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Home/ Questions/Q 5944465
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T16:32:26+00:00 2026-05-22T16:32:26+00:00

I have already tried to do it using the following code: NSURL *phoneURL =

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I have already tried to do it using the following code:

NSURL *phoneURL = [[NSURL alloc] initWithString:@"tel:**21*<destination_number>#"];
[[UIApplication sharedApplication] openURL:phoneURL];

but it doesn’t work and sometimes the app crashs.

It works for me using a normal number (like 19765432221)…

Anybody?

Thanks!

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  1. Editorial Team
    Editorial Team
    2026-05-22T16:32:27+00:00Added an answer on May 22, 2026 at 4:32 pm

    I’m not sure what *21# means, but if it doesn’t work the way you’re doing it, it’s most certainly because Apple doesn’t allow it. There’s no other (public) API method for doing this kind of thing.

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