是否可以使用openURL打开设置应用程序？

Question

我知道一个应用程序可以使用此代码启动其他应用程序：[[UIApplication sharedApplication] openURL:appUrl];。我知道打开safari和邮件的URL方案，但是我做了一些搜索，并没有发现任何关于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中被弃用。

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

Answer 2

您可以打开设置应用程序试试这个（只能从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]]; 

较低的其他版本（小于iOS8上）其不可能以编程方式打开设置应用。

Answer 3

开放设置应用程式编程是可能仅从的iOS 8.所以，从使用以下代码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) 
}