使用StreamProvider的Flutter提供者MultiProvider

时间:2019-07-31 02:11:49

标签: flutter stream provider flutter-provider

我正在使用提供程序(提供者:3.0.0 + 1)开发Flutter应用。我正在将MultiProvider与控制器一起使用。但是我总是遇到错误。

下面是我的代码

main.dart

Widget build(BuildContext context) {
return MultiProvider(
  providers: [
    StreamProvider<RecipeStreamService>.value(value: RecipeStreamService().controllerOut)
  ],
  child: MaterialApp(
        debugShowCheckedModeBanner: false,
        title: 'Home Food',
        routes: {
          '/register': (BuildContext context) => RegisterPage(),
          '/login': (BuildContext context) => LoginPage()
        },
        theme: ThemeData(
          primaryColor: Colors.lightBlue[300],
          accentColor: Colors.green[300],
          textTheme: TextTheme(
            headline: TextStyle(fontSize: 42.0, fontWeight: FontWeight.bold, color: Colors.black54),
            title: TextStyle(fontSize: 22.0, fontStyle: FontStyle.italic),
            body1: TextStyle(fontSize: 18.0),
            button: TextStyle(fontSize: 20.0,fontWeight: FontWeight.normal, color: Colors.white)
          )
        ),
        home: HomePage(title: 'Home'),
      ),
  );
}

RecipeStreamService.dart

class RecipeStreamService {
   List<Recipe> _recipes = <Recipe>[];

   final _controller = StreamController<List<Recipe>>.broadcast();
   get controllerOut => _controller.stream;
   get controllerIn => _controller.sink;

   RecipeStreamService() {
      getRecipes();
   }

   addNewRecipe(Recipe recipe) {
     _recipes.add(recipe);
    controllerIn.add(_recipes);
  }

  getRecipes() async{
     List<Map<String, dynamic>> result = await ApiService().getRecipes();
     List<Recipe> data = result.map((data) => Recipe.fromMap(data)).toList();
     data.map((f) => addNewRecipe(f));
 }

 void dispose() {
   _controller.close();
 }
}

但是我总是收到此错误:

type '_BroadcastStream<List<Recipe>>' is not a subtype of type 'Stream<RecipeStreamService>'
I/flutter (16880): When the exception was thrown, this was the stack:
I/flutter (16880): #0      MyApp.build (package:app_recipe/main.dart:20:80)

行:main.dart中的20:80是(RecipeStreamService()。controllerOut)

******更新******

将Multiprovider更改为以下代码

 providers: [
    StreamProvider<List<Recipe>>.value(value: RecipeStreamService().controllerOut)
  ],

也可以在HomePage.dart中使用它

final recipeService = Provider.of<List<Recipe>>(context);

现在,配方服务总是以空值出现

谢谢

1 个答案:

答案 0 :(得分:0)

看到错误的原因是,由于StreamProvider<T>传递给T type参数与您返回{{ 1}}

在这种情况下,您要流式传输create 类型的值。
因此,您还应该将其传递给通用类型:

List<Recipe>

StreamProvider<List<Recipe>>.value(value: stream) 是您的stream流。


如果List<Recipe>为您返回Provider.of<List<Recipe>>(context),则表示您尚未向流中添加值。

如果要在流发出值之前使用null以外的其他内容,则可以传递null

initialValue