How to use tel: with * (star, asterisk) or # (hash, pound) on iOS?

Question

I'm trying to initiate a call on the iPhone with the tel url that has a * in it. It properly brings up the call dialog but drops back to safari when you click call.

<a href="tel:123*12">Test</a>

Answer 1

This documentation from Apple should be helpful:

To prevent users from maliciously redirecting phone calls or changing the behavior of a phone or account, the Phone application supports most, but not all, of the special characters in the tel scheme. Specifically, if a URL contains the * or # characters, the Phone application does not attempt to dial the corresponding phone number.

UPDATE (Jan 2, 2018): The information mentioned here may be outdated. Please refer to new documentation if Apple has relaxed these rules in their newer SDKs. Refer to Husam's answer.

Answer 2

iOS11 now allows us to call number with * or #

---

Swift Example code

    let number = "*111*12345#"
    let app = UIApplication.shared
    if let encoded = number.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed) {
        let u = "tel://\(encoded)"
        if let url = URL(string:u) {
            if app.canOpenURL(url) {
                app.open(url, options: [:], completionHandler: { (finished) in

                })
                return
            }
        }
    }

Answer 3

The approved answer is not correct, at least anymore. I've tested both as a web page and in app being able to dial using the special character # and *. What you do have to do if you wish to use those characters in either instance though is to encode them.

In HTML, # becomes %23 and * does not need to be escaped

If using Swift, you can encode your link and press it from an app using this function:

//number format example (commas are pauses): 123-456-7890,,555#,1234
fileprivate func callNumber(_ number: String) {
    let dialString = "telprompt://" + number
    if let escapedDialString = dialString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed) {
      if let dialURL = URL(string: escapedDialString) {
        UIApplication.shared.openURL(dialURL)
      }
    }
  }