Open Phone Settings Programmatically in iOS9

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

How to programmatically open the WIFI settings in Objective-C on iOS 10

This works fine on iOS 10,

Go to Targets --> (Application) --> Info --> URL Types --> +

In the URL Schemes write

prefs

See the image,

Then add the following code,

-(void)openWifiSettings{
    if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:@"prefs:root=WIFI"]]) {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=WIFI"]];
    } else {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"App-Prefs:root=WIFI"]];
    }
}

going to settings programmatically from local notification iOS 9

Try this:

  if let appSettings = NSURL(string: UIApplicationOpenSettingsURLString) {
      UIApplication.sharedApplication().openURL(appSettings)
  }

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