How to open Location services screen from setting screen?

Question

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

Answer 1

Location services App-Prefs:root=Privacy&path=LOCATION worked for me. When I tested on a device and not a simulator.

I won't list the things I tried that did not work, it's a long list.

Usage example that assumes either location services are disabled or permission is denied or not determined:

if !CLLocationManager.locationServicesEnabled() {
    if let url = URL(string: "App-Prefs:root=Privacy&path=LOCATION") {
        // If general location settings are disabled then open general location settings
        UIApplication.shared.openURL(url)
    }
} else {
    if let url = URL(string: UIApplicationOpenSettingsURLString) {
        // If general location settings are enabled then open location settings for the app
        UIApplication.shared.openURL(url)
    }
}

Answer 2

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 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

Swift 4.2

Go straight to YOUR app's settings like this. Don't forget to put in your bundle identifier -

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

Answer 5

SWIFT 4 tested:

Only way to avoid getting rejected and open Location Preferences of own app is:

if let bundleId = Bundle.main.bundleIdentifier,
   let url = URL(string: "\(UIApplication.openSettingsURLString)&path=LOCATION/\(bundleId)") {
        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

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 9

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 10

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)
}

Answer 11

🚨🚨

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