'flutter dynamic links open web page when app is not installed

So, I want to redirect user to my web app created in flutter with specific item id when user don't have mobile application installed

Here is my code for create dynamic link url:

Future createDynamicLink(
BuildContext context,
String title,
String image,
String postId,
) async {
bool short = false;
final DynamicLinkParameters parameters = DynamicLinkParameters(
uriPrefix: 'https://helpmechoose.page.link',
link: Uri.parse('https://helpmechoose.page.link/dynamiclink/?id=$postId'),
androidParameters: AndroidParameters(
  packageName: 'com.app.helpmechoose',
  minimumVersion: 0,
),
socialMetaTagParameters:
    SocialMetaTagParameters(title: title, imageUrl: Uri.parse(image)),
dynamicLinkParametersOptions: DynamicLinkParametersOptions(
  shortDynamicLinkPathLength: ShortDynamicLinkPathLength.short,
),
iosParameters: IosParameters(
  bundleId: 'com.app.ioshelpmechoose',
  minimumVersion: '0',
  ),
);

 Uri url;
 if (short) {
  final ShortDynamicLink shortLink = await parameters.buildShortLink();
url = shortLink.shortUrl;
} else {
Uri longLink = await parameters.buildUrl();
final ShortDynamicLink shortDynamicLink =
    await DynamicLinkParameters.shortenUrl(
        Uri.parse(longLink.toString() + "&ofl=https://helpmechoose.app/"));
url = shortDynamicLink.shortUrl;
print(url);
}

return url.toString();
}


Solution 1:[1]

just add => fallbackUrl: Uri.parse(''),

androidParameters: AndroidParameters(
  packageName: 'com.app.helpmechoose',
  minimumVersion: 0,
  fallbackUrl: Uri.parse('https://helpmechoose.page.link/dynamiclink/?id=$postId'),
),

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 HamZa