How to programatically open Settings/Preferences window in a macOS SwiftUI App

Question

I have created a simple macOS-only SwiftUI app and added a Settings screen as per Apple's instructions:

import SwiftUI

@main
struct MyApp: App {
    var body: some Scene {
        WindowGroup {
            RootView()
        }
        Settings {
            SettingsView()
        }
    }
}

It works: the Preferences option shows up in the app menu. Now, I would also like to open this settings window programatically, e.g. when a button is clicked. Is there a way to achieve this?

I was hoping there would be a method in NSApplication, similar to the About window, but there doesn't seem to be one.

score 39 · Answer 1 · edited Jun 19 '23 at 07:24

39

By inspecting the app menu, I found this solution:

NSApp.sendAction(Selector(("showPreferencesWindow:")), to: nil, from: nil)

I assume that it may not pass App Review for using a private API, though, so less shady alternatives are still welcome.

Update: This has actually passed App Review, so I had marked it as the answer.

Update 2: For a macOS 13 solution, see Manny's answer.

Update 3: For macOS 14 with SwiftUI, you must use the new SettingsLink view documented here. NSApp.sendAction no longer works.

edited Jun 19 '23 at 07:24

greybeard

2,249
8
30
66

answered Dec 18 '20 at 11:56

Joshua

1,349
15
26

3

Can confirm in 2022, works like a charm – Ross Moody May 15 '22 at 18:05
4

I'm on macOS Ventura and this doesn't work anymore – Hieu Dinh Oct 15 '22 at 06:31
1

I got it working thanks to https://developer.apple.com/forums/thread/711940?answerId=726331022#726331022 – Hieu Dinh Oct 15 '22 at 06:35
1

On macOS Sonoma DB1, that would not work anymore, you will get an error "Please use SettingsLink for opening the Settings scene.", so you need to use SettingsLink – outcoldman Jun 12 '23 at 16:00
@outcoldman, If you post this as an answer, I‘ll gladly mark it as correct. – Joshua Jun 13 '23 at 16:50

score 30 · Answer 2 · answered Jun 29 '22 at 14:45

On macOS 13+ versions the action name has been renamed to Settings. So to update Joshua's answer, here is the solution to use:

NSApp.sendAction(Selector(("showSettingsWindow:")), to: nil, from: nil)

I wonder if there is a settings scene id we can use with the new openWindow environment method. :)

score 9 · Accepted Answer · answered Mar 12 '23 at 10:17

9

On macOS 13+, I use this solution:

if #available(macOS 13, *) {
    NSApp.sendAction(Selector(("showSettingsWindow:")), to: nil, from: nil)
} else {
    NSApp.sendAction(Selector(("showPreferencesWindow:")), to: nil, from: nil)
}

answered Mar 12 '23 at 10:17

Alex

124
1
4

score 0 · Answer 4 · answered Jul 18 '23 at 14:59

My take on the answers which have already been given is to probe the application delegate for selector response. This has worked well for me so far:

extension NSApplication {
    
    func openSettings() {
        guard let delegate = NSApplication.shared.delegate else { return }
        
        // macOS 13 Ventura
        var selector = Selector(("showSettingsWindow:"));
        if delegate.responds(to: selector) {
            delegate.perform(selector, with: nil, with: nil);
            return;
        }
        
        // macOS 12 Monterrey
        selector = Selector(("showPreferencesWindow:"));
        if delegate.responds(to: selector) {
            delegate.perform(selector, with: nil, with: nil);
            return;
        }
    }
    
}

How to programatically open Settings/Preferences window in a macOS SwiftUI App

4 Answers4