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.

Swift 4,

private func callNumber(phoneNumber:String) {    if let phoneCallURL = URL(string: "telprompt://\(phoneNumber)") {        let application:UIApplication = UIApplication.shared        if (application.canOpenURL(phoneCallURL)) {            if #available(iOS 10.0, *) {                application.open(phoneCallURL, options: [:], completionHandler: nil)            } else {                // Fallback on earlier versions                 application.openURL(phoneCallURL as URL)            }        }    }}