CODEWITHSUNDEEP
swifturlphone-number
Robert Dresler
Robert Dresler

Reputation: 11140

Can't construct URL for phone call in Swift

I'm trying to ask for phone call programmatically but I'm not able to construct URL from my nine-digit phone number. When I try it with for example 999999999 phone number, it works, it asks for call

@IBAction func callButtonPressed(_ sender: UIButton) {
    askForCall(to: "999999999")
}

func askForCall(to number: String) {
    guard let url = URL(string: "tel:\(number)"), UIApplication.shared.canOpenURL(url) else { return }
    UIApplication.shared.open(url)
}

but when I use real phone number 736XXXXXX it shows nothing.

Note: when I try it without canOpenUrl it doesn’t work so I guess problem is with constructing URL from my real number

Any ideas?

Upvotes: 1

Views: 655

Answers (2)

Francesco Destino
Francesco Destino

Reputation: 349

You should type "tel://" + number and not tel:\(number)

EDIT 2

Try something like this

func call(phoneNumber: String) {
    if let url = URL(string: phoneNumber) {
        if #available(iOS 10, *) {
            UIApplication.shared.open(url, options: [:],
                                  completionHandler: {
                                    (success) in
                                    print("Open \(phoneNumber): \(success)")
            })
        } else {
            let success = UIApplication.shared.openURL(url)
            print("Open \(phoneNumber): \(success)")
        }
    }
} 


let number = "736XXXXXX"

let phoneNumber = "tel://\(number)"
call(phoneNumber: phoneNumber)

Try with that number to see if it's a bigger problem than the simple code :)

Upvotes: 2

Lê Minh
Lê Minh

Reputation: 56

You need to add the scheme 'tel' into your info.plist

<key>LSApplicationQueriesSchemes</key>
      <string>tel</string>

Then normal use:

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

Goodluck

Upvotes: 0

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