未安装应用程序时颤振动态链接打开网页

Question

所以，当用户没有安装移动应用程序时，我想将用户重定向到我在 flutter 中创建的具有特定项目 ID 的网络应用程序

这是我创建动态的代码 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();
}

Answer 1

只需添加 => fallbackUrl: Uri.parse(''),

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

未安装应用程序时颤振动态链接打开网页

flutter dynamic links open web page when app is not installed

dart

firebase

flutter

firebase-dynamic-links