How to use openURL for making a phone call in Swift?

Question

I have converted the code for making a phone call from Objective-C to Swift, but in Objective-C, we can set the type of the URL that we like to open (e.g. telephone, SMS, web) like this:

@"tel:xx"
@"mailto:info@example.es"
@"http://stackoverflow.com"
@"sms:768number"

The code in Swift is:

UIApplication.sharedApplication().openURL(NSURL(string : "9809088798")

I read that have not released any scheme parameter for tel:, but I don't know if Swift can detect if the string is for making a phone call, sending email, or opening a website. Or may I write:

(string : "tel//:9809088798")

?

I don't see what the problem is. Did you try it? Did it work? And please do not delete and repost questions. — Mick MacCallum, Jun 16 '14 at 20:04
My problem is that I want know if this is correct for call: UIApplication.sharedApplication().openURL(NSURL(string : "9809088798") Because have only the emulator for try it — user3739367, Jun 16 '14 at 20:06

Lou Franco · Accepted Answer · 2015-08-06T20:47:07.637

82

I am pretty sure you want:

UIApplication.sharedApplication().openURL(NSURL(string: "tel://9809088798")!)

(note that in your question text, you put tel//:, not tel://).

NSURL's string init expects a well-formed URL. It will not turn a bunch of numbers into a telephone number. You sometimes see phone-number detection in UIWebView, but that's being done at a higher level than NSURL.

EDIT: Added ! per comment below

edited Aug 06 '15 at 20:47

answered Jun 16 '14 at 20:05

Lou Franco

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Just a tiny correction UIApplication.sharedApplication().openURL(NSURL(string: "tel://9809088798")!) you need to use the ! after the NSURL because it returns an optional. – brduca Aug 06 '15 at 11:21
this will directly start calling on the number (ofcourse after 10.3 iOS will ask permission to call) but what if I need to open the iPhone dialer with filled a number ?? Can I achieve this? – Suryakant Sharma Dec 11 '17 at 12:18
Awesome Answer sir :) – steveSarsawa Jul 23 '19 at 08:56

Zorayr · Answer 2 · 2015-05-22T16:53:37.637

19

A self-contained solution in Swift:

private func callNumber(phoneNumber:String) {
  if let phoneCallURL:NSURL = NSURL(string:"tel://\(phoneNumber)") {
    let application:UIApplication = UIApplication.sharedApplication()
    if (application.canOpenURL(phoneCallURL)) {
      application.openURL(phoneCallURL);
    }
  }
}

Now, you should be able to use callNumber("7178881234") to make a call.

edited May 22 '15 at 16:53

answered Apr 25 '15 at 19:04

Zorayr

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The problem is that your solution does not return to the app after the phone call has been finished on iOS7. – confile Jun 19 '15 at 13:50
Hm, curious if there is a solution that does do that.. might be a restriction on iOS. – Zorayr Jun 19 '15 at 19:12

score 8 · Answer 3 · edited Nov 13 '15 at 03:23

8

For Swift in iOS:

var url:NSURL? = NSURL(string: "tel://9809088798")
UIApplication.sharedApplication().openURL(url!)

edited Nov 13 '15 at 03:23

Pang

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answered Nov 13 '14 at 06:41

Vinod Joshi

7,094
46
49

this will directly start calling on the number (ofcourse after 10.3 iOS will ask permission to call) but what if I need to open the iPhone dialer with filled a number ?? Can I achieve this? – Suryakant Sharma Dec 11 '17 at 12:18

score 7 · Answer 4 · answered Mar 31 '16 at 23:14

You need to remember to remove the whitespaces or it won't work:

if let telephoneURL = NSURL(string: "telprompt://\(phoneNumber.stringByReplacingOccurrencesOfString(" ", withString: ""))") {
        UIApplication.sharedApplication().openURL(telelphoneURL)
    }

"telprompt://" will prompt the user to call or cancel while "tel://" will call directly.

score 6 · Answer 5 · answered Sep 24 '15 at 14:08

@ confile:

The problem is that your solution does not return to the app after the phone call has been finished on iOS7. – Jun 19 at 13:50

&@ Zorayr

Hm, curious if there is a solution that does do that.. might be a restriction on iOS.

use

UIApplication.sharedApplication().openURL(NSURL(string: "telprompt://9809088798")!)

You will get a prompt to Call/Cancel but it returns to your application. AFAIK there is no way to return (without prompting)

score 5 · Answer 6 · answered Jun 07 '15 at 22:09

You must insert "+"\ is another way

private func callNumber(phoneNumber:String) {
  if let phoneCallURL:NSURL = NSURL(string:"tel://"+"\(phoneNumber)") {
    let application:UIApplication = UIApplication.sharedApplication()
    if (application.canOpenURL(phoneCallURL)) {
      application.openURL(phoneCallURL);
    }
  }
}

score 5 · Answer 7 · answered Dec 23 '16 at 16:54

5

Small update to Swift 3

UIApplication.shared.openURL(NSURL(string: "telprompt://9809088798")! as URL)

answered Dec 23 '16 at 16:54

Paula Hasstenteufel

680
8
17

score 3 · Answer 8 · answered Feb 28 '17 at 12:22

The following code snippet can tell if the SIM is there or not and if the device is capable of making the call and if ok then it'll make the call

  var info = CTTelephonyNetworkInfo()
    var carrier = info.subscriberCellularProvider
    if carrier != nil && carrier.mobileNetworkCode == nil || carrier.mobileNetworkCode.isEqual("") {
       //SIM is not there in the phone
    }
    else if UIApplication.sharedApplication().canopenURL(NSURL(string: "tel://9809088798")!)
    {
    UIApplication.sharedApplication().openURL(NSURL(string: "tel://9809088798")!)
    }
    else    
    {
      //Device does not have call making capability  
    }

NDCoder · Answer 9 · 2021-02-21T23:23:30.177

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

score 0 · Answer 10 · answered May 25 '17 at 06:15

For swift 3

if let phoneCallURL:URL = URL(string:"tel://\(phoneNumber ?? "")") {
            let application:UIApplication = UIApplication.shared
            if (application.canOpenURL(phoneCallURL)) {
                application.open(phoneCallURL, options: [:], completionHandler: nil);
            }
        }

score 0 · Answer 11 · answered Apr 26 '18 at 10:00

For swift 4:

func call(phoneNumber: String) {
    if let url = URL(string: phoneNumber) {
        if #available(iOS 10, *) {
            UIApplication.shared.open(url, options: [:],
                                      completionHandler: {
                                        (success) in
                                        print("Open \(phoneNumber): \(success)")
            })
        } else {
            let success = UIApplication.shared.openURL(url)
            print("Open \(phoneNumber): \(success)")
        }
    }
}

Then, use the function:

let phoneNumber = "tel://+132342424"
call(phoneNumber: phoneNumber)

score 0 · Answer 12 · answered Oct 02 '20 at 05:02

0

Swift 4 and above

let dialer = URL(string: "tel://5028493750")
    if let dialerURL = dialer {
        UIApplication.shared.open(dialerURL)
}

answered Oct 02 '20 at 05:02

Akbar Khan

1,442
13
17

How to use openURL for making a phone call in Swift?

12 Answers12

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

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