How to open Location services screen from setting screen?

Question

I want to open location service screen programmatically to turn on service.

Answer 1

I have tried all the above answers,it's not working on iOS11..it just opens settings page and not the app settings .. Finally this works..

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

Swift 4.2:

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

Refer: https://developer.apple.com/documentation/uikit/uiapplicationopensettingsurlstring?language=swift

Answer 2

Swift 4.2

Go straight to YOUR app's settings like this:

if let bundleId = Bundle.main.bundleIdentifier,
    let url = URL(string: "\(UIApplication.openSettingsURLString)&path=LOCATION/\(bundleId)") 
{
    UIApplication.shared.open(url, options: [:], completionHandler: nil)
}

Answer 3

You can open it directly like using below code,

But first set URL Schemes in Info.plist's URL Type Like:

Then write below line at specific event:

In Objective - C :

[[UIApplication sharedApplication] openURL:
 [NSURL URLWithString:@"prefs:root=LOCATION_SERVICES"]];

In Swift :

UIApplication.sharedApplication().openURL(NSURL(string: "prefs:root=LOCATION_SERVICES")!)

Hope this will help you.

Answer 4

Do you want to be safe? use UIApplicationOpenSettingsURLString, which will open the app settings, without deep-link.

Using App-prefs your app will be rejected, as many sub comments said. https://github.com/mauron85/cordova-plugin-background-geolocation/issues/394

Answer 5

Swift 5+
Easy Way Direct Your said application page will open

if let BUNDLE_IDENTIFIER = Bundle.main.bundleIdentifier,
    let url = URL(string: "\(UIApplication.openSettingsURLString)&path=LOCATION/\(BUNDLE_IDENTIFIER)") {
    UIApplication.shared.open(url, options: [:], completionHandler: nil)
}

Answer 6

First:

Add URL

Go to Project settings --> Info --> URL Types --> Add New URL Schemes

See image below:

Second:

Use below code to open Location settings:

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=LOCATION_SERVICES"]];

referred from: https://stackoverflow.com/a/35987082/5575752

Answer 7

Step 1: Click on project name >> target>> info >> url Types

Step 2:

-(IBAction)openSettingViewToEnableLocationService:(id)sender
{
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=LOCATION_SERVICES"]];
}

Answer 8

If you set locationManager.startUpdatingLocation() and you have disabled on your iphone, it automatically show you an alertView with the option to open and activated location.

Answer 9

This solution works for me, but I didn't know App Store will reject my app or not

if let bundleId = Bundle.main.bundleIdentifier,
    let url = URL(string: "\(UIApplication.openSettingsURLString)&root=Privacy&path=LOCATION/\(bundleId)") {
    UIApplication.shared.open(url, options: [:], completionHandler: nil)
}

Answer 10

Actually there's much simpler solution to that. It'll show your app settings with loction services/camera access, etc.:

func showUserSettings() {
    guard let urlGeneral = URL(string: UIApplicationOpenSettingsURLString) else {
        return
    }
    UIApplication.shared.open(urlGeneral)
}

Answer 11

Eureka: Forget about URLs. Use the method requestWhenInUseAuthorization from the CLLocationManager code to open the native prompt message. This is what Tinder and other apps do. Took me a while to figure it out.

This will open a prompt to take you EXACTLY to the Location Services screen:

let locationManager = CLLocationManager()
locationManager.requestWhenInUseAuthorization()

Answer 12

After adding prefs as a url type, use the following code to go directly to the location settings of an application.

if let url = URL(string: "App-prefs:root=LOCATION_SERVICES") {
     UIApplication.shared.open(url, options: [:], completionHandler: nil)
}