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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.

Peter Haddad
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A.khaled
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  • well, after some search I found this useful answer https://stackoverflow.com/questions/38251166/how-get-extra-parameter-from-dynamic-link-using-firebase – A.khaled Apr 29 '18 at 22:54

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