android - 在继续下一页之前强制用户登录

Question

美好的一天，我正在创建一个应用程序，要求用户使用Google登录。我已经完成了一个教程(使用Firebase)，并成功地设法使登录系统正常工作。

我的问题:用户可以通过单击“使用Google登录”，然后单击屏幕侧面以退出登录来绕过登录。



假设:在login-page.dart文件中:我的_signInButton()函数onPressed()在断言用户已登录时不正确。

需要帮助:如果用户未登录，请重写/修改我的当前代码以将用户发送回主屏幕。仅当用户通过身份验证后，才允许用户继续操作。

login-page.dart文件:

class LoginPage extends StatefulWidget {
  @override
  _LoginPageState createState() => _LoginPageState();
}

class _LoginPageState extends State<LoginPage> {
  @override
  Widget build(BuildContext context) {
    return Scaffold(
      body: Container(
        color: Colors.white,
        child: Center(
          child: Column(
            mainAxisSize: MainAxisSize.max,
            mainAxisAlignment: MainAxisAlignment.center,
            children: <Widget>[
              Image(image: AssetImage("images/localhourlogo.png"), height: 80.0),
              SizedBox(height: 50),
              _signInButton(),
            ],
          ),
        ),
      ),
    );
  }

  Widget _signInButton() {
    return OutlineButton(
      splashColor: Colors.grey,
        onPressed: () {                           //****problem area?*****
          signInWithGoogle().whenComplete(() {
            Navigator.of(context).push(
              MaterialPageRoute(
                builder: (context) {
                  return MyTabs();
                },
              ),
            );
          }
        );
      },
      shape: RoundedRectangleBorder(borderRadius: BorderRadius.circular(40)),
      highlightElevation: 0,
      borderSide: BorderSide(color: Colors.grey),
      child: Padding(
        padding: const EdgeInsets.fromLTRB(0, 10, 0, 10),
        child: Row(
          mainAxisSize: MainAxisSize.min,
          mainAxisAlignment: MainAxisAlignment.center,
          children: <Widget>[
            Image(image: AssetImage("images/google_logo.png"), height: 35.0),
            Padding(
              padding: const EdgeInsets.only(left: 10),
              child: Text(
                'Sign in with Google',
                style: TextStyle(
                  fontSize: 20,
                  color: Colors.grey,
                ),
              ),
            )
          ],
        ),
      ),
    );
  }
}

sign-in.dart文件:

final FirebaseAuth _auth = FirebaseAuth.instance;
final GoogleSignIn googleSignIn = GoogleSignIn();

Future<String> signInWithGoogle() async {
  final GoogleSignInAccount googleSignInAccount = await googleSignIn.signIn();
  final GoogleSignInAuthentication googleSignInAuthentication =
  await googleSignInAccount.authentication;

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

  final AuthResult authResult = await _auth.signInWithCredential(credential);
  final FirebaseUser user = authResult.user;

  assert(!user.isAnonymous);
  assert(await user.getIdToken() != null);

  final FirebaseUser currentUser = await _auth.currentUser();
  assert(user.uid == currentUser.uid);

  return 'signInWithGoogle succeeded: $user';
}

void signOutGoogle(context) async{
  await googleSignIn.signOut();
  Navigator.popUntil(context, ModalRoute.withName("/"));
  print("User Sign Out");
}

附加信息:
这是我遵循的教程:https://medium.com/flutter-community/flutter-implementing-google-sign-in-71888bca24ed

非常感谢您提供的所有帮助。

Answer 1

您需要稍微重构代码。在signInWithGoogle()返回字符串是检查登录是否成功的一种不好的方式。但是，现在，我将使代码相等，仅对其进行一些更改以修复错误。在您的实际代码中，您总是返回一个字符串“succeeded”，并且您没有检查是否发生了问题。因此，我添加了一个try catch bloc。

try {
    GoogleSignInAccount user = await googleSignIn.signIn();
    //The rest of your code
   return 'Succeed';
} catch (error) {
    return 'Error';
}

在login-page.dart处，我在onPressed()处更改_signInButton():

onPressed: () async{
    String result = await signInWithtGoogle();
    if(result == 'Succeed'){
        //Send the user to the tab page like you did
    }else{
        //Reload login page
    }
}