Phone call button in Swift 3 doesn't work

Question

I need my app make a phone call when user taps a button. I found some tutorials and people there say use this code

let numberUrl = URL(string: "tel://12345678")!
UIApplication.shared.open(numberUrl, options: [:], completionHandler: nil)

But it just does not work. Was it change in Swift 3?

Possible duplicate of [how to open an URL in Swift3](https://stackoverflow.com/questions/39546856/how-to-open-an-url-in-swift3) — Kiran Sarvaiya, Nov 17 '17 at 12:52

Nimisha Ranipa · Accepted Answer · 2017-11-17T11:10:35.190

5

UPDATE

Try this code for Call. Code Works for all iOS versions

Swift 2.0

 let url = URL(string: "tel://123456789")!
 UIApplication.shared.openURL(url)

Swift 3.0

 let url = URL(string: "tel://123456789")!
 UIApplication.shared.open(url)

edited Nov 17 '17 at 11:10

answered Dec 24 '16 at 07:21

Nimisha Ranipa

8,304
2
14
29

No its working fine for me even not display warning for deprecated. – Nimisha Ranipa Dec 26 '16 at 03:42
No its not deprecated. cmd+click on open(url), it will navigate to documentation, there it shows its available in ios10 – Apogee Nov 17 '17 at 12:41

Cœur · Answer 2 · 2016-12-28T14:17:42.107

1

According to your comment, you want to deal with the deprecation warning.

if #available(iOS 10, *) {
    UIApplication.shared.open(numberUrl)
} else {
    UIApplication.shared.openURL(numberUrl)
}

edited Dec 28 '16 at 14:17

answered Dec 25 '16 at 02:23

Cœur

37,241
25
195
267

score 0 · Answer 3 · answered Dec 24 '16 at 03:48

try this way in your real device, simulator doesn't place a call.

let phoneUrl = "tel://45657657" //your url
      if let url = URL(string: phoneUrl){
           if #available(iOS 10.0, *) {
              UIApplication.shared.open(url, options: [:])
            } else {
              // Fallback on earlier versions
            }
       }

score -1 · Answer 4 · answered Nov 17 '17 at 12:33

-1

Try this...

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

answered Nov 17 '17 at 12:33

Rishu Gupta

43
1
14

Yes, check the condition that checks for if application can open url or not – Rishu Gupta Nov 17 '17 at 12:40
Please stop upvoting each other's stuff. And please only post useful answers. That's all I'm saying. Be honest, be fair, and post good content that has not already been posted a thousand times. – Eric Aya Nov 17 '17 at 12:45
Yes, @KiranSarvaiya. You and Rishu should flag as duplicate *instead* of posting the same stuff again. – Eric Aya Nov 17 '17 at 12:49

Phone call button in Swift 3 doesn't work

4 Answers4