openURL 回调? [英] openURL Callback?

问题描述

用户拒绝定位，我将用户发送到设置"应用上的定位设置:

User refuses location, I send user to Location settings on Settings app:

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

用户从设置"应用授权位置，返回到我的应用，左上角带有 返回应用

User authorizes location from Settings app, returns to my app with Back to App on top left

我怎么知道他回到了应用程序?viewDidAppear 不起作用

How do I know that he came back to the app? viewDidAppear doesn't work

推荐答案

您可以通过检查AppDelegate的方法轻松检测:-

You can easily detect by checking AppDelegate's method:-

func applicationWillEnterForeground(application: UIApplication!) {

或者通过NSNotificationCenter在你的视图控制器的viewDidLoad()中注册一个通知:-

Or by registering a notification in your view controller's viewDidLoad() by NSNotificationCenter:-

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


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

这篇关于openURL 回调?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！