I am developing an app for a cab and have to implement the call facility.I am done with this calling feature but it is showing the dialed number in dialer, but I am in need to hide the dial number from the dialer.we are doing this just to make female passenger safe and comfortable while having the ride and after the ride.
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1User has to be able to see where does his device calling. There is no way, and should be no way to hide a number, as the user is paying money for the call. If you want not to show target number, use VoIP to deliver call to some backend, and from there - use some call gateway to call to the phone number. – Vladyslav Matviienko Jul 11 '17 at 08:25
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As @VladMatvienko said it is not possible. The activity for calling has the complete controll until it is finished and your programm has the controll again. That is a design to prevent missusing the dialing function. – Nico Jul 11 '17 at 08:27
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Create custom dialer for this purpose, where you can customize UI of dialer with some unrelated UX. http://eagle.phys.utk.edu/guidry/android/simpleDialer.html – Pankaj Kumar Jul 11 '17 at 08:28
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@Santosh Yadav i think you can use Twillio for this its provides some functionalities like not showing user number ..its provides fake number.. – Dileep Patel Jul 11 '17 at 08:40
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While calling Intent, Use Intent.ACTION_CALL. It will directly call at particular number without showing dialer to user.
Pallavi Tapkir
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it will show the phone app with the phone number it is calling. – Vladyslav Matviienko Jul 11 '17 at 08:28
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I don't want anything, I'm not the question owner, but he wants the phone number not to be visible to the user. – Vladyslav Matviienko Jul 11 '17 at 08:31
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I don't want to show the user which number is calling. i have already gone through Intent.ACTION_CALL but it will bring the phone app in front showing which no. is connecting. – Santosh Yadav Jul 11 '17 at 08:32
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Use Intent.ACTION_CALL it will directly call
Intent intent = new Intent(Intent.ACTION_CALL);
intent.setData(Uri.parse("tel:100"));
startActivity(intent);
