates get initialState => LoginUninitializedState();

  @override
  Stream<LoginListingStates> mapEventToState(
      LoginListingStates currentState, LoginListingEvent event) async* {
    if (event is LoginEvent) {
      yield LoginFetchingState();
      try {
        final loginInfo = await loginRepository.fetchLoginToPage(
            event.loginInfoModel.username, event.loginInfoModel.password);
        yield LoginFetchedState(userInfo: loginInfo);
      } catch (_) {
        yield LoginErrorState();
      }
    }
  }
}

和其他 class 如果您想查看主题

AppApiProvider class:

class AppApiProvider {
  final successCode = 200;

  Future<UserInfo> fetchLoginToPage(String username, String password) async {
    final response = await http.get(Constants.url + "/api/v1/getPersons");
    final responseString = jsonDecode(response.body);
    if (response.statusCode == successCode) {
      print(responseString);
      return UserInfo.fromJson(responseString);
    } else {
      throw Exception('failed to get information');
    }
  }
}

LoginEvent:

class LoginEvent extends LoginListingEvent {
  final LoginInfoModel loginInfoModel;

  LoginEvent({@required this.loginInfoModel}) : assert(loginInfoModel != null);
}

LoginInfoModel:

class LoginInfoModel {
  String username;
  String password;

  LoginInfoModel({this.username, this.password});
}

final testLogin = LoginInfoModel(username:'exmaple',password:'text');

Answer 1

无需从 context 访问 loginListingBloc，因为它存在于当前 class 而不是小部件树中。

变化：

BlocProvider.of<LoginListingBloc>(context).dispatch(LoginEvent(loginInfoModel: testLogin));

至：

_loginListingBloc.dispatch(LoginEvent(loginInfoModel: testLogin));

Answer 2

对于所有来这里查看错误消息的人：确保您始终指定类型并且不要省略它们：

BlocProvider<YourBloc>(
    create: (context) => YourBloc()
    child: YourWidget()
);

还有

BlocProvider.of<YourBloc>(context);

BlocProvider.of() 使用不包含 CLASS 类型的 Bloc 的上下文调用

BlocProvider.of() called with a context that does not contain a Bloc of type CLASS

flutter

bloc