【发布时间】:2019-04-16 14:55:01
【问题描述】:
我有一个包含 ListView 的有状态小部件 MyList。在点击任何列表图块时,新屏幕(MyScreen)打开,返回后(通过按左上角的返回按钮)滚动位置丢失,因为 MyList 被释放并且 initState 再次运行。如何防止 MyList 被销毁?
class MyList extends StatefulWidget {
@override
_MyListState createState() => new _MyListState();
}
class _MyListState extends State<MyList> {
List<String> items = new List.generate(20, (index) => 'Hello $index');
@override
Widget build(BuildContext context) {
return new Scaffold(
body: new Scrollbar(
child: new ListView.builder(
itemBuilder: (context, index) {
return new ListTile(
title: Text(items[index] + ' index $index'),
onTap: () {
Navigator.push(
context,
new MaterialPageRoute(
builder: (BuildContext context) => new MyScreen(index),
));
},
);
},
itemCount: items.length,
),
),
);
}
}
这是我的屏幕:
class MyScreen extends StatefulWidget {
final int indx;
MyScreen(this.indx);
_TaskDetailState createState() => new _TaskDetailState();
}
class _TaskDetailState extends State<MyScreen> {
@override
void initState() {
super.initState();
}
Widget build(context) {
return Scaffold(
appBar: AppBar(
elevation: 0.0,
title: Text('yoba'),
),
body: Text('yoba ${widget.indx}'),
);
}
}
【问题讨论】:
标签: flutter