某些电话号码的openURL失败

Question

I'm trying to make a call programmatically to a number which has double extension.

When I try to run this,

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel:2133776478,213#,213#"]]]

I get "No" as the response. But when I type this number on WhatsApp, I'm allowed to make a call with this number.

Is there a better to achieve this? Am I missing something here?

Answer 1

According to the documentation ,

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

I would try using the escaped version of the telephone number first (using NSString:stringByAddingPercentEncodingWithAllowedCharacters: ), and if that fails, I would try removing the , and # characters.

Answer 2

A tel: URL with a single # works, but the presence of two # characters is an invalid URL as far as NSURL is concerned because it looks like the URL contains two anchor segments. This results in a nil NSURL.

If you percent-encode your string then the # characters simply become part of the URL string and it all works:

NSString *baseString = @"tel:2133776478,213#,213#";
NSString *encodedString = [baseString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
if (encodedString != nil) {
    NSURL *url = [NSURL URLWithString:encodedString];
    if (url != nil) {
        [[UIApplication sharedApplication] openURL:url];
    }
}