Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
menu search
person
Welcome To Ask or Share your Answers For Others

Categories

Is it possible to fix an object in the screen that stays fixed regardless of scrolling?

Something similar to CSS position fixed.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
1.5k views
Welcome To Ask or Share your Answers For Others

1 Answer

You can absolutely position a child of a Stack widget using the Positioned widget.

The minimal example below places the red box above the list view, by placing the child in a Positioned widget after the ListView in the Stack's children.

List<String> todos = [...];
return new Stack(
  children: <Widget>[
    new ListView(
     children: todos
       .map((todo) => new ListTile(title: new Text(todo)))
       .toList(),
     ),
     new Positioned(
       left: 30.0,
       top: 30.0,
       child: new Container(
         width: 100.0,
         height: 80.0,
         decoration: new BoxDecoration(color: Colors.red),
         child: new Text('hello'),
        )
      ),
   ],
);

And here it is inside of a Scaffold body. If you add more items you'll find that the list scrolls without moving the red box.

an absolutely positioned example


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
...