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)