如何从另一个屏幕在Flutter中重置表单？

例如，使用aGlobalKey<FormState>访问表单状态和reset()字段。考虑到这一点，您可以：

1. 将第二个屏幕的构造函数传递给第一个表单字段键，并同时重置两者（当您完成对服务器的数据处理后）

class Screen1 extends StatefulWidget {
  _Screen1State createState() => _Screen1State();
}

class _Screen1State extends State<Screen1> {
  final _formKeyScreen1 = GlobalKey<FormState>();

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text('First screen'),
      ),
      body: Center(
        child: Form(
          key: _formKeyScreen1,
          child: Column(
            children: <Widget>[
              TextFormField(),
              TextFormField(),
              TextFormField(),
              RaisedButton(
                child: Text('Navigate to second screen...'),
                onPressed: () => Navigator.of(context).push(MaterialPageRoute(
                      builder: (BuildContext context) => Screen2(_formKeyScreen1),
                    )),
              )
            ],
          ),
        ),
      ),
    );
  }
}

class Screen2 extends StatefulWidget {
  final GlobalKey<FormState> firstScreenFormKey;
  Screen2(this.firstScreenFormKey);
  _Screen2State createState() => _Screen2State();
}

class _Screen2State extends State<Screen2> {
  final _formKeyScreen2 = GlobalKey<FormState>();

  void _processData() {
    // Process your data and upload to server
    _formKeyScreen2.currentState?.reset();
    widget.firstScreenFormKey?.currentState?.reset();
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text('Second screen'),
      ),
      body: Center(
        child: Form(
          key: _formKeyScreen2,
          child: Column(
            children: <Widget>[
              TextFormField(),
              TextFormField(),
              TextFormField(),
              RaisedButton(
                child: Text('Submit data'),
                onPressed: () => _processData(),
              )
            ],
          ),
        ),
      ),
    );
  }
}

2. 或者，当第二个屏幕从堆栈中弹出时，您可以重置第一个屏幕窗体，这应该是IMO，因为我相信您不是在处理后从第二个屏幕访问第一个屏幕窗体数据，因此，目的，返回时重置就足够了。在这种情况下，您将获得以下结果

class Screen1 extends StatefulWidget {
  _Screen1State createState() => _Screen1State();
}

class _Screen1State extends State<Screen1> {
  final _formKeyScreen1 = GlobalKey<FormState>();

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text('First screen'),
      ),
      body: Center(
        child: Form(
          key: _formKeyScreen1,
          child: Column(
            children: <Widget>[
              TextFormField(),
              TextFormField(),
              TextFormField(),
              RaisedButton(
                child: Text('Navigate to second screen...'),
                onPressed: () {
                  Navigator.of(context)
                      .push(MaterialPageRoute(
                        builder: (BuildContext context) => Screen2(),
                      ))
                      .then((_) => _formKeyScreen1.currentState.reset());
                },
              )
            ],
          ),
        ),
      ),
    );
  }
}

class Screen2 extends StatefulWidget {
  _Screen2State createState() => _Screen2State();
}

class _Screen2State extends State<Screen2> {
  final _formKeyScreen2 = GlobalKey<FormState>();

  void _processData() {
    // Process your data and upload to server
    _formKeyScreen2.currentState?.reset();
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text('Second screen'),
      ),
      body: Center(
        child: Form(
          key: _formKeyScreen2,
          child: Column(
            children: <Widget>[
              TextFormField(),
              TextFormField(),
              TextFormField(),
              RaisedButton(
                child: Text('Submit data'),
                onPressed: () => _processData(),
              )
            ],
          ),
        ),
      ),
    );
  }
}