是否可以使用openURL打開設置應用程序？

Question

我知道一個應用程序可以使用此代碼啓動其他應用程序：[[UIApplication sharedApplication] openURL:appUrl];。我知道打開safari和郵件的URL方案，但是我做了一些搜索，並沒有發現任何關於settings.app的方案。

Answer 1

您可以在iOS 5.0 - 5.0.1版本中使用它。它在iOS 5.1中被棄用。

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs://"]]; 

Answer 2

您可以打開設置應用程序試試這個（只能從iOS8上起）。

如果您使用的斯威夫特：

UIApplication.sharedApplication().openURL(NSURL(string: UIApplicationOpenSettingsURLString)) 

如果您正在使用的Objective-C

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:UIApplicationOpenSettingsURLString]]; 

較低的其他版本（小於iOS8上）其不可能以編程方式打開設置應用。

Answer 3

開放設置應用程式編程是可能僅從的iOS 8.所以，從使用以下代碼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

夫特4版本：的

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