Firebase Dynamic Link pass custom parameters to iOS and android - firebase

Question

I have a custom function in python to build the dynamic link:
def generate_dynamic_link(link, title=None, image=None, description=None, short=True, timeout=10):
api_url = FIREBASE_DYNAMIC_LINK_API_URL
domain = DYNAMIC_LINK_DOMAIN
apn = APP_APN
isi = APP_ISI
ibi = APP_IBI
payload = {
"dynamicLinkInfo": {
"domainUriPrefix": domain,
"link": link,
"androidInfo": {
"androidPackageName": apn,
},
"iosInfo": {
"iosBundleId": ibi,
"iosAppStoreId": isi
},
"socialMetaTagInfo": {
"socialTitle": title,
"socialDescription": description,
"socialImageLink": image
}
},
"suffix": {
"option": "SHORT" if short else "UNGUESSABLE"
}
response = requests.post(api_url, json=payload, timeout=timeout)
data = response.json()
if not response.ok:
raise Exception(data)
return data['shortLink']
I want to pass two parameters to the android and ios app. How can I Do that?
Example:
?type=user&username=testuser

Answer 1

I wrote my first Medium article about this (it’s not a great tutorial) but it shows how to do this. You are correct with how you pass data using ?yourDataHere at the end of your link.
https://augustkimo.medium.com/simple-flutter-sharing-and-deeplinking-open-apps-from-a-url-be56613bdbe6
Then you can handle the deep links by calling the function below. Pretty much you can get the link used to open the app, then get the data from that URL/link string
//ADD THIS FUNCTION TO HANDLE DEEP LINKS
Future<Null> initUniLinks()async{
try{
Uri initialLink = await getInitialUri();
print(initialLink);
var dataFromLink = initialLink.toString().split(‘?’)[1];
print(dataFromLink);
} on PlatformException {
print('platfrom exception unilink');
}
}