从远程通知打开 ViewController

Question

当我的应用收到远程通知时，我尝试打开特定的 ViewController。

让我展示我的项目架构。 这是我的故事板：

当我收到通知时，我想打开一个“SimplePostViewController”，所以这是我的 appDelegate ：

var window: UIWindow?
var navigationVC: UINavigationController?

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool {
    let notificationTypes: UIUserNotificationType = [UIUserNotificationType.Alert, UIUserNotificationType.Badge, UIUserNotificationType.Sound]
    let pushNotificationSettings = UIUserNotificationSettings(forTypes: notificationTypes, categories: nil)
    let storyboard = UIStoryboard(name: "Main", bundle: nil)

    self.navigationVC = storyboard.instantiateViewControllerWithIdentifier("LastestPostsNavigationController") as? UINavigationController
    application.registerUserNotificationSettings(pushNotificationSettings)
    application.registerForRemoteNotifications()
    return true
}

func application(application: UIApplication, didReceiveRemoteNotification userInfo: [NSObject : AnyObject]) {
    if let postId = userInfo["postId"] as? String {
        print(postId)

        let api = EVWordPressAPI(wordpressOauth2Settings: Wordpress.wordpressOauth2Settings, site: Wordpress.siteName)

        api.postById(postId) { post in
            if (post != nil) {
                self.navigationVC!.pushViewController(SimplePostViewController(), animated: true)
            } else {
                print("An error occurred")
            }
        }

    }
}

我在应用程序启动时保存我的 UINavigationViewController，并在收到通知时尝试推送一个新的 SimplePostViewController。但什么也没有发生。 我放置了断点并看到我的 pushViewController 方法已到达，但不是我的 SimplePostViewController 的 ViewWillAppear。

我还使用“whats new”视图添加执行我的 segue，但也没有任何反应。

解决方案：

for child in (self.rootViewController?.childViewControllers)! {
    if child.restorationIdentifier == "LastestPostsNavigationController" {
       let lastestPostsTableViewController = (child.childViewControllers[0]) as! LastestPostsTableViewController
       let simplePostVC = (self.storyboard?.instantiateViewControllerWithIdentifier("PostViewController"))! as! PostViewController

       simplePostVC.post = post
       lastestPostsTableViewController.navigationController?.pushViewController(simplePostVC, animated: true)
    }
 }

我用：

child.childViewControllers[0]

因为在我的示例中我只有一个孩子。

Answer 1

我创建了一个带有本地通知而不是远程通知的示例项目，以便于显示功能，但它应该像在应用程序委托 didreceiveremote 通知中设置窗口的根视图控制器一样简单。

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool {

    // Subscribe for notifications - assume the user chose yes for now
    application.registerUserNotificationSettings(UIUserNotificationSettings(forTypes: [.Alert, .Badge, .Sound], categories: nil))

    return true
}


func applicationDidEnterBackground(application: UIApplication) {
    //Crete a local notification
    let notification = UILocalNotification()
    notification.alertBody = "This is a fake notification"
    notification.fireDate  = NSDate(timeIntervalSinceNow: 2)
    UIApplication.sharedApplication().scheduleLocalNotification(notification)
}

func application(application: UIApplication, didReceiveLocalNotification notification: UILocalNotification) {
    let sb = UIStoryboard(name: "Main", bundle: nil)
    let otherVC = sb.instantiateViewControllerWithIdentifier("otherVC") as! OtherViewController
    window?.rootViewController = otherVC;
}

func application(application: UIApplication, didReceiveRemoteNotification userInfo: [NSObject : AnyObject]) {
    //Your code here
}

` 您需要担心管理视图层次结构并将需要从通知用户数据发送的任何内容发送给它。

在我的示例中，当您关闭在查看几秒钟后触发的应用程序时，我会创建一个本地通知。如果您随后从通知中启动应用程序，它将打开“其他视图控制器”，在您的情况下为“SimplePostViewController”。

另外，请确保您在 didFinishLaunchWithOptions 中注册远程通知。

Github 非常简单的示例：https://github.com/spt131/exampleNotificationResponse