1

when creating a firebase dynamic link manually i want the link to contain a variable as a parameter to deal with it when the link is clicked, i tried to do this: 

StringBuilder link= new StringBuilder("https://qn937.app.goo.gl/?link=https://www.teblya.com/&apn=com.example.abdo.foodproject");

link.append("/?token="+itemٍ.getId()+"/");
Intent shareIntent = new Intent();
shareIntent.setAction(Intent.ACTION_SEND);

shareIntent.putExtra(Intent.EXTRA_TEXT, link.toString());
shareIntent.setType("text/plain");
startActivity(Intent.createChooser(shareIntent, getResources().getText(R.string.app_name)));

but the problem is that it opens the deep link in the browser instead of my app. does anyone know any approach i can sent this item.getId() parameter with the link? thanks in advance.

Improve this question
1
Related questions
13
How get extra parameter from dynamic link using Firebase?
6
Pass multiple params via Firebase Dynamic links Android
0
Passing parameters in firebase dynamic link not working
Load 7 more related questions Show fewer related questions

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.