Swift 3 Open Link

Question

I'm trying to run this function when a button is tapped:

@IBAction func openLink(_ sender: UIButton) {
    let link1 = "https://www.google.com/#q="
    let link2 = birdName.text!
    let link3 = link2.replacingOccurrences(of: " ", with: "+") //EDIT
    let link4 = link1+link3
    guard
        let query = link4.addingPercentEncoding( withAllowedCharacters: .urlQueryAllowed),
        let url = NSURL(string: "https://google.com/#q=\(query)")
        else { return }
    UIApplication.shared.openURL(URL(url))
}

However, the last line is flagged as "cannot call value of non-function type "UIApplication". This syntax is from here, so I'm not sure whats going on.

That causes the app to crash when the button is pressed "unexpectedly found nil while unwrapping an optional value" — Alex, Sep 09 '16 at 02:02
Make sure you are using the proper url link. You might have to percent escape your query string — Leo Dabus, Sep 09 '16 at 02:35
I've tried the link on a browser, and the link is good. What is percent escape? — Alex, Sep 09 '16 at 14:25
I've edited my code in, could you let me know what I'm doing wrong? I'm getting a "argument labels do not match any available overlords" on UIApplication... line — Alex, Sep 12 '16 at 23:58
try Swift 3 syntax `UIApplication.shared.open(url, options: [:], completionHandler: nil)` — Leo Dabus, Sep 13 '16 at 00:33
BTW Looks like you are adding https://google.com/#q twice to your string. and adding percent to the whole link instead of just the query string — Leo Dabus, Sep 13 '16 at 00:35
Okay syntax is updated. I'm getting the same error as before on let url = NSURL(query) — Alex, Sep 13 '16 at 00:46
should be `let url = URL(string: "https://google.com/#q=\(query)")` URL not NSURL — Leo Dabus, Sep 13 '16 at 00:47
Possible duplicate of [OpenURL in iOS10](http://stackoverflow.com/questions/38964264/openurl-in-ios10) — KSigWyatt, Jan 24 '17 at 22:00

Leo Dabus · Accepted Answer · 2017-02-24T12:28:49.503

10

Use guard to unwrap the textfield text property, replacing the occurrences, add percent encoding to the result and create an URL from the resulting string:

Try like this:

guard
    let text = birdName.text?.replacingOccurrences(of: " ", with: "+"),
    let query = text.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed),
    let url = URL(string: "https://google.com/#q=" + query)
else { return }
if #available(iOS 10.0, *) {
    UIApplication.shared.open(url)
} else {
    UIApplication.shared.openURL(url)
}

edited Feb 24 '17 at 12:28

answered Sep 13 '16 at 00:56

Leo Dabus

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I really appreciate people like you helping people like us with our not so educated bugs and questions. You're literally helping us make a living! – mfaani Sep 13 '16 at 18:27

Swift 3 Open Link

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