Flutter Bloc Rx dart combineLatest2 combine 功能未运行

Question

我正在编写一个颤振应用程序并使用 bloc 库。我有一个 bloc 和一个 cubit，在每个状态中都有一个我需要从 firestore 获取的其他一些文档的 id 列表。可能有一些重叠，并且已经获取了一些文档，所以我想从两个状态中获取 id 列表，比较它们，然后只去 firestore 获取那些存在于一个状态但没有另一个状态的 id。

我为此设置了一个新的肘：

class CircleRecipesCubit extends Cubit<CircleRecipesState> {
  CircleRecipesCubit({
    @required RecipesBloc recipesBloc,
    @required CirclesCubit circlesCubit,
  })  : assert(
          recipesBloc != null,
          circlesCubit != null,
        ),
        _recipesBloc = recipesBloc,
        _circlesCubit = circlesCubit,
        super(CircleRecipesInitial());

  final RecipesBloc _recipesBloc;
  final CirclesCubit _circlesCubit;

  StreamSubscription _recipesSubscription;
  StreamSubscription _circlesSubscription;

  Future<void> getCircleRecipes() async {
    // get a list of recipes the user already has loaded
    List<String> userRecipesIds;
    _recipesSubscription = _recipesBloc.stream.listen((RecipesState event) {
      if (event is RecipesLoaded) {
        userRecipesIds = event.recipes.map((e) => e.id).toList();
        print('*');
        print(userRecipesIds);
        print('*');
      }
    });

    // get a list of recipes in the circles
    List<String> circleRecipeIds;
    _circlesSubscription = _circlesCubit.stream.listen((CirclesState event) {
      if (event is CirclesLoaded) {
        circleRecipeIds = event.circles.fold([],
            (previousValue, element) => [...previousValue, ...element.recipes]);

        print('|');
        print(circleRecipeIds);
        print('|');
        // List<String> circleOnlyRecipeIds = circleRecipeIds;
        // circleRecipeIds.removeWhere((e) => userRecipesIds.contains(e));
        // print(circleOnlyRecipeIds);
      }
    });

    // reduce the list of recipes to a set of only circle recipes
    //TODO

    //------- Try with RX dart

    Stream<RecipesState> recipesStream = _recipesBloc.stream;
    Stream<CirclesState> circlesStream = _circlesCubit.stream;
    Rx.combineLatest2(recipesStream, circlesStream, (
      RecipesState recipesState,
      CirclesState circlesState,
    ) {
      print("This doesn't print!");
      print(recipesState);
      print(circlesState);

      if (recipesState is RecipesLoaded) {
        userRecipesIds = recipesState.recipes.map((e) => e.id).toList();
        print('*');
        print(userRecipesIds);
        print('*');
      }
      if (circlesState is CirclesLoaded) {
        circleRecipeIds = circlesState.circles.fold([],
            (previousValue, element) => [...previousValue, ...element.recipes]);

        print('|');
        print(circleRecipeIds);
        print('|');
        // List<String> circleOnlyRecipeIds = circleRecipeIds;
        // circleRecipeIds.removeWhere((e) => userRecipesIds.contains(e));
        // print(circleOnlyRecipeIds);

      }
      // fetch the set of recipes
    });
  }

  @override
  Future<void> close() {
    _recipesSubscription.cancel();
    _circlesSubscription.cancel();
    return super.close();
  }
}

所以上面是我的cubit - 它听recipesBloc 和circlesCubit。 getCiricleRecipes() 函数中的前两个表达式只是为了证明它的连接正确 - 当它运行这些打印语句时，会打印我希望它从另一个 bloc 和另一个 cubit 得到的 id。

我需要两者的最新值同时比较它们 - 所以我认为 rx.combinelatest2 会很好。我给它从集团和腕尺的溪流。但是即使事情看起来“连线”正确，组合器功能甚至都不会运行。

非常感谢任何帮助。

Answer 1

确保两个流都已发出至少一项。

combineLatest documentation 状态：

在所有流都发出至少一项之前，流不会发出。

由于第一个区块（您订阅_circlesCubit）打印出来，那么_recipesBloc 很可能是这里的罪魁祸首。