How to open your app's settings (inside the Settings app) with Swift (iOS 11)?
Oops, it seems it works in iOS 11.4.1:
if let url = URL(string:UIApplicationOpenSettingsURLString) {
if UIApplication.shared.canOpenURL(url) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
}
How to open your app in Settings iOS 11
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)
}
}
How do I open phone settings when a button is clicked?
Using UIApplication.openSettingsURLString
Update for Swift 5.1
override func viewDidAppear(_ animated: Bool) {
let alertController = UIAlertController (title: "Title", message: "Go to Settings?", preferredStyle: .alert)
let settingsAction = UIAlertAction(title: "Settings", style: .default) { (_) -> Void in
guard let settingsUrl = URL(string: UIApplication.openSettingsURLString) else {
return
}
if UIApplication.shared.canOpenURL(settingsUrl) {
UIApplication.shared.open(settingsUrl, completionHandler: { (success) in
print("Settings opened: \(success)") // Prints true
})
}
}
alertController.addAction(settingsAction)
let cancelAction = UIAlertAction(title: "Cancel", style: .default, handler: nil)
alertController.addAction(cancelAction)
present(alertController, animated: true, completion: nil)
}
Swift 4.2
override func viewDidAppear(_ animated: Bool) {
let alertController = UIAlertController (title: "Title", message: "Go to Settings?", preferredStyle: .alert)
let settingsAction = UIAlertAction(title: "Settings", style: .default) { (_) -> Void in
guard let settingsUrl = URL(string: UIApplicationOpenSettingsURLString) else {
return
}
if UIApplication.shared.canOpenURL(settingsUrl) {
UIApplication.shared.open(settingsUrl, completionHandler: { (success) in
print("Settings opened: \(success)") // Prints true
})
}
}
alertController.addAction(settingsAction)
let cancelAction = UIAlertAction(title: "Cancel", style: .default, handler: nil)
alertController.addAction(cancelAction)
present(alertController, animated: true, completion: nil)
}
Opening the Settings app from another app
As mentioned by Karan Dua this is now possible in iOS8 using UIApplicationOpenSettingsURLString
see Apple's Documentation.
Example:
Swift 4.2
UIApplication.shared.open(URL(string: UIApplication.openSettingsURLString)!)
In Swift 3:
UIApplication.shared.open(URL(string:UIApplicationOpenSettingsURLString)!)
In Swift 2:
UIApplication.sharedApplication().openURL(NSURL(string:UIApplicationOpenSettingsURLString)!)
In Objective-C
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];
Prior to iOS 8:
You can not. As you said this has been covered many times and that pop up asking you to turn on location services is supplied by Apple and not by the App itself. That is why it is able to the open the settings application.
Here are a few related questions & articles:
is it possible to open Settings App using openURL?
Programmatically opening the settings app (iPhone)
How can I open the Settings app when the user presses a button?
iPhone: Opening Application Preferences Panel From App
Open UIPickerView by clicking on an entry in the app's preferences - How to?
Open the Settings app?
iOS: You’re Doing Settings Wrong
Opening the Settings app from another app
As mentioned by Karan Dua this is now possible in iOS8 using UIApplicationOpenSettingsURLString
see Apple's Documentation.
Example:
Swift 4.2
UIApplication.shared.open(URL(string: UIApplication.openSettingsURLString)!)
In Swift 3:
UIApplication.shared.open(URL(string:UIApplicationOpenSettingsURLString)!)
In Swift 2:
UIApplication.sharedApplication().openURL(NSURL(string:UIApplicationOpenSettingsURLString)!)
In Objective-C
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];
Prior to iOS 8:
You can not. As you said this has been covered many times and that pop up asking you to turn on location services is supplied by Apple and not by the App itself. That is why it is able to the open the settings application.
Here are a few related questions & articles:
is it possible to open Settings App using openURL?
Programmatically opening the settings app (iPhone)
How can I open the Settings app when the user presses a button?
iPhone: Opening Application Preferences Panel From App
Open UIPickerView by clicking on an entry in the app's preferences - How to?
Open the Settings app?
iOS: You’re Doing Settings Wrong
Opening app's notification settings in the settings app
Updated 8 Dec, 2021:
This method will open Settings > Your App. It will show all available privacy toggles like camera, photos, notifications, cellular data, etc.
After a comment from @Mischa below, tested and updated the answer to this (more succinct):
if let appSettings = URL(string: UIApplication.openSettingsURLString), UIApplication.shared.canOpenURL(appSettings) {
UIApplication.shared.open(appSettings)
}
Previous answer:
I found the answer to this question (albeit helpful) has a bit too much assumed logic. Here is a plain and simple Swift 5 implementation if anyone else stumbles upon this question:
if let bundleIdentifier = Bundle.main.bundleIdentifier, let appSettings = URL(string: UIApplication.openSettingsURLString + bundleIdentifier) {
if UIApplication.shared.canOpenURL(appSettings) {
UIApplication.shared.open(appSettings)
}
}
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