Question

这是我的代码，我不知道如何将此listOfTiles传递给我的StatefulWidget，您能帮我描述一下它的工作原理吗？

        body: new ListView.builder(
          itemBuilder: (BuildContext context, int index) {
            return new StuffInTilesState(listOfTiles[index]);//i want to pass this
          },
          itemCount: listOfTiles.length,
        ),
      ),
    );
  }
}
 
class StuffInTiles extends StatefulWidget{
  @override
  StuffInTilesState createState() => StuffInTilesState();//i know i need to change this, but i dont know how
}
 
class StuffInTilesState extends State<StuffInTiles> {
  final MyTile myTile;
  StuffInTilesState(this.myTile);//this is constructor, shuld this also be changed?
  final _controller = TextEditingController();
  String name = "";

如果您想查看我的工作代码：https://pastebin.pl/view/c4dbc2af如果您想看到我的工作代码：https://pastebin.pl/view/83f9cad0（https://codeshare.io/GLLm66）

Answer 1

您需要使用窗口小部件类的构造函数，而不是状态类。您可以使用 widget.YouProperty

访问状态类中的值

class StuffInTiles extends StatefulWidget {
  final MyTile myTile;

  const StuffInTiles(this.myTile);
  
  @override
  _StuffInTilesState createState() => _StuffInTilesState();
}

class _StuffInTilesState extends State<StuffInTiles> {
  @override
  Widget build(BuildContext context) {
    return Container(child: 
      Text(widget.myTile),);
  }
}

Flutter，如何将参数传递给构造函数

1 个答案: