Make a phone call programmatically

Question

How can I make a phone call programmatically on iPhone? I tried the following code but nothing happened:

NSString *phoneNumber = mymobileNO.titleLabel.text;
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];

For code in swift you can follow this answer http://stackoverflow.com/a/29869456/3950397 — Sahil Kapoor, Apr 27 '15 at 10:36

score 219 · Answer 1 · answered Apr 10 '12 at 12:33

219

To go back to original app you can use telprompt:// instead of tel:// - The tell prompt will prompt the user first, but when the call is finished it will go back to your app:

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

answered Apr 10 '12 at 12:33

Craig Mellon

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10

I have a question: will the apps that use telprompt:// be rejected by Apple? – Smeegol Jul 23 '12 at 08:46
8

I know this is old but I did find a couple people claiming their apps were approved using telprompt: "I've submited my app that uses telprompt with shared application without the uiwebkit and has been successfully approved by apple. answered January 19, 2013 Pablo Alejandro Junge" – Mark McCorkle May 15 '13 at 14:16
2

telprompt:// is undocumented and should therefore never be relied upon. Things may changed, For example Apple could decide a given URL scheme is something they need exclusively or simply do not wish to allow, then your app will break. – DevC Jul 10 '14 at 08:31
@DevC: Then what is alternative to have phone call feature in our app? – BaSha Jan 21 '15 at 05:46
@BaSha use `tel://` not `telprompt://` – DevC Jan 21 '15 at 10:46
without backslash also works..i.e we can use **tel:** and **telprompt:** – Durai Amuthan.H Feb 28 '17 at 11:45

score 192 · Accepted Answer · edited May 18 '21 at 14:31

192

Probably the mymobileNO.titleLabel.text value doesn't include the scheme //

Your code should look like this:

ObjectiveC

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

Swift

if let url = URL(string: "tel://\(mymobileNO.titleLabel.text))") {
    UIApplication.shared.open(url)
}

edited May 18 '21 at 14:31

gfpacheco

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answered Feb 08 '11 at 05:04

Cristian Radu

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3

There's no chance we could go back to original application with this approach. – Artem Oboturov Oct 30 '11 at 10:58
4

I think Mellon's approach is better as it prompts the user first, and then returns to your app after the call. Just make sure to use canOpenURL: and fallback to the tel: prompt since telprompt is not official and could be removed in a future iOS version. – joel.d Jan 10 '14 at 00:56
How can I make phone call from apple watch. – Ahad Khan Nov 22 '15 at 07:16
1

Can I turn on speaker programmatically for call originated this way in ios? – Uniruddh Feb 26 '16 at 08:58
even **tel:1234567890** also works just like how **tell://1234567890** it works – Durai Amuthan.H Feb 28 '17 at 11:44

score 25 · Answer 3 · edited Dec 27 '18 at 07:02

Merging the answers of @Cristian Radu and @Craig Mellon, and the comment from @joel.d, you should do:

NSURL *urlOption1 = [NSURL URLWithString:[@"telprompt://" stringByAppendingString:phone]];
NSURL *urlOption2 = [NSURL URLWithString:[@"tel://" stringByAppendingString:phone]];
NSURL *targetURL = nil;

if ([UIApplication.sharedApplication canOpenURL:urlOption1]) {
    targetURL = urlOption1;
} else if ([UIApplication.sharedApplication canOpenURL:urlOption2]) {
    targetURL = urlOption2;
}

if (targetURL) {
    if (@available(iOS 10.0, *)) {
        [UIApplication.sharedApplication openURL:targetURL options:@{} completionHandler:nil];
    } else {
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wdeprecated-declarations"
        [UIApplication.sharedApplication openURL:targetURL];
#pragma clang diagnostic pop
    }
}

This will first try to use the "telprompt://" URL, and if that fails, it will use the "tel://" URL. If both fails, you're trying to place a phone call on an iPad or iPod Touch.

Swift Version :

let phone = mymobileNO.titleLabel.text
let phoneUrl = URL(string: "telprompt://\(phone)"
let phoneFallbackUrl = URL(string: "tel://\(phone)"
if(phoneUrl != nil && UIApplication.shared.canOpenUrl(phoneUrl!)) {
  UIApplication.shared.open(phoneUrl!, options:[String:Any]()) { (success) in
    if(!success) {
      // Show an error message: Failed opening the url
    }
  }
} else if(phoneFallbackUrl != nil && UIApplication.shared.canOpenUrl(phoneFallbackUrl!)) {
  UIApplication.shared.open(phoneFallbackUrl!, options:[String:Any]()) { (success) in
    if(!success) {
      // Show an error message: Failed opening the url
    }
  }
} else {
    // Show an error message: Your device can not do phone calls.
}

Kundan · Answer 4 · 2015-09-16T05:27:48.420

10

The answers here are perfectly working. I am just converting Craig Mellon answer to Swift. If someone comes looking for swift answer, this will help them.

 var phoneNumber: String = "telprompt://".stringByAppendingString(titleLabel.text!) // titleLabel.text has the phone number.
        UIApplication.sharedApplication().openURL(NSURL(string:phoneNumber)!)

edited Sep 16 '15 at 05:27

answered Sep 15 '15 at 14:41

Kundan

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you can also use tel:// in place of telprompt:// – Kundan Sep 19 '15 at 05:47
let phoneNumber: String = "telprompt://\(titleLabel.text)" UIApplication.shared.openURL(URL(string:phoneNumber)!) – tspentzas Mar 04 '17 at 23:51

score 9 · Answer 5 · answered Sep 17 '14 at 16:48

If you are using Xamarin to develop an iOS application, here is the C# equivalent to make a phone call within your application:

string phoneNumber = "1231231234";
NSUrl url = new NSUrl(string.Format(@"telprompt://{0}", phoneNumber));
UIApplication.SharedApplication.OpenUrl(url);

score 6 · Answer 6 · edited Dec 28 '16 at 10:34

6

Swift 3

let phoneNumber: String = "tel://3124235234"
UIApplication.shared.openURL(URL(string: phoneNumber)!)

edited Dec 28 '16 at 10:34

Tiago Almeida

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answered Nov 17 '16 at 15:20

Jorge Costa

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score 4 · Answer 7 · answered Jul 26 '17 at 12:20

In Swift 3.0,

static func callToNumber(number:String) {

        let phoneFallback = "telprompt://\(number)"
        let fallbackURl = URL(string:phoneFallback)!

        let phone = "tel://\(number)"
        let url = URL(string:phone)!

        let shared = UIApplication.shared

        if(shared.canOpenURL(fallbackURl)){
            shared.openURL(fallbackURl)
        }else if (shared.canOpenURL(url)){
            shared.openURL(url)
        }else{
            print("unable to open url for call")
        }

    }

score 3 · Answer 8 · answered May 07 '15 at 20:11

3

The Java RoboVM equivalent:

public void dial(String number)
{
  NSURL url = new NSURL("tel://" + number);
  UIApplication.getSharedApplication().openURL(url);
}

answered May 07 '15 at 20:11

EntangledLoops

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Charlie S · Answer 9 · 2016-12-19T15:42:04.717

3

Tried the Swift 3 option above, but it didnt work. I think you need the following if you are to run against iOS 10+ on Swift 3:

Swift 3 (iOS 10+):

let phoneNumber = mymobileNO.titleLabel.text       
UIApplication.shared.open(URL(string: phoneNumber)!, options: [:], completionHandler: nil)

edited Dec 19 '16 at 15:42

answered Dec 07 '16 at 14:41

Charlie S

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score 2 · Answer 10 · edited Jun 20 '20 at 09:12

2

Swift

if let url = NSURL(string: "tel://\(number)"), 
    UIApplication.sharedApplication().canOpenURL(url) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
}

edited Jun 20 '20 at 09:12

Community

1
1

answered May 20 '19 at 06:53

Sazzad Hissain Khan

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score 1 · Answer 11 · edited May 27 '16 at 08:10

1

let phone = "tel://\("1234567890")";
let url:NSURL = NSURL(string:phone)!;
UIApplication.sharedApplication().openURL(url);

edited May 27 '16 at 08:10

Dovydas Šopa

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answered May 27 '16 at 06:57

Himali Shah

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score 1 · Answer 12 · answered Dec 15 '18 at 09:01

'openURL:' is deprecated: first deprecated in iOS 10.0 - Please use openURL:options:completionHandler: instead

in Objective-c iOS 10+ use :

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

score 0 · Answer 13 · answered Apr 26 '21 at 23:07

Use openurl.

For making a call in swift 5.1, just use the following code: (I have tested it in Xcode 11)

let phone = "1234567890"
if let callUrl = URL(string: "tel://\(phone)"), UIApplication.shared.canOpenURL(callUrl) {
     UIApplication.shared.open(callUrl)
}

Edit: For Xcode 12.4, swift 5.3, just use the following:

UIApplication.shared.open(NSURL(string: "tel://555-123-1234")! as URL)

Make a phone call programmatically

13 Answers13

Swift

Make sure that you import UIKit, or it will say that it cannot find UIApplication in scope.

Linked

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