Open Settings app from another app programmatically in iPhone
Solution 1
Good news :
You can open settings apps programmatically like this (works only from iOS8 onwards).
If you are using Swift 3.0:
UIApplication.shared.open(URL(string: UIApplicationOpenSettingsURLString)!)
If you are using Objective-C:
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];
For other lower versions (less than iOS8) its not possible to programatically open the settings app.
Solution 2
As others answered, you cannot open the Settings from your app.
However You can solve the situation, like I have done:
Output a message that Location services must be enabled explaining why, and show the path in that message:
"Settings->Privacy->LocationServices"
Solution 3
Opening settings apps programmatically is possible only from iOS 8. So, use the following code...
if([CLLocationManager locationServicesEnabled]&&
[CLLocationManager authorizationStatus] != kCLAuthorizationStatusDenied)
{
//...Location service is enabled
}
else
{
if([[[UIDevice currentDevice] systemVersion] floatValue] < 8.0)
{
UIAlertView* curr1=[[UIAlertView alloc] initWithTitle:@"This app does not have access to Location service" message:@"You can enable access in Settings->Privacy->Location->Location Services" delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];
[curr1 show];
}
else
{
UIAlertView* curr2=[[UIAlertView alloc] initWithTitle:@"This app does not have access to Location service" message:@"You can enable access in Settings->Privacy->Location->Location Services" delegate:self cancelButtonTitle:@"OK" otherButtonTitles:@"Settings", nil];
curr2.tag=121;
[curr2 show];
}
}
- (void)alertView:(UIAlertView *)alertView clickedButtonAtIndex:(NSInteger)buttonIndex
{
if (alertView.tag == 121 && buttonIndex == 1)
{
//code for opening settings app in iOS 8
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];
}
}
Solution 4
Till iOS 5.0 it was possible to open settings
via the URL schema
, i.e
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"My Settings URL"]];
This has been deprecated from iOS 5.1 onwards.
Solution 5
Here is a Swift2 version that worked for me including an Alert that instructs the user in what to do when the settings opens.
func initLocationManager() {
locationManager = CLLocationManager()
locationManager.delegate = self
locationManager.desiredAccuracy = kCLLocationAccuracyBest
locationManager.requestAlwaysAuthorization()
// If there isn't a Lat/Lon then we need to see if we have access to location services
// We are going to ask for permission to use location if the user hasn't allowed it yet.
let status = CLLocationManager.authorizationStatus()
if(status == CLAuthorizationStatus.NotDetermined || status == CLAuthorizationStatus.Denied) {
//println(locationManager)
// check that locationManager is even avaiable. If so, then ask permission to use it
if locationManager != nil {
locationManager.requestAlwaysAuthorization()
//open the settings to allow the user to select if they want to allow for location settings.
let alert = UIAlertController(title: "I Can't find you.", message: "To let my App figure out where you are on the map click 'Find Me' and change your location to 'Always' and come back to MyMobi.", preferredStyle: UIAlertControllerStyle.Alert)
alert.addAction(UIAlertAction(title: "No Thanks", style: UIAlertActionStyle.Default, handler:nil))
alert.addAction(UIAlertAction(title: "Find Me", style: UIAlertActionStyle.Default, handler: {
(alert: UIAlertAction!) in
UIApplication.sharedApplication().openURL(NSURL(string: UIApplicationOpenSettingsURLString)!)
}))
self.presentViewController(alert, animated: true, completion: nil)
}
}
}
Comments
-
Edward Sagayaraj about 4 years
I have to open settings app from my app if gps is not enabled in iPhone. I have used the following code. It works well in iOS simulator but it does not work in iPhone. May I know is there any problem in this code.
if (![CLLocationManager locationServicesEnabled]) { int (*openApp)(CFStringRef, Boolean); void *hndl = dlopen("/System/Library/PrivateFrameworks/SpringBoardServices.framework/SpringBoardServices"); openApp = (int(*)(CFStringRef, Boolean)) dlsym(hndl, "SBSLaunchApplicationWithIdentifier"); openApp(CFSTR("com.apple.Preferences"), FALSE); dlclose(hndl); }
-
Oscar Gomez about 10 yearsThis is still true unfortunately =(... no idea why this got depreacted on iOS 5.1
-
Dipesh KC almost 10 yearsdoes this fallback gracefully for older versions?
-
nurnachman almost 10 yearsCan it open a specific settings screen?
-
Yatheesha almost 10 years@nume currently its not possible.
-
William Denniss over 9 yearsto make your code work on iOS 7, first check that
UIApplicationOpenSettingsURLString
exists like this: stackoverflow.com/a/25884389/72176 -
Tejas K over 8 years@nurne - No, the above code will open your app's settings.
-
Krešimir Prcela over 8 yearsIt is possible to open any settings screen on any iOS 7,8,9. Check this answer: stackoverflow.com/a/34024467/475978
-
user3404693 over 7 years@YatheeshaBL no more working in iOS 10. Any workaround?
-
Khadija Daruwala over 7 years@YatheeshaBL This solution is not working for iOS 10 . I am working on Xcode 8, swift 3. Any alternate solution?
-
Markus over 6 yearsIS POSSIBLE TO ACCESS? Storage & iCloud Usage -> Manage Storage
-
funct7 about 6 years
options:
and the trailingcompletionHandler:
are unnecessary