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, we

Answer 1

@ 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)

Answer 2

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

Answer 3

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)

Answer 4

Small update to Swift 3

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

Answer 5

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);
            }
        }

Answer 6

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.