123456789" - error "(nulll)"

Asked Jun 08 '16 at 13:10

Active Jun 08 '16 at 17:22

Viewed 2,022 times

I am new to iOS development and I ran into a problem where I tried multiple solutions but couldn't fix it.

So what I want to do is, when I click on a button, make a phone call.

I already added this to my info.plist:

<key>LSApplicationQueriesSchemes</key>
<array>
    <string>telprompt</string>
    <string>tel</string>
</array>

And this is my code for making the phone call:

func doCall() {
    let url: NSURL = NSURL(string: "tel://1234567890")!
    if UIApplication.sharesApplication().canOpenURL(url) {
        UIApplication.sharesApplication().openURL(url)
    } esle {
        print("Call failed")
    }
}

So when I actually click on button, the following error shows up in the console:

-canOpenURL: failed for URL: "tel://1234567890" - error: "(null)"

And when I remove theif statement I am not getting that error but nothing happens. Also there is no phone app on the simulator, maybe that's the problem but I don't know either if that is resolvable.

edited Jun 08 '16 at 17:22

rmaddy

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asked Jun 08 '16 at 13:10

Jonh Doe

4

are trying this in a simulator ? This only works in a device that has the ability to call. – Teja Nandamuri Jun 08 '16 at 13:17
@TejaNandamuri yes i am trying it in the simulator, i don't own a apple device either, so i won't be able to test it or is there another way? – Jonh Doe Jun 08 '16 at 13:28
4

There is no another way. You cannot test some api like call,camera... without a real device. – Teja Nandamuri Jun 08 '16 at 13:29
By the way, tel: Uris don't have the // just `tel:123456789` – Paulw11 Jun 08 '16 at 13:41

canOpenURL: failed for URL: "tel://123456789" - error "(nulll)"

0 Answers0