title: Firebase.google.com Firebase Dynamic Links in iOS ( Swift )
body: <p>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)?<br>
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  for eg :- <a href="https://q3tyj.app.goo.gl/abcd" rel="nofollow">https://q3tyj.app.goo.gl/abcd</a>     </p>
  
  <p>My URL Scheme is ‘q3tyj.app.goo.gl’   in iOS app in Url Types.  </p>
  
  <p>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  <a href="https://q3tyj.app.goo.gl/abcd" rel="nofollow">https://q3tyj.app.goo.gl/abcd</a>  in safari, , I am not able to open the application.  </p>
  
  <p>please also explain me how to get “link” parameter (which associated with dynamic link) from dynamic link in iOS app ( Swift ) .</p>
  
  <p>  I followed steps which were mentioned in Firebase.google.com for iOS ( Swift ).but its not working.</p>
  
  <p>Thanks,
  Nirav Virpara</p>