OSX Swift open URL in default browser

Question

How to open a URL in the system default browser by using Swift as programming language and OSX as platform?

I found a lot with UIApplication like:

UIApplication.sharedApplication().openURL(NSURL(string: object.url))

but this works just on iOS and not on OSX

And the Launch Services, I found has no examples for swift and there is a lot deprecated for OSX 10.10.

I guess it's because we're supposed to use the new Extensions instead... — Michele De Pascalis, Nov 02 '14 at 22:31

Leo Dabus · Accepted Answer · 2019-04-13T17:02:40.333

158

Swift 3 or later

import Cocoa

let url = URL(string: "https://www.google.com")!
if NSWorkspace.shared.open(url) {
    print("default browser was successfully opened")

}

edited Apr 13 '19 at 17:02

answered Nov 03 '14 at 01:01

Leo Dabus

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Technically this needs an "if let url = ...", since URL(string:) returns a URL?. Otherwise, great, thank you very much! – Apollo Grace Jan 12 '23 at 13:06
@ApolloGrace well if you know the string will never mutate and it is a valid url you can safely force unwrap the initializer result `URL(string: "https://www.google.com")!` therefore in this case `url` is not optional. `url` type is `URL` – Leo Dabus Jan 12 '23 at 17:26

score 56 · Answer 2 · edited Aug 23 '19 at 13:24

56

For MacOS, you can use this:

let url = URL(string: "https://www.stackoverflow.com")!
NSWorkspace.sharedWorkspace().openURL(url))

For iOS, you can use the following:

let url = NSURL(string: "https://google.com")!
UIApplication.sharedApplication().openURL(url)

You have to unwrap NSURL.

edited Aug 23 '19 at 13:24

Leo Dabus

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answered Apr 27 '15 at 19:34

Connor Knabe

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10

Why does this have 23 upvotes? Off topic to the question and should be removed. – xandermonkey Dec 10 '16 at 21:19
3

This is misleading, as it does not work for the platform used in the question. – KoCMoHaBTa Nov 09 '17 at 21:20
6

Blame Google that indexed this page this way and all searches for iOS lead hare ... and it's useful :) – Lachezar Todorov Feb 12 '18 at 16:11
If the URL is nil because the string couldn't be a valid URL, this code crashes. You need to be careful in Swift when you're using exclamation points, or the whole program crashes. – Kaydell Feb 21 '18 at 23:12
1

@Kaydell If you're using a string literal or any compile-time constant to create the url, then force-unwrapping it is perfectly fine. In these cases, there's no point in adding any error-handling code; if the value is nil, then that just indicates you made a typo. – Peter Schorn Jul 02 '20 at 03:01

score 15 · Answer 3 · edited Jun 12 '18 at 00:57

15

macOS:

NSWorkspace.sharedWorkspace().openURL(NSURL(string: "https://google.com")!)

iOS:

UIApplication.sharedApplication().openURL(NSURL(string: "https://google.com")!)

edited Jun 12 '18 at 00:57

pkamb

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answered Feb 17 '16 at 06:16

Alvin George

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64

Good reference for earlier versions of Swift (I'm on 2.3) – Stefano Buora Aug 24 '17 at 15:54
2

Please describe that the second code block is for iOS – Lachezar Todorov Feb 12 '18 at 16:11

Dylan · Answer 4 · 2016-11-01T15:44:11.737

11

When using Swift 3, you can open a webpage in the default browser using the following:

NSWorkspace.shared().open(NSURL(string: "https://google.com")! as URL)

In the accepted answer above, you can also check a URL using Swift 3 by inputting the following:

if let checkURL = NSURL(string: "https://google.com") {
    if NSWorkspace.shared().open(checkURL as URL) {
        print("URL Successfully Opened")
    }
} else {
    print("Invalid URL")
}

I hope that this information helps whomever it applies to.

edited Nov 01 '16 at 15:44

answered Oct 18 '16 at 15:34

Dylan

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As this is Swift 3, I don't think NSURL is necessary any more - just use URL: NSWorkspace.shared().open(URL(string: "https://google.com")!) and if let checkURL = URL(string: "https://google.com") { if NSWorkspace.shared().open(checkURL) { print("URL Successfully Opened") } } else { print("Invalid URL") } – gepree May 03 '17 at 13:14

score 8 · Answer 5 · answered Jan 02 '17 at 18:06

Just a bonus. If you want to open a URL in a specific browser(even other client who can handle that URL), here is the Swift 3 code tested on Xcode 8.2.1 and macOS 10.12.2.

/// appId: `nil` use the default HTTP client, or set what you want, e.g. Safari `com.apple.Safari`
func open(url: URL, appId: String? = nil) -> Bool {
  return NSWorkspace.shared().open(
    [url],
    withAppBundleIdentifier: appId,
    options: NSWorkspaceLaunchOptions.default,
    additionalEventParamDescriptor: nil,
    launchIdentifiers: nil
  )
}

hcarrasko · Answer 6 · 2019-05-22T04:59:48.290

8

For Swift 5, Xcode 10 and MAC OS:

NSWorkspace.shared.open(NSURL(string: "http://www.lichess.org")! as URL)

edited May 22 '19 at 04:59

answered May 22 '19 at 03:09

hcarrasko

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4

Just use URL, not both URL and NSURL. For example: `if let url = URL(string: "https://www.lichess.org") { NSWorkspace.shared.open(url) }` – gepree Aug 01 '19 at 15:23

score 7 · Answer 7 · answered Jan 04 '18 at 20:08

7

xCode 9 update

let url = URL(string: "https://www.google.com")!

UIApplication.shared.open(url, options: [:], completionHandler: nil)

answered Jan 04 '18 at 20:08

gannonbarnett

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6

This is for iOS and not macOS – Grumme Aug 03 '18 at 13:26

score 4 · Answer 8 · answered Oct 31 '18 at 08:01

4

MacOS Xcode 10 Swift 4.2 update

NSWorkspace.shared.open(URL(string: "https://www.google.com")!)

answered Oct 31 '18 at 08:01

Xavier Lasne

41
1

OSX Swift open URL in default browser

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