How to open your app in Settings iOS 11

Question

It seems that Apple has moved a lot of the app configurations to the App path with iOS 11, how to open the app path programmatically in Settings? I tried "App-Prefs:root=\(Bundle.main.bundleIdentifier!)" but this doesn't seem to work.

Please note that my question is specific to: How to open the app path in settings: NOT how to open the settings

Damien · Accepted Answer · 2021-01-11T16:25:25.250

63

Here is the code you're looking for, I guess:

if let url = URL(string: UIApplicationOpenSettingsURLString) {
    if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    }
}

And in addition, the updated version for swift 5 :

if let url = URL(string: UIApplication.openSettingsURLString) {
    if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    }
}

edited Jan 11 '21 at 16:25

answered Sep 26 '17 at 09:39

Damien

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1

Note: UIApplication.shared.open(url, options: [:], completionHandler: nil) is iOS 10+. – Andrea Leganza May 30 '18 at 06:50
1

The call to `canOpenURL` is not needed - the subsequent call to `open` just fails (it doesn't raise an exception or crashes the app). – Theo Dec 18 '18 at 13:39
Use `UIApplication.shared.openURL(url)` when targeting prior to IOS 10. – Mohamed Salah Jul 26 '19 at 07:23
This opens your app in the Settings. Great! – coolcool1994 Mar 14 '20 at 21:25
if let url = URL.init(string: UIApplication.openSettingsURLString) { UIApplication.shared.open(url, options: [:], completionHandler: nil) } – FontFamily Mar 26 '22 at 23:27

Alessandro Francucci · Answer 2 · 2019-03-05T08:55:17.927

16

Swift 4.2, iOS 12

Opening just the settings is possible with the function below:

extension UIApplication {

    ...

    @discardableResult
    static func openAppSettings() -> Bool {
        guard
            let settingsURL = URL(string: UIApplication.openSettingsURLString),
            UIApplication.shared.canOpenURL(settingsURL)
            else {
                return false
        }

        UIApplication.shared.open(settingsURL)
        return true
    }
}

Usage: UIApplication.openAppSettings()

But be careful to NOT use "non-public URL scheme", such as: prefs:root= or App-Prefs:root, because otherwise your app will be rejected. This happened to me recently since I was trying to have a deeplink to the wifi section in the settings.

edited Mar 05 '19 at 08:55

answered Oct 22 '18 at 10:39

Alessandro Francucci

1,528
17
25

2

You should probably name that `UIApplication.openAppSettings()` (notice the `s`) for compliance with `openSettingsURLString` – Markus Rautopuro Nov 06 '18 at 10:24
3

This take's you to app specific settings if they exist (like Siri usage, or contacts). IF they don't exist it takes you to the settings app and whatever page was last open. What it does not do, and what nobody has been able to figure out, is how to always open the Settings app, at the home screen, if your app has a settings page. – Kyle Browning Jan 02 '19 at 19:24
Swift 4.2 UIApplicationOpenSettingsURLString was renamed to UIApplication.openSettingsURLString – maslovsa Feb 28 '19 at 09:14

score 4 · Answer 3 · answered Aug 21 '18 at 10:12

And if you want to make it work for both, older and newer iOS-versions, then do:

if let url = URL(string:UIApplicationOpenSettingsURLString) {
    if UIApplication.shared.canOpenURL(url) {
        if #available(iOS 10.0, *) {
            UIApplication.shared.open(url, options: [:], completionHandler: nil)
        } else {
            UIApplication.shared.openURL(url)
        }
    }
}

score 2 · Answer 4 · edited Mar 08 '18 at 03:08

2

openURL has been deprecated since iOS 10, so I would advise you to use:

if let url = URL(string:UIApplicationOpenSettingsURLString) {
    if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.open(url, options: [:], completionHandler: { success in
            log.debug("Open app settings success: \(success)")
        })
    }
}

edited Mar 08 '18 at 03:08

Pang

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answered Oct 10 '17 at 11:10

skornos

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Hi. I didnt find other string that open wifi setting screen – kemdo Mar 26 '18 at 20:13

score 1 · Answer 5 · answered Nov 19 '20 at 07:25

UIApplicationOpenSettingsURLString has been renamed to UIApplication.openSettingsURLString.

if let url = URL(string: UIApplication.openSettingsURLString) {
    if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    }
}

score 0 · Answer 6 · answered Jan 05 '22 at 07:47

SWift 5

In some case we can not open App's setting after trying all the above. To solve this problem 100% just make sure these two step are followed

Step 1. Right click on the project file -> Add New File -> Add Settings.Bundle in project and edit according to your requirements.

Step 2. Now add some code in your buttons action.

if let url = URL(string: UIApplication.openSettingsURLString) {
    if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    }
}

Note: Using "prefs:Root" is forbidden by apple and your app will be rejected. So, avoid using this api.

How to open your app in Settings iOS 11

6 Answers6

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