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Home/ Questions/Q 8434731
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T06:44:14+00:00 2026-06-10T06:44:14+00:00

I used this code to make phone call: NSString *phoneNumber = [@tel:// stringByAppendingString:mymobileNO.titleLabel.text]; [[UIApplication

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I used this code to make phone call:

NSString *phoneNumber = [@"tel://" stringByAppendingString:mymobileNO.titleLabel.text];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];

and when the call ends, it does not take user back to the app. However, if I show a website in UIWebView and there is a phone number in the website(ie UIWebView recognises it), and if I tap that phone number link to make phone call, I will be taken back to the app when the call finishes.

My preliminary thinking is that the UIWebView does something internally like a deep link to the Phone app then another deep link inside the deep link to take the user back to the app. But I’m not sure. Is there a way to implement this feature?

Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-10T06:44:16+00:00Added an answer on June 10, 2026 at 6:44 am

    You need to use the telprompt URL, not tel.

    So:

    NSString *phoneNumber = [@"telprompt://" stringByAppendingString:mymobileNO.titleLabel.text];
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
    

    This will also give the user a confirmation box before calling the number.

    Edit:

    This question covers the same issue.

    Edit 2:

    For those wondering if this URL will result in rejection from the App Store, the answer is generally no. The greater risk is that Apple will suddenly stop supporting the telprompt scheme. As explained by this article, there is a slightly ‘safer’ way of implementing telprompt with UIWebView (which uses telprompt internally, even if you call it with tel). The relevant code from the article shows how using the documented tel scheme can still give you the effect of telprompt:

    + (void)callWithString:(NSString *)phoneString {
        [self callWithURL:[NSURL URLWithString:[NSString     
            stringWithFormat:@"tel:%@",phoneString]]];
    }
    
    + (void)callWithURL:(NSURL *)url {
        static UIWebView *webView = nil;
        static dispatch_once_t onceToken;
        dispatch_once(&onceToken, ^{  
            webView = [UIWebView new];
        });
        [webView loadRequest:[NSURLRequest requestWithURL:url]];
    }
    

    (Code taken from the article reference in my second edit)

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