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