Call to a number, which contain #. (IPhone SDK)

Question

I need to make call to a number, that start with #. For example phone number in Russia looks like +79123817711 and I need to call #79123817711. I\'m trying this:

<         


        

Answer 1

I tried iOS 10. But it was not working. But from iOS 12 now its working. do not know how.

NSString *mobileNocleanedString = @"*454*200#";

NSURL *phoneUrl = [NSURL URLWithString:[NSString  stringWithFormat:@"tel:%@",mobileNocleanedString]];

if ([[UIApplication sharedApplication] canOpenURL:phoneUrl]) {
    [[UIApplication sharedApplication] openURL:phoneUrl];
}

Answer 2

just replace * by %2a and # by %23

Answer 3

iOS SDK (Apple URL scheme reference) tells: To prevent users from maliciously redirecting phone calls or changing the behavior of a phone or account, the Phone application supports most, but not all, of the special characters in the tel scheme. Specifically, if a URL contains the * or # characters, the Phone application does not attempt to dial the corresponding phone number.

So, no luck.

Answer 4

I have achieved this by encoding the ussd and pass it in the url and it works just fine

Here is an example:

extension String {
      /// Calling ussd number
      func callUSSD() {
          guard let urlString = self.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed) else { return }
          guard let url = URL(string: "tel://\(urlString)") else { return }
         UIApplication.shared.open(url, options: [:])
    }
}

Usage:

@IBAction func callTapped(_ sender: UIButton) {
    "*1124#".callUSSD()        
}