在 iPhone 中以编程方式从另一个应用程序打开设置应用程序

Question

如果 iPhone 中未启用 gps，我必须从我的应用程序中打开设置应用程序。我使用了以下代码。它在 iOS 模拟器中运行良好，但在 iPhone 中无法运行。我可以知道这段代码有什么问题吗？

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 及更高版本）。

如果您使用的是 Swift 3.0：

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

如果你使用的是 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

正如其他人回答的那样，您无法从您的应用中打开“设置”。

但是你可以像我一样解决这个问题：

输出必须启用位置服务的消息，解释原因，并在该消息中显示路径：

“设置->隐私->定位服务”

Answer 3

只能从 iOS 8 以编程方式打开设置应用程序。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]];
 }
}

Answer 4

在 iOS 5.0 之前，可以通过 URL schema 打开 settings，即

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"My Settings URL"]];

这已从 iOS 5.1 开始弃用。

Answer 5

这是一个适用于我的 Swift2 版本，其中包括一个提醒用户在设置打开时该做什么的提示。

func initLocationManager() {
    locationManager = CLLocationManager()
    locationManager.delegate = self
    locationManager.desiredAccuracy = kCLLocationAccuracyBest
    locationManager.requestAlwaysAuthorization()


// If there isn't a Lat/Lon then we need to see if we have access to location services
// We are going to ask for permission to use location if the user hasn't allowed it yet.
let status = CLLocationManager.authorizationStatus()
if(status == CLAuthorizationStatus.NotDetermined || status == CLAuthorizationStatus.Denied)  {

    //println(locationManager)

    //  check that locationManager is even avaiable.  If so, then ask permission to use it
    if locationManager != nil {
        locationManager.requestAlwaysAuthorization()

        //open the settings to allow the user to select if they want to allow for location settings.
        let alert = UIAlertController(title: "I Can't find you.", message: "To let my App figure out where you are on the map click 'Find Me' and change your location to 'Always' and come back to MyMobi.", preferredStyle: UIAlertControllerStyle.Alert)
        alert.addAction(UIAlertAction(title: "No Thanks", style: UIAlertActionStyle.Default, handler:nil))
        alert.addAction(UIAlertAction(title: "Find Me", style: UIAlertActionStyle.Default, handler: {
            (alert: UIAlertAction!) in
            UIApplication.sharedApplication().openURL(NSURL(string: UIApplicationOpenSettingsURLString)!)
        }))
        self.presentViewController(alert, animated: true, completion: nil)


    }
}
}

Answer 6

openURL 在 iOS10.0 中已弃用：请改用 openURL:options:completionHandler

let url = URL(string: UIApplicationOpenSettingsURLString)!
UIApplication.shared.open(url, options: [:]) { success in }