Opening the Settings app from another app

Question

Okay, I know that there are many question about it, but they are all from many time ago.

So. I know that it is possible because the Map app does it.

In the Map app if I turn off the localization for this app, it send me a message, and if I press okay, the "Settings App" will be open. And my question is, how is this possible? How can I open the "Setting app" from my own app?

Basically I need to do the same thing, if the user turn off the location for my app, then I'll show him a message saying something that will open the "Setting app"

Answer 1

You can use the below code for it.

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

Answer 2

iOS 10 update

Apple changed the method to open async on the main thread. However, from now it is only possible to open the app settings in native settings.

[[UIApplication sharedApplication] openURL:url options:@{} completionHandler:nil];

iOS 9 update

It is now possible to go directly to sub-settings menu. However, a URL scheme has to be created. It can be done using two ways:

1. XCode - You will find it in Target, Info, URL Scheme. Then, just type prefs.
2. Directly adding to *-Info.plist. Add the following: <key>CFBundleURLTypes</key> <array> <dict> <key>CFBundleTypeRole</key> <string>Editor</string> <key>CFBundleURLSchemes</key> <array> <string>prefs</string> </array> </dict> </array>

Then the code:

Swift

UIApplication.sharedApplication().openURL(NSURL(string:"prefs:root=General&path=Keyboard")!)

Objective-c

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=General&path=Keyboard"]];

Answer 3

Tested with iOS 10. Working

NSArray* urlStrings = @[@"prefs:root=WIFI", @"App-Prefs:root=WIFI"];
for(NSString* urlString in urlStrings){
NSURL* url = [NSURL URLWithString:urlString];
if([[UIApplication sharedApplication] canOpenURL:url]){
    [[UIApplication sharedApplication] openURL:url];
    break;
}
}

Happy Coding :)

Answer 4

Add this to your class,

 public class func showSettingsAlert(title:String,message:String,onVC viewController:UIViewController,onCancel:(()->())?){
            YourClass.show2ButtonsAlert(onVC: viewController, title: title, message: message, button1Title: "Settings", button2Title: "Cancel", onButton1Click: {
                if let settingsURL = NSURL(string: UIApplicationOpenSettingsURLString){
                    UIApplication.sharedApplication().openURL(settingsURL)
                }
                }, onButton2Click: {
                    onCancel?()
            })
        }

 public class func show2ButtonsAlert(onVC viewController:UIViewController,title:String,message:String,button1Title:String,button2Title:String,onButton1Click:(()->())?,onButton2Click:(()->())?){
            dispatch_async(dispatch_get_main_queue()) {
                let alert : UIAlertController = UIAlertController(title: title, message: message, preferredStyle: UIAlertControllerStyle.Alert)

                alert.addAction(UIAlertAction(title: button1Title, style:.Default, handler: { (action:UIAlertAction) in
                    onButton1Click?()
                }))

                alert.addAction(UIAlertAction(title: button2Title, style:.Default, handler: { (action:UIAlertAction) in
                    onButton2Click?()
                }))

                viewController.presentViewController(alert, animated: true, completion: nil)
            }
        }

Call like this,

YourClass.showSettingsAlert("App would like to access camera", message: "App would like to access camera desc", onVC: fromViewController, onCancel: {
  print("canceled")
})

Answer 5

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

Answer 6

In Swift 3 all I needed is this (here for example redirect to my app notifications):

if let url = URL(string: "App-Prefs:root=NOTIFICATIONS_ID&path=your app bundleID") {
    if #available(iOS 10.0, *) {
        UIApplication.shared.open(url, completionHandler: .none)
    } else {
        // Fallback on earlier versions
    }
}

Source: phynet gist.

This worked with me only when settings is in background. It will redirect you to your app notification settings but if settings wasn't running in the background it will just redirect you to notification settings in general.