在flutter中尝试使用widget.string访问字符串时获取null值

Question

我正在尝试获取使用电子邮件和密码登录的firebase auth异常。在我的应用程序中，所有的身份验证服务都在另一个文件AuthService.dart中的AuthService类中，我想将异常从那里转移到另一个文件中的logingreen类

这是我的authservice.dart文件中的类

异常在这里起作用，我可以将其打印到控制台。

class AuthService {
//  final FirebaseAuth _auth = FirebaseAuth.instance;
//  UserModel _userFromFirebaseUser(User user) {
//    return user != null ? UserModel(uid: user.uid) : null;
}
// Stream<UserModel> get user {
//   return _auth.authStateChanges().map(_userFromFirebaseUser);
// }

Future loginWithEmailpasswd(String email, String password) async {
try {
return await _auth.signInWithEmailAndPassword(
email: email, password: password);
} on FirebaseAuthException catch (e) {
print(e.toString());
StudentLoginScreen(
message: e.toString(), // passing  the exception
);
}
}

这是我的登录绿色

class StudentLoginScreen extends StatefulWidget {
final Function toggleView;
final String message;
StudentLoginScreen({this.toggleView, this.message});
@override
_StudentLoginScreenState createState() => _StudentLoginScreenState();
}
final _formkey = GlobalKey<FormState>();
class _StudentLoginScreenState extends State<StudentLoginScreen> {
String email = '';
String password = '';
String _message;
final AuthService _authService = AuthService();
//SharedPreferences usertype;
@override
Widget build(BuildContext context) {
Size size = MediaQuery.of(context).size;
return Scaffold(
backgroundColor: HexColor(studentPrimaryColour),
body: SafeArea(
child: SingleChildScrollView(
child: Form(
key: _formkey,
child: Column(
children: <Widget>[
showAlert(),       // i want to show the exception message in this widget
SizedBox(
height: 25.0,
),
// HeadingText(
//   text: 'Login',
//   size: 60.0,
//   color: Colors.white,
// ),
// RoundedInputField(
//   hintText: "Email",
//   validator: (val) =>
//       val.isEmpty ? 'Oops! you left this field empty' : null,
//   onChanged: (val) {
//     email = val;
//   },
// ),
// RoundedInputField(
//   hintText: "Password",
//   validator: (val) =>
//       val.isEmpty ? 'Oops! you left this field empty' : null,
//   boolean: true,
//   onChanged: (val) {
//     password = val;
//   },
// ),
Container(
child: FlatButton(
color: Colors.white,
onPressed: () async {
if (_formkey.currentState.validate()) {
dynamic result = await _authService
.loginWithEmailpasswd(email, password);
if (result != null) {
print('logged in');
} else {
print('error logging in');
setState(() {
_message = widget.message;
});
print(_message);          // its returning a null value
}
}
},
child: Text(
'login',
style: GoogleFonts.montserrat(
color: HexColor(studentPrimaryColour),
fontSize: 20),
),
),
),
),
),
],
),
),
),
),
);
}
Widget showAlert() {
if (_message != null) {  //nothing happens here since _message is always null even if an exception
return Container(      // is thrown in authservice
color: Colors.amber,
padding: EdgeInsets.all(8.0),
child: Row(
children: <Widget>[
Icon(Icons.error_outline_rounded),
Expanded(
child: AutoSizeText(
_message,
maxLines: 3,
))
],
),
);
}
return SizedBox(
height: 0,
);
}
}

我错过了什么重要的东西吗？我相对来说是个新手，我发现很难将值从一个类传递到另一个类。

Answer 1

尝试返回异常，而不是在屏幕中传递它

Future loginWithEmailpasswd(String email, String password) async {
try {
return await _auth.signInWithEmailAndPassword(
email: email, password: password);
} on FirebaseAuthException catch (e) {
print(e.toString());
return e;
}

//屏幕登录

onPressed: () async {
if (_formkey.currentState.validate()) {
dynamic result = await _authService
.loginWithEmailpasswd(email, password);
if (result is bool) {

if (result) {
print('logged in');
}}
else {
print('error logging in');
setState(() {
_message =result;
});
print(result);        
}
}
},