@override
Widget build(BuildContext context) {
return Scaffold(
appBar: AppBar(
title: Text("Home Page"),
),
body: FutureBuilder<HomePageStuffResponse>(
future: getHomePageStuff(),
builder: (context, snap) {
if(snap.hasError) {
return ErrorWidget('Error occurred while fetching data');
}
if(snap.hasData) {
return Center(
child: RaisedButton(
onPressed: () {},
child: Text('Go back!'),
),
);
}
}
),
);
}
}
Future<HomePageStuffResponse> getHomePageStuff() async {
Tokens token = await DBProvider.db.getToken();
//Accessing the token here throws an NPE
var accessToken = token.accessToken;
debugPrint("token = " + accessToken);
final response = await http.get(..);
if (response.statusCode == 200) {
debugPrint("FETCH SUCCESS");
return stuff;
} else {
throw Exception('Failed to fetch home page stuff');
}
}
我已经看到了对字典的每个列表求和的一种pythonic方法,但是有一种对字典中的列表的所有对应元素求和的pythonic方法吗?
dict = {1:[1,2,3,4],2:[5,5,5,3],3:[5,6,7,8]}
目前,我正在使用double for循环。