How to Open Settings App Programmatically

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 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 to open Setting page from my application

You can't use internal strings in order to open stuff, that's why your original code was rejected from the AppStore. Mainly, it's dangerous because those strings aren't documented and might change in the future without warning, breaking your application.

Sadly, you can't open any specific setting screen (you're only allowed to open your app settings), so you will have to stick to your new code (the one using UIApplicationOpenSettingsURLString), or replace it with instructions to the user to open the WiFi settings in the device.

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.

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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.
}

Opening Android Settings programmatically

I used the code from the most upvoted answer:

startActivityForResult(new Intent(android.provider.Settings.ACTION_SETTINGS), 0);

It opens the device settings in the same window, thus got the users of my android application (finnmglas/Launcher) for android stuck in there.

The answer for 2020 and beyond (in Kotlin):

startActivity(Intent(Settings.ACTION_SETTINGS))

It works in my app, should also be working in yours without any unwanted consequences.

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