swift3 - canOpenURL: failed for URL tel

Question

I want to intent a call , this is my code :

if let urlMobile = NSURL(string: "tel://076938483"), UIApplication.shared.canOpenURL(urlMobile as URL) {

                if #available(iOS 10.0, *) {
                    UIApplication.shared.open(urlMobile as URL, options: [:], completionHandler: nil)
                }
                else {
                    UIApplication.shared.openURL(urlMobile as URL)
                }
            }

I'm using swift 3 to do so but I get this error:

-canOpenURL: failed for URL: "tel://09178883828" - error: "The operation couldn’t be completed. (OSStatus error -10814.)"

any idea to do so ?

Answer 1

Your code works perfectly fine. Run it on an actual device if your using Simulator. You can't simulate a call on a Mac/MacBook.

Please have a look at Simulator Hardware Actions in Apple documentation.

Answer 2

Some LSApplicationQueriesScheme do not work on Simulator. Error Code -10814 is for kLSApplicationNotFoundErr. Simulator can't launch Dial Pad for Telephone. So run it on iPhone device.

This worked for me!!!

Code Should be

if let url = NSURL(string: "tel://\(yourNumber)"), UIApplication.shared.canOpenURL(url as URL) {
  UIApplication.shared.open(url as URL, options: [:], completionHandler: nil)
}

Answer 3

网址应该是：

if let urlMobile = NSURL(string: "tel:///076938483"), UIApplication.shared.canOpenURL(urlMobile as URL) {

Answer 4

let phonenumber = "076938483"
guard let url = URL(string: "tel://\(phonenumber )") else {
            return
            
        }
        if #available(iOS 10.0, *) {
            UIApplication.shared.open(url)
        } else {
            UIApplication.shared.openURL(url)
        }

In the else part of the URL generation, print anything, if it is getting printed, you need to check the format of the phone number.

Answer 5

@IBAction func Call(_ sender: Any) {

    let busPhone = "7355535586"
    if let url = URL(string: "tel://\(busPhone)"), UIApplication.shared.canOpenURL(url) {
        if #available(iOS 10, *) {
            UIApplication.shared.open(url)
        } else {
            UIApplication.shared.openURL(url)
        }
    }    

Answer 6

@objc func callBtn() {
       let userPhone = String((phoneNum.filter {!" \n\t\r".contains($0)}))
       if let url = URL(string: "tel://\(phoneNum)"), UIApplication.shared.canOpenURL(url) {
           DispatchQueue.main.async {
               UIApplication.shared.open(url)
           }
        }
    }

Note : This code can work when you run with the real device not simulator. And don't forget add LSApplicationQueriesSchemes in your info.plist.