How do I receive query parameters from Firebase Dynamic Links using react-native-firebase?

Question

I've followed the instructions from the official react-native-firebase documentation and everything, as per the instructions, works fine.

It is my understanding that I should be able to create a link like the following in the Firebase Dynamic Links console: https://myapp.page.link/offer?offerid=123456 and be able to receive the offerid parameter in my app. However, when I click that link (in the iOS simulator, Android emulator, and on a real device) the app opens, but I just get a link with the following data:

Received initial link {"minimumAppVersion": null, "url": "https://mysite.co/offers"}

There are no query string parameters attached. Am I missing something in the Firebase Dynamic Link configuration? Or in my implementation of react-native-firebase that was not covered in the documentation linked above? Or is this actually just not possible?

Answer 1

I think you are confusing deep links with dynamic links. Dynamic links handle your deep links to content depending upon the platform.

Example you have a deep link: https://example.com/offer?offerid=123456 . You wish to open it in your app depending upon the platform.

Suppose you created a domain for dynamic link from console: https://myapp.page.link

Now you can create a dynamic link which opens the deep linked content on your app. The dynamic link can take in only specific parameters as mentioned here . However you can pass your parameters to your deep link,

https://myapp.page.link/?link=https%3A%2F%2Fexample.com%2Foffer%3Fofferid%3D123456&apn=com.example.app&ibi=com.example.app

The above dynamic link opens the deep link in your android app with package name com.example.app and ios app with bundle ID com.example.app .