将数据从另一个 class 导入到 Stateful Widget

Question

我在一个屏幕上对用户进行身份验证，然后将他们推送到另一个屏幕，但在另一个屏幕上，变量不可访问，其中包含我通过变量传递的用户详细信息。 在 Home() 中无法访问detailsUser ； 即使我已经提到要推送另一页。 它进行身份验证并且不会将我带到另一个屏幕。

class SignInScreen extends StatefulWidget {
  @override
  _SignInScreenState createState() => _SignInScreenState();
}

class _SignInScreenState extends State<SignInScreen> {

  final FirebaseAuth _firebaseAuth = FirebaseAuth.instance;
  final GoogleSignIn _googleSignIn = GoogleSignIn();

  Future<FirebaseUser> _signIn(BuildContext context) async{
    Fluttertoast.showToast(
        msg: 'Signed In'
    );

    final GoogleSignInAccount googleUser = await _googleSignIn.signIn();
    final GoogleSignInAuthentication googleAuth = await googleUser.authentication;

    final AuthCredential credential = GoogleAuthProvider.getCredential(
        idToken: googleAuth.accessToken,
        accessToken: googleAuth.idToken,
    );

    FirebaseUser userDetails = (await _firebaseAuth.signInWithCredential(credential)) as FirebaseUser;
    ProviderDetails providerInfo = new ProviderDetails(userDetails.providerId);

    List<ProviderDetails> providerData = List <ProviderDetails>();
    providerData.add(providerInfo);

    UserDetails details = UserDetails(
      userDetails.providerId,
      userDetails.displayName,
      userDetails.photoUrl,
      userDetails.email,
      providerData);

    Navigator.push(context, MaterialPageRoute(builder: (context) => ThemeConsumer(child: Home(detailsUser: details))));

    return userDetails;
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold();

class Home extends StatefulWidget {
  final UserDetails detailsUser;

  Home({Key key, @required this.detailsUser}) : super(key: key);

  @override
  _HomeState createState() => _HomeState();
}

class _HomeState extends State<Home> {

  @override
  Widget build(BuildContext context) {
    return Scaffold(
     body: Center(
     child: Text(
     detailsUser.username //detailsUser is not accessible here, why?
)));

Answer 1

要从该小部件的State class访问Widget字段，您可以使用widget.fieldName （正如 Patel Pinkal 评论的那样）

这是一个更熟悉 flutter 中的小部件的链接

Answer 2

正如 Patel Pinkal 在评论中所说：

Use this "widget.detailsUser.username" 

代替

"detailsUser.username"