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")
})

UIApplicationOpenSettingsURLString这只会在你之前允许任何权限的情况下起作用。例如位置，照片，联系人，推送通知访问。因此，如果您没有用户的此类权限:

如果是iOS 10或以上版本，

它会打开设置，但随后会崩溃。原因是，你的应用设置里什么都没有。

下面的代码将在iOS设置中打开你的应用程序设置。

NSURL *url = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
if ([[UIApplication sharedApplication] canOpenURL:url]) {
    [[UIApplication sharedApplication] openURL:url];
}

由于设备不可用，我无法在iOS < 10上检查这一点。

此外，我可以从一些要点中找到下面的代码，它在iOS 10上也能正常工作。但我不确定这是否会得到苹果审查团队的批准。

https://gist.github.com/johnny77221/bcaa5384a242b64bfd0b8a715f48e69f

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")
})

添加到接受的答案:(来自苹果开发者文档)当你打开这个字符串(openSettingsURLString)构建的URL时，系统启动设置应用程序，并显示应用程序的自定义设置(如果有的话)。如果你想打开应用的设置，创建你自己的settings。bundle。

斯威夫特4

我更喜欢以更安全的方式打开设置，

if let settingUrl = URL(string:UIApplicationOpenSettingsURLString) {

    UIApplication.shared.open(settingUrl)
}
else {
    print("Setting URL invalid")
}

您可以使用下面的代码。

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