Hey, the best approach would be to open the widget in the designated browser. Whilst it is possible to render a webview, this is not ideal as some providers like Google may reject authentication.
To open the url in swift:
if let url = URL(string: urlSting), UIApplication.shared.canOpenURL(url) {
UIApplication.shared.open(url)
}
In kotlin:
Intent browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
startActivity(browserIntent);
In react native:
Linking.canOpenURL(url).then(supported => {
if (supported) {
Linking.openURL(url);
}
}
And in flutter:
final Uri url = Uri.parse(url);
await launchUrl(url)