Open Settings App from Another App Programmatically in Iphone

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

How to open Settings programmatically like in Facebook app?

You can't, there is no API call to do this.

Only system dialogs, dialogs from Apple Frameworks, can open the settings app.
In iOS 5 there was a app url scheme to open the system dialog but Apple removed it later.

---

With the coming of iOS 8 you can open the settings dialog on your apps page.

if (&UIApplicationOpenSettingsURLString != NULL) {
   NSURL *url = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
    [[UIApplication sharedApplication] openURL:url];
}
else {
  // Present some dialog telling the user to open the settings app.
}

Open Settings app from another app programmatically in iPhone

Good news :

You can open settings apps programmatically like this (works only from iOS8 onwards).

If you are using Swift 3.0:

UIApplication.shared.open(URL(string: UIApplicationOpenSettingsURLString)!)

If you are using Objective-C:

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];

For other lower versions (less than iOS8) its not possible to programatically open the settings app.

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 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 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)
}

---

Related Topics