从远程通知打开 ViewController

Question

我尝试在我的应用捕捉到远程通知时打开特定的 ViewController。

让我展示一下我的项目架构。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，看到我的 pushViewController 方法已经到达，但不是我的 SimplePostViewController.

的 ViewWillAppear

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

解法：

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)
    }
 }

我使用：

child.childViewControllers[0]

因为我的示例中只有一个 child。

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 window 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
}

` 您需要担心管理您的视图层次结构并向它发送您需要从通知用户数据发送的任何内容。

在我的示例中，我创建了一个本地通知，当您关闭在查看几秒后触发的应用程序时。如果您随后从通知中启动该应用程序，它将打开 "other view controller"，在您的情况下将是 "SimplePostViewController"。

此外，请确保您在 didFinishLaunchWithOptions 中注册了远程通知。

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

从远程通知打开 ViewController

Open a ViewController from remote notification

push-notification

apple-push-notifications

appdelegate

swift