Google Sign-In With Flutter: Error Code -4 - ios

Question

I currently try to implement google_sign_in package in Flutter (https://pub.dartlang.org/packages/google_sign_in).
For this, I followed the example of their repository (https://github.com/flutter/plugins/blob/master/packages/google_sign_in/lib/google_sign_in.dart).
In that example in "initState" is a call signInSilently.
#override
void initState() {
super.initState();
_googleSignIn.onCurrentUserChanged.listen((GoogleSignInAccount account) {
setState(() {
_currentUser = account;
loggedIn = true;
});
});
loggedIn = false;
_googleSignIn.signInSilently();
}
I tried this code in iOS. On my first App Start, it worked well. But since I logged out I get an error here all the time I restart my app.It is the following PlatformException:
PlatformException(sign_in_required, com.google.GIDSignIn, The operation couldn’t be completed. (com.google.GIDSignIn error -4.))
I found in question Google Sign-In Error -4 that the error code is because of a missing Auth in Keychain.
The solution while swift programming is to call the method * hasAuthInKeychain* before the try to signInSilently. My problem is that the GoogleSignIn class in the flutter package has no function named like this.
Is there another call I need to run with this package to be sure I can try a silent log in? Or am I doing something wrong to get this message or is there even the possibility of catching this error?
Edit
I tried Marcel's solution, too. Somehow it is not catching the PlatfromException.
I do not know if this will help: signInSilently() is calling a method in which there is a the following call (google_sign_in.dart, line 217):
await channel.invokeMethod(method)
In platform_channel.dart there is a call
codec.decodeEnvelope(result);
The platform exception gets thrown in here.
if (errorCode is String && (errorMessage == null || errorMessage is String) && !buffer.hasRemaining)
throw PlatformException(code: errorCode, message: errorMessage, details: errorDetails);
else
throw const FormatException('Invalid envelope');
Edit 2
Since I just run my app and not started it in debug mode it somehow works again without throwing an exception. I do not know how this affects the code and why I got this exception. I can also run the code in debug mode again.
Since then I had the exception once again. Again I restarted android studio and runned the application once without debug mode.

Answer 1

You could just check if the sign in failed by handling the PlatformException like this:
void _setUpGoogleSignIn() async {
try {
final account = await _googleSignIn.signInSilently();
print("Successfully signed in as ${account.displayName}.");
} on PlatformException catch (e) {
// User not signed in yet. Do something appropriate.
print("The user is not signed in yet. Asking to sign in.");
_googleSignIn.signIn();
}
}

Answer 2

This is one way to catch the error and run _googleSignIn.signIn();
GoogleSignInAccount googleSignInAccount = await googleSignIn
.signInSilently(suppressErrors: false)
.catchError((dynamic error) async {
GoogleSignInAccount googleSignInAccount =
await _googleSignIn.signIn();
});

Answer 3

In my case, I did not want the user to see the login window automatically. In this case I changed from signIn to signOut. This way, I send the user to another view with an explanatory message and a login button.
GoogleSignInAccount googleSignInAccount = await googleSignIn
.signInSilently(suppressErrors: false)
.catchError((dynamic error) async {
GoogleSignInAccount googleSignInAccount = await _googleSignIn.signOut();
return googleSignInAccount;
});