Flutter 如何将 void 函数放入单独的 dart 文件的单独页面中

下面是你如何做到这一点的，就像这样:

正如我所 comments 的，创建一个接受Function的separate StatefulWidget:

class Display extends StatefulWidget {
  final String selected;
  final Function onSelected;
  const Display({Key? key, required this.selected, required this.onSelected})
      : super(key: key);

  @override
  State<Display> createState() => _DisplayState();
}

class _DisplayState extends State<Display> {
  // --> Using `late` since we're initializing it later on - in initState.
  late Widget tacOpDisplayLeft;

  void display(String selected) {
    switch (selected) {
      case 'option 1':
        tacOpDisplayLeft = Container(
          child: Column(
            children: [
              Button3(
                onTap: (val) {
                  widget.onSelected(val);
                },
              ),
              Button5(
                onTap: (val) {
                  widget.onSelected(val);
                },
              ),
            ],
          ),
        );
        break;

      default:
        tacOpDisplayLeft = Container();
        break;
    }
  }

  @override
  void initState() {
    tacOpDisplayLeft = Container();
    super.initState();
  }

  @override
  Widget build(BuildContext context) {
    return tacOpDisplayLeft;
  }
}

现在，您可以在separate个类中调用它:

class MyApp extends StatefulWidget {
  const MyApp({Key? key}) : super(key: key);

  @override
  State<MyApp> createState() => _MyAppState();
}

class _MyAppState extends State<MyApp> {
  var _tacOpSelected = 'option 1';
  @override
  Widget build(BuildContext context) {
    return Display(
      selected: 'option 1',
      onSelected: (val) {
        setState(() {
          _tacOpSelected = val;
        });
      },
    );
  }
}