在iPhone中以编程方式从另一个应用程序打开设置应用

Question

如果iPhone中没有启用gps,我必须从我的应用程序打开设置应用程序.我使用了以下代码.它在iOS模拟器中运行良好,但在iPhone中不起作用.我可以知道此代码中是否有任何问题.

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);
    }

Answer 1

好消息 :

您可以像这样以编程方式打开设置应用程序(仅适用于iOS8以上版本).

如果您使用的是Swift 3.0:

UIApplication.shared.open(URL(string: UIApplicationOpenSettingsURLString)!)

如果您使用的是Objective-C:

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

对于其他较低版本(低于iOS8),无法以编程方式打开设置应用程序.

Answer 2

正如其他人所回答的那样,您无法从应用中打开"设置".

但是你可以解决这个问题,就像我做的那样:

输出一条消息,必须启用位置服务来解释原因,并在该消息中显示路径:

"设置 - >与隐私> LocationServices"

Answer 3

以编程方式打开设置应用程序只能从iOS 8开始.因此,请使用以下代码...

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]];
 }
}