Calling a phone number in swift

Just try:

if let url = NSURL(string: "tel://\(busPhone)") where UIApplication.sharedApplication().canOpenURL(url) {
  UIApplication.sharedApplication().openURL(url)
}

assuming that the phone number is in busPhone.

NSURL's init(string:) returns an Optional, so by using if let we make sure that url is a NSURL (and not a NSURL? as returned by the init).

For Swift 3:

if let url = URL(string: "tel://\(busPhone)"), UIApplication.shared.canOpenURL(url) {
    if #available(iOS 10, *) {
        UIApplication.shared.open(url)
    } else {
        UIApplication.shared.openURL(url)
    }
}

We need to check whether we're on iOS 10 or later because:

'openURL' was deprecated in iOS 10.0

A self contained solution in iOS 10, Swift 3 :

private func callNumber(phoneNumber:String) {

  if let phoneCallURL = URL(string: "tel://\(phoneNumber)") {

    let application:UIApplication = UIApplication.shared
    if (application.canOpenURL(phoneCallURL)) {
        application.open(phoneCallURL, options: [:], completionHandler: nil)
    }
  }
}

You should be able to use callNumber("7178881234") to make a call.