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）。

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（小于 iOS8 ），无法以编程方式打开设置应用。

Answer 2

在其他人回答时，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 - ＆GT;与隐私＆GT; LocationServices”

Answer 3

只能从iOS 8开始以编程方式打开设置应用。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

直到iOS 5.0，可以通过settings打开URL schema，即

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

从iOS 5.1起，这已被弃用。

Answer 5

这是一个适合我的Swift2版本，包括一个警告，指示用户在设置打开时该怎么做。

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

在iOS10.0中不推荐使用openURL：请使用openURL：options：completionHandler而不是

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

在iPhone中以编程方式从另一个应用程序打开设置应用

6 个答案: