Firebase.google.com Firebase Dynamic Links in iOS ( Swift )

Question

How to Open an iOS App using Firebase Dynamic Links and Pass or get Parameters To an App Via Custom URL Scheme in iOS(swift)?


for eg :- https://q3tyj.app.goo.gl/abcd 


My URL Scheme is ‘q3tyj.app.goo.gl’ in iOS app in Url Types.


If I type q3tyj.app.goo.gl in safari, I am able to open the application. But if I type q3tyj.app.goo.gl with some extra parameter like https://q3tyj.app.goo.gl/abcd in safari, , I am not able to open the application.



please also explain me how to get “link” parameter (which associated with dynamic link) from dynamic link in iOS app ( Swift ) .



I followed steps which were mentioned in Firebase.google.com for iOS ( Swift ).but its not working.

Thanks, Nirav Virpara

Answer 1

It's hard to know exactly where your error is, but a few things to help you debug:

1. If you go to https://q3tyj.app.goo.gl/apple-app-site-association, you should see some JSON that points to your app. If you don't see this, make sure you've entered your team ID and your App Store ID in the project settings in the Firebase console

2. Make sure you're using your Bundle ID, not the shortlink domain, as your custom URL scheme in Xcode

3. Make sure you've enabled Associated Domains in the Capabilities tab of your Xcode project, and your domain looks like applinks:q3tyj.app.goo.gl

4. Universal Links (and, therefore, Dynamic Links) generally don't work if you type them directly into the Safari address bar. Instead, try typing the URL into an app like Notes and then clicking on them from there.

Good luck!