I want to open location service screen programmatically to turn on service.
I want to open location service screen programmatically to turn on service.
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)
}
}
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.
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)!)
Refer:https://developer.apple.com/documentation/uikit/uiapplicationopensettingsurlstring?language=swift
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)
}
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)
}
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
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"]];
}
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)
}
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)
}