openURL回调？

Question

用户拒绝位置，我将用户发送到“设置”应用中的位置设置：

UIApplication.sharedApplication().openURL(NSURL(string:"prefs:root=LOCATION_SERVICES")!)

用户从“设置”应用中授权位置，返回到我的应用，左上角为Back to App

我怎么知道他回到了应用程序？ viewDidAppear无效

Answer 1

您可以通过检查AppDelegate's方法轻松检测： -

func applicationWillEnterForeground(application: UIApplication!) {

或者在NSNotificationCenter的视图控制器的viewDidLoad（）中注册通知： -

NSNotificationCenter.defaultCenter().addObserver(self, selector: "applicationWillEnterForeground", name: UIApplicationWillEnterForegroundNotification, object: nil)


//calling selector method
 func applicationWillEnterForeground() {
            println("did enter foreground")
        }

Answer 2

尝试UIApplicationDelegate applicationDidBecomeActive。

Answer 3

在viewController中使用NSNotificationCenter

在viewDidLoad

上调用此方法

- (void)registerAppEnterForeground
{
    [[NSNotificationCenter defaultCenter]
     addObserver:self
     selector:@selector(applicationWillEnterForeground)
     name:UIApplicationWillEnterForegroundNotification
     object:nil];
}


-(void)applicationWillEnterForeground
{
 /*HANDLE YOUR CODE HERE*/
}