flutter - 导航器使用 pushNamed 传递参数

Question

之前可能有人问过，但我找不到，但你如何将参数传递给命名路由？

这就是我建立路线的方式

Widget build(BuildContext context) {
    return new Navigator(
      initialRoute: 'home/chooseroom',
      onGenerateRoute: (RouteSettings settings) {
        WidgetBuilder builder;
        switch (settings.name) {
          case 'home/chooseroom':
            // navigates to 'signup/choose_credentials'.
            builder = (BuildContext _) => new ChoosePage();
            break;
          case 'home/createpage':
            builder = (BuildContext _) => new CreateRoomPage();
            break;
          case 'home/presentation':
            builder = (BuildContext _) => new Presentation();
            break;
          default:
            throw new Exception('Invalid route: ${settings.name}');
        }
        return new MaterialPageRoute(builder: builder, settings: settings);
      },
    );

这就是你的称呼 Navigator.of(context).pushNamed('home/presentation')

但是如果我的小部件是new Presentation(arg1, arg2, arg3)怎么办？

标签： flutter

Answer 1

不需要onGenerateRoute。只需使用

var exampleArgument = 'example string';

Navigator.pushNamed(
    context,
    '/otherscreen',
    arguments: {'exampleArgument': exampleArgument},
);

并提取参数如下：

@override
Widget build(BuildContext context) {
    final arguments = (ModalRoute.of(context)?.settings.arguments ?? <String, dynamic>{}) as Map;

    print(arguments['exampleArgument']);

    return Scaffold(...);
}