Open Setting App from my applicaiton in iPhone iOS 7

Question

I want to open iphone default setting application form my application, Please any one tell me what I have to do. OR is there any code to open it.

I try following code but it dont works.in iOS 7.1.

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

Thanks in Advance.

This works in iOS 5.0 and below only – Jayprakash Dubey Aug 08 '14 at 11:42 — Jayprakash Dubey, Aug 08 '14 at 11:42

score 2 · Answer 1 · answered Nov 10 '14 at 14:35

2

For iOS8 and later use :

// Send the user to the Settings for this app
//iOS 8 only
NSURL *settingsURL = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
[[UIApplication sharedApplication] openURL:settingsURL];

answered Nov 10 '14 at 14:35

Husam

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The question is specifically for iOS 7. – Ben Lachman May 04 '15 at 19:51
Also, OP asked for how to go to the default settings page, whereas the method provided here brings you to an app-specific settings page. – Lev Jun 24 '15 at 12:27
1

I successfully submitted an app using this method last month. – matt.writes.code May 20 '16 at 16:46

score 0 · Answer 2 · edited May 23 '17 at 12:16

0

Since iOS 5.1, there is no official way to open Settings via App.

Here to Go.

edited May 23 '17 at 12:16

Community

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answered Apr 04 '14 at 06:05

Rushabh

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I need to open it in ios7. – Jignesh Patel Apr 04 '14 at 06:09
but you can't do it below iOS 5.1 version :) – Rushabh Apr 04 '14 at 06:11
are you getting deprecated warning using your code. – Rushabh Apr 04 '14 at 06:25
1

This changed in iOS 8. However for iOS 7 there doesn't seem to be a public way to do this. – Ben Lachman May 04 '15 at 19:51

score 0 · Answer 3 · answered Sep 18 '14 at 07:55

You can use private API to open Settings App from your application. But i'm not sure if it will get accepted from apple.

void (*openApp)(CFStringRef, Boolean);
void *hndl = dlopen("/System/Library/PrivateFrameworks/SpringBoardServices.framework/SpringBoardServices", 0);
openApp = dlsym(hndl, "SBSLaunchApplicationWithIdentifier");
openApp(CFSTR("com.apple.Preferences"), FALSE);

score 0 · Answer 4 · answered Jan 22 '18 at 20:26

0

Swift 4

let url = URL(string: UIApplicationOpenSettingsURLString)
    if UIApplication.shared.canOpenURL(url!) {
       UIApplication.shared.openURL(url!)
    }

answered Jan 22 '18 at 20:26

Faisal Alneela

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Open Setting App from my applicaiton in iPhone iOS 7

4 Answers4