当应用程序退出时显示远程通知的uiviewcontroller

我试图在接收远程推送通知时显示特定的viewcontroller。我已经添加了我所有的代码添加到方法didReceiveRemoteNotification：当应用程序退出时显示远程通知的uiviewcontroller

func application(application: UIApplication, didReceiveRemoteNotification userinfo: [NSObject: AnyObject])

我已经添加以下代码：

func application(application: UIApplication, didReceiveRemoteNotification userInfo: [NSObject : AnyObject]) { 
    let code = (userInfo["aps"] as! [String: AnyObject]) 
    // Call to retrieve blog 
    if let blog = code["b"] as? NSNumber { 
     let blogId = blog as! Int 

     // Show blog from notification 
     let mainStoryboard: UIStoryboard = UIStoryboard(name: "Main",bundle: nil) 
     var controller = mainStoryboard.instantiateViewControllerWithIdentifier("blogCtrl") as! BlogController 
     controller.blogId = blogId 
     var rootController = mainStoryboard.instantiateViewControllerWithIdentifier("navCtrl1") as! UINavigationController 
     self.window?.rootViewController = rootController 
     rootController.pushViewController(controller, animated: true) 
     self.window?.makeKeyAndVisible() 
    } 
    if let tonic = code["t"] as? NSNumber { 
     let mainStoryboard: UIStoryboard = UIStoryboard(name: "Main",bundle: nil) 
     var controller = mainStoryboard.instantiateViewControllerWithIdentifier("tonicDetail") as! TonicDetailController 
     controller.tonicId = tonic as! Int 
     var rootController = mainStoryboard.instantiateViewControllerWithIdentifier("navCtrl1") as! UINavigationController 
     self.window?.rootViewController = rootController 
     rootController.pushViewController(controller, animated: true) 
     self.window?.makeKeyAndVisible() 
    } 
    if let gin = code["g"] as? NSNumber { 
     let mainStoryboard: UIStoryboard = UIStoryboard(name: "Main",bundle: nil) 
     var controller = mainStoryboard.instantiateViewControllerWithIdentifier("GinDetail") as! GinDetailController 
     controller.ginId = gin as! Int 
     var rootController = mainStoryboard.instantiateViewControllerWithIdentifier("navCtrl1") as! UINavigationController 
     self.window?.rootViewController = rootController 
     rootController.pushViewController(controller, animated: true) 
     self.window?.makeKeyAndVisible() 
    } 
}

当应用程序在后台一切正常，但是当应用程序被quited，并我收到一个远程通知，它只启动应用程序。如果应用程序之前被退出，是否有可以调用的方法？

来源

2015-05-21 Hannes Van den Berghe

如果退出应用程序，在此期间你有远程通知，然后didFinishLaunchingWithOptions是在AppDelegate中第一种方法，其触发上启动应用程序，请检查您是否已经收到任何通知或不并据此执行您的操作后，你的应用程序启动。

你必须寻找这种方法在应用程序委托： -

-(BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions

现在检查你的应用是否已接收到推送通知或不

UILocalNotification *notification = [launchOptions objectForKey:UIApplicationLaunchOptionsRemoteNotificationKey]; 
    if (notification) { 
     NSLog(@"app received a notification %@",notification); 
     [self application:application didReceiveRemoteNotification:(NSDictionary*)notification]; 
    }else{ 
     NSLog(@"app did not receive a notification"); 
    }

来源

2015-05-21 13:49:25 Vizllx

在第一线你的代码创建了一个UILocalNotification，但是在第四行中，你将它用作NSDictionary。它是否正确？因为当我试着你的代码我有一个错误： 'var notification = launchOptions？[UIApplicationLaunchOptionsRemoteNotificationKey] as？ UILocalNotification 如果通知！= nil {“有一个通知”） self.application（application，didReceiveRemoteNotification：notification as！NSDictionary）}' Error说：'NSDictionary不能隐式转换为[NSObject：AnyObject] ]' –

@HannesVandenBerghe而不是'UILocalNotification'使用'NSDictionary'将工作 – Aanabidden

@Aababidden这确实是解决方案！谢谢！ –

当应用程序退出时显示远程通知的uiviewcontroller

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