Can't create URL from string with '#' in the middle

Question

Issue explained through code.

    let u = "tel://*111*11111111111#" //Works perfectly on iOS11 and later
    let u = "tel://*111#11111111111#" //Doesn't work, Can't create URL

    let app = UIApplication.shared
    if let url = URL(string:u) {
        if app.canOpenURL(url) {
            app.open(url, options: [:], completionHandler: { (finished) in 
            })
        }
    }

** If I encode the string then canOpenURL() fails !

Phone numbers with hash or asterisk characters may not be allowed in urls: stackoverflow.com/questions/4660951/… — paulvs
– paulvs, Commented Nov 1, 2017 at 19:48

Charles Srstka · Accepted Answer · 2017-11-01 19:47:24Z

1

Just use URLComponents. Handles percent escapes automatically for you and everything.

var comps = URLComponents()

comps.scheme = "tel"
comps.host = "*111#11111111111#"

print(comps.url!) // prints "tel://*111%2311111111111%23"

answered Nov 1, 2017 at 19:47

Charles Srstka

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Community · Accepted Answer · 2020-06-20 09:12:55Z

0

Problem fixed,

Number should be encoded separately without tel://

let u = "tel://*111#11111111111#"
if let encoded = u.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed) {
    let u = "tel://\(encoded)"
    if let url = URL(string:u) {
        if app.canOpenURL(url) {
            app.open(url, options: [:], completionHandler: { (finished) in
                
            })
        }
    }
}

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Nov 1, 2017 at 19:46

Husam

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4 Comments

paulvs Over a year ago

Does the call work even though you're percent-escaping the hash? The final URL will be tel://tel%3A%2F%2F*111%2311111111111%23.

Husam Over a year ago

Yes worked, the call prompt decoded the url automatically

Charles Srstka Over a year ago

IMO, URLComponents is a simpler and cleaner way to do this. URLComponents is also more up to date, since it conforms to RFC 3986, unlike URL(string:) which the documentation states to conform to older RFCs.

rmaddy Over a year ago

Why does the initial value of u contain tel://? Just encode actual number, not the scheme. And tel URLs do not use the //.

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