来自@Yatheeshaless的回答:

你可以在iOS8中以编程方式打开设置应用程序，但在早期版本的iOS中不能。

迅速:

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

斯威夫特4:

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

Swift 4.2 (BETA):

if let url = NSURL(string: UIApplication.openSettingsURLString) as URL? {
    UIApplication.shared.open(url, options: [:], completionHandler: nil)
}

objective - c:

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

正如Karan Dua提到的，这现在可以在iOS8中使用UIApplicationOpenSettingsURLString，参见苹果的文档。

例子:

斯威夫特4.2

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

在Swift 3中:

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

在Swift 2中:

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

在objective - c中

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

iOS 8之前:

你不能。正如你所说，这已经被报道了很多次，弹出窗口要求你打开位置服务是由苹果公司提供的，而不是由应用程序本身。这就是为什么它能够打开设置应用程序。

以下是一些相关的问题和文章:

是否可以使用openURL打开设置应用程序?

以编程方式打开设置应用程序(iPhone)

当用户按下按钮时，如何打开设置应用程序?

iPhone:从应用程序打开应用程序首选项面板

通过单击应用程序首选项中的一个条目打开UIPickerView -如何?

打开设置应用程序?

iOS:你的设置是错误的

你可以在iOS 5.0及更高版本上使用这个:这个不再有效。

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

在iOS 10上测试。工作

NSArray* urlStrings = @[@"prefs:root=WIFI", @"App-Prefs:root=WIFI"];
for(NSString* urlString in urlStrings){
NSURL* url = [NSURL URLWithString:urlString];
if([[UIApplication sharedApplication] canOpenURL:url]){
    [[UIApplication sharedApplication] openURL:url];
    break;
}
}

快乐编码:)

斯威夫特3:

guard let url = URL(string: UIApplicationOpenSettingsURLString) else {return}
if #available(iOS 10.0, *) {
  UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
  // Fallback on earlier versions
  UIApplication.shared.openURL(url)
}

把这个加到你的课上，

 public class func showSettingsAlert(title:String,message:String,onVC viewController:UIViewController,onCancel:(()->())?){
            YourClass.show2ButtonsAlert(onVC: viewController, title: title, message: message, button1Title: "Settings", button2Title: "Cancel", onButton1Click: {
                if let settingsURL = NSURL(string: UIApplicationOpenSettingsURLString){
                    UIApplication.sharedApplication().openURL(settingsURL)
                }
                }, onButton2Click: {
                    onCancel?()
            })
        }

 public class func show2ButtonsAlert(onVC viewController:UIViewController,title:String,message:String,button1Title:String,button2Title:String,onButton1Click:(()->())?,onButton2Click:(()->())?){
            dispatch_async(dispatch_get_main_queue()) {
                let alert : UIAlertController = UIAlertController(title: title, message: message, preferredStyle: UIAlertControllerStyle.Alert)

                alert.addAction(UIAlertAction(title: button1Title, style:.Default, handler: { (action:UIAlertAction) in
                    onButton1Click?()
                }))

                alert.addAction(UIAlertAction(title: button2Title, style:.Default, handler: { (action:UIAlertAction) in
                    onButton2Click?()
                }))

                viewController.presentViewController(alert, animated: true, completion: nil)
            }
        }

像这样调用，

YourClass.showSettingsAlert("App would like to access camera", message: "App would like to access camera desc", onVC: fromViewController, onCancel: {
  print("canceled")
})