Calling a phone number in swift
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)
}
}
}
}
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