Flutter 类型“Future<dynamic>”不是“Stream<PostModel>?”类型的子类型

如何解决Flutter 类型“Future<dynamic>”不是“Stream<PostModel>?”类型的子类型

我正在尝试使用 bloc 和 rxdart 创建一个列表，但我收到一个错误类型“未来”不是“流”类型的子类型？我正在使用 rxdart：^0.27.1。只想将数组响应放入列表中。

我的列表屏幕

 Column(
      children: [
        StreamBuilder(
          stream: bloc.fetchPosts(),builder: (context,AsyncSnapshot<PostModel> snapshot) {
            if (snapshot.hasData) {
              return buildList(snapshot);
            } else if (snapshot.hasError) {
              return Text(snapshot.error.toString());
            }
            return Center(child: CircularProgressIndicator());
          },),],

PostBloc 集团 使用流时可能出现错误，但我找不到错误。

 class PostBloc {
    final _repository = PostRepository();
    final _postsFetcher = PublishSubject<PostModel>();

    Stream<PostModel> get allPosts => _postsFetcher.stream;

    fetchPosts() async {
    PostModel itemModel = await _repository.fetchAllPosts();
    _postsFetcher.sink.add(itemModel);
     }

     dispose() {
    _postsFetcher.close();
     }
    }

final bloc = PostBloc();

帖子存储库

    class PostRepository {
    final postApiProvider = PostApiProvider();
    Future<PostModel> fetchAllPosts() => postApiProvider.getPostList();
    }

发布 ApiProvider

class PostApiProvider {
  Client client = Client();

  Future<PostModel> getPostList() async {
    print("entered");
    final response = await client
        .get(Uri.parse("https://jsonplaceholder.typicode.com/posts"));
    print(response.body.toString());
    if (response.statusCode == 200) {

      return PostModel.fromJson(json.decode(response.body));
    } else {
      // If that call was not successful,throw an error.
      throw Exception('Failed to load post');
    }
  }
}

发布模型

class Post {
  late int userId;
  late int id;
  late String title;
  late String body;

  Post(result) {
    userId = result['userId'];
    id = result['id'];
    title = result['title'];
    body = result['body'];
  }

  int get getUserID => userId;

  int get getID => id;

  String get getTitle => title;

  String get getBody => body;

}

class PostModel {
  List<Post> results = [];

  PostModel.fromJson(Map<String,dynamic> parsedJson) {
    List<Post> temp = [];
    for (int i = 0; i < parsedJson.length; i++) {
      Post result = Post(parsedJson[i]);
      temp.add(result);
    }
    results = temp;
  }

  List<Post> get getResults => results;
}

示例响应

[
    {
        "userId": 1,"id": 1,"title": "sunt aut facere repellat provident occaecati excepturi optio reprehenderit","body": "quia et suscipit\nsuscipit recusandae consequuntur expedita et cum\nreprehenderit molestiae ut ut quas totam\nnostrum rerum est autem sunt rem eveniet architecto"
    },{
        "userId": 1,"id": 2,"title": "qui est esse","body": "est rerum tempore vitae\nsequi sint nihil reprehenderit dolor beatae ea dolores neque\nfugiat blanditiis voluptate porro vel nihil molestiae ut reiciendis\nqui aperiam non debitis possimus qui neque nisi nulla"
    }
]

解决方法

BLoC 的基本概念是首先要借助 sink 将数据放入流中，然后借助 Stream 才能从流中获取数据。

在您的代码中，您将接收器函数用作 StreamBuilder's 流。这是不正确的。你应该像下面这样调用流，

  StreamBuilder(
      stream: bloc.allPosts,/// THIS IS STREAM OF DATA
      builder: (context,AsyncSnapshot<PostModel> snapshot) {
        if (snapshot.hasData) {
          return buildList(snapshot);
        } else if (snapshot.hasError) {
          return Text(snapshot.error.toString());
        }
        return Center(child: CircularProgressIndicator());
      },);

在 initState 或任何你想要的地方调用你的接收器函数

bloc.fetchPosts(); /// THIS WILL TRIGGER YOUR API

第二个问题是模型，它是关于数据类型的。您从 API 获得的响应是​​ List<dynamic>，并且您已声明 Map<String,dynamic>。这是不正确的。定义如下

class PostModel {
  List<Post> results = [];

  PostModel.fromJson(List<dynamic> parsedJson) {
    List<Post> temp = [];
    for (int i = 0; i < parsedJson.length; i++) {
      Post result = Post(parsedJson[i]);
      temp.add(result);
    }
    results = temp;
  }

  List<Post> get getResults => results;
} 

最后一件事，您忘记在 fetchPosts() 函数中返回模型

Future<PostModel> fetchPosts() async {
    PostModel itemModel = await _repository.fetchAllPosts();
    _postsFetcher.sink.add(itemModel);
    return itemModel;
  }