如何在StatelessWidget的任何函数中获取上下文?

问题描述 投票:0回答:1

我们希望在某些异步处理(例如网络进程)之后显示AlertDialog。

当从外部类调用'showAlertDialog()'时,我想在没有上下文的情况下调用它。有什么好方法吗?

class SplashPage extends StatelessWidget implements SplashView {
  BuildContext _context;
  @override
  Widget build(BuildContext context) {
    this._context = context;
    ...
  }

我已经考虑过上述方法,但是我担心附带问题。

帮助

我当前的代码

class SplashPage extends StatelessWidget implements SplashView {

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      backgroundColor: MyStoreColors.eats_white1_ffffff,
      body: Center(
        child: new SvgPicture.asset('assets/ic_splash.svg'),
      ),
    );
  }

  @override
  void showAlertDialog() {

    showDialog<void>(
      context: /*How to get context?*/,
      builder: (BuildContext context) {
        return AlertDialog(
          title: Text('Not in stock'),
          content: const Text('This item is no longer available'),
          actions: <Widget>[
            FlatButton(
              child: Text('Ok'),
              onPressed: () {
                Navigator.of(context).pop();
              },
            ),
          ],
        );
      },
    );
  }

  @override
  void moveToHomeContainer() {
  }

  @override
  void moveToLoginContainer() {
  }
}

flutter statelesswidget flutter-alertdialog
1个回答
0
投票

您应在异步事件完成后触发重建,将小部件转换为StatefulWidget并调用setState()或使用诸如Bloc之类的状态管理解决方案。

例如,使用`StatefulWidget,您的代码将如下所示:

class SplashPage extends StatefulWidget {
  @override
  State<SplashPage> createState() => _SplashPageState();
}

class _SplashPageState extends State<SplashPage> implements SplashView {

  bool _asynOpDone = false;

  ///
  void _onAsynOpDone() => setState(() => _asyncOpDone = true);

  @override
  Widget build(BuildContext context) {
    if (_asyncOpDone) showAlertDialog(context);

    return Scaffold(
      ...,
      ///
    );
  }

  @override
  void showAlertDialog(BuildContext context) {
    showDialog<void>(
      context: context,
      builder: ...,
    );
  }
}
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