is it possible to open Settings App using openURL? - iphone

Question

I know an app can launch other apps by using this code: [[UIApplication sharedApplication] openURL:appUrl];. And I know the scheme of URL to open safari and mail, but I did some searches and found nothing about the scheme of settings.app.

Answer 1

You can open settings apps programmatically try this(works only from iOS8 onwards).
If you are using Swift:
UIApplication.sharedApplication().openURL(NSURL(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 settings app.

Answer 2

You can use this in iOS versions 5.0 - 5.0.1. It was then deprecated in iOS 5.1.
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:#"prefs://"]];

Answer 3

Opening settings apps programmatically is possible only from iOS 8. So, use the following code from http://code-ios.blogspot.in/2014/10/opening-settings-app-from-another-app.html
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
{
NSLog(#"buttonIndex:%d",buttonIndex);
if (alertView.tag == 121 && buttonIndex == 1)
{
//code for opening settings app in iOS 8
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]];
}
}

Answer 4

Swift 4 version:
if let url = URL(string: UIApplicationOpenSettingsURLString) {
UIApplication.shared.openURL(url)
}