CODEWITHSUNDEEP
swiftnsurlxcode6openurl
user3739367
user3739367

Reputation: 4461

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

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")

?

Upvotes: 46

Views: 63200

Answers (13)

Jensie
Jensie

Reputation: 781

I prefer deferring the URL creation to the built-ins like:

var components = URLComponents()
components.scheme = "tel"
components.path = "1234567890"
print(components.url!) //Prints out: "tel:1234567890"

Which can then be used in UIApplication.shared.openURL

Upvotes: 0

NDCoder
NDCoder

Reputation: 93

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.

Upvotes: 1

Akbar Khan
Akbar Khan

Reputation: 2415

Swift 4 and above

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

Upvotes: 0

Nil
Nil

Reputation: 139

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)

Upvotes: 0

Arshad
Arshad

Reputation: 906

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);
            }
        }

Upvotes: 0

Durai Amuthan.H
Durai Amuthan.H

Reputation: 32320

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  
    }

Upvotes: 3

Paula Hasstenteufel
Paula Hasstenteufel

Reputation: 720

Small update to Swift 3

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

Upvotes: 5

Robert Wagstaff
Robert Wagstaff

Reputation: 2674

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.

Upvotes: 8

Vinod Joshi
Vinod Joshi

Reputation: 7862

For Swift in iOS:

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

Upvotes: 8

Glenn
Glenn

Reputation: 2806

@ 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)

Upvotes: 6

Lou Franco
Lou Franco

Reputation: 89232

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

Upvotes: 82

Mauricio Lucon
Mauricio Lucon

Reputation: 53

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);
    }
  }
}

Upvotes: 5

Zorayr
Zorayr

Reputation: 24962

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.

Upvotes: 19

Related Questions