通过推送通知打开应用程序时打开特定的ViewController

本文介绍了通过推送通知打开应用程序时打开特定的ViewController的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

顾名思义，当用户通过滑动推送通知启动应用程序时，如何打开特定的ViewController?

As the title implies: how do I open a specific viewcontroller when a user starts the app by swiping on a push notification?

谢谢！

推荐答案

执行以下操作，

在您的应用主视图控制器的"viewDidLoad"方法中添加观察者，

in your apps main view controller "viewDidLoad" method add an observer,

NSNotificationCenter.defaultCenter().addObserver(self, selector: "SomeNotificationAct:", name: "SomeNotification", object: nil)

还要添加一个方法

func SomeNotificationAct(notification: NSNotification){ dispatch_async(dispatch_get_main_queue()) { self.performSegueWithIdentifier("NotificationView", sender: self) } }

并在您的应用的Appdelegate类的"didReceiveRemoteNotification"方法中添加以下代码

and add the below code in your "didReceiveRemoteNotification" method of Appdelegate class of your app

func application(application: UIApplication, didReceiveRemoteNotification userInfo: [NSObject : AnyObject]){ NSNotificationCenter.defaultCenter().postNotificationName("SomeNotification", object:nil, userInfo:someData) }