在处置方法中访问 BlocProvider.of<Bloc>(context)

Question

有人知道我该怎么做吗？

我的代码：

@override
void dispose() {
final FiltersBloc filtersBloc = 
BlocProvider.of<FiltersBloc>(context);
super.dispose();
}

错误是：

flutter:         BlocProvider.of() called with a context that does not contain a Bloc of type FiltersBloc.
flutter:         No ancestor could be found starting from the context that was passed to
flutter: BlocProvider.of<FiltersBloc>().
flutter:
flutter:         This can happen if:
flutter:         1. The context you used comes from a widget above the BlocProvider.
flutter:         2. You used MultiBlocProvider and didn't explicity provide the BlocProvider types.
flutter:
flutter:         Good: BlocProvider<FiltersBloc>(builder: (context) => FiltersBloc())
flutter:         Bad: BlocProvider(builder: (context) => FiltersBloc()).
flutter:
flutter:         The context used was: FiltersDrawer(dirty, state: _FiltersDrawerState#86e8a)

另外，如果我遵循错误代码并改用final filtersBloc = BlocProvider<FiltersBloc>(builder: (context) => FiltersBloc())，则无法再调用filtersBloc.dispatch()。

我知道对于 initState，我们可以改didChangeDependencies。但是我找不到等价物来处理。

任何帮助将不胜感激。谢谢！

Answer 1

BlockProvider 需要初始化上下文。您可以在屏幕上初始化它Widget build()

late FiltersBloc filtersBloc;
@override
Widget build(BuildContext context){
filtersBloc = BlocProvider.of<FiltersBloc>(context);
return ...
}

。然后在dispose()上调用所需的方法

@override
void dispose() {
filtersBloc.dispatch();
super.dispose();
}

Answer 2

一个更详细的小方法：

声音为零安全，您可以使用@Omatt答案。

不健全的空安全，可以使用以下方法：

在类中声明一个静态变量，并在init或build中分配它，如下所示：

class MyAwesomeWidget extends StatefulWidget {
const MyAwesomeWidget({
Key key,
});
@override
_MyAwesomeWidgetState createState() => _MyAwesomeWidgetState();
}
class _MyAwesomeWidgetState extends State<MyAwesomeWidget> {
FiltersBloc filtersBloc;
@override
void initState() {
super.initState();
new Future.delayed(Duration.zero, () {
filterBloc = BlocProvider.of<FiltersBloc>(context);
});
}
@override
void dispose() {
filterBloc.myFunction();
super.dispose();
}
}