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Swift UIApplication.sharedApplication().openURL for Telephone

edited December 2014 in iOS Devlopment

Hi everyone, I am just trying to make phone call like this. However, I can't call for second phone number. When I call, I got this error. May I know what is wrong?

fatal error: unexpectedly found nil while unwrapping an Optional value

            let phone = "tel://1818,,,1,,,344#,,,959533788" //This one ok
            let phone = "tel://18005333333,,,1#,,,421#,,,959533788" //This one not okay
            let url:NSURL = NSURL(string:phone)!;
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