如何从快速4.2中的通知中传递url以便在webview中打开

Question

到目前为止，在我的委托中，我已经成功地从通知中获得了URL，我遇到的问题是如何将url传递给我的viewcontroller.swift，以便在webview中打开url。我已经尝试过堆栈中的几个示例，但它们似乎不能用于快速4.2。有谁可以帮我？对不起，我对斯威夫特不熟悉。

AppDelegate.swift

func application(_ application: UIApplication, didReceiveRemoteNotification userInfo: [AnyHashable : Any], fetchCompletionHandler completionHandler: @escaping (UIBackgroundFetchResult) -> Void) {

    let data = userInfo as! [String: AnyObject]

    let state = UIApplication.shared.applicationState

    if state == .background {
        // background
        print("==== Active Running ====")
        if let aps = data["aps"] {
            let url = aps["url"]
        }
    }
    else if state == .inactive {
        // inactive
        print("==== Inactive Running ====")
        if let aps = data["aps"] {
            let url = aps["url"]
        }
    }
    else if state == .active {
        // foreground
        print("==== Foreground Running ====")
        if let aps = data["aps"] {
            let url = aps["url"]
        }
    }
}

Answer 1

如果您要从头开始初始化该视图控制器，您只需在应用程序委托中这样做，在这里您可以通过以下方式获得URL：

        let storyBoard = UIStoryboard(name: "Main", bundle: nil)
        let vc = storyBoard.instantiateViewController(withIdentifier: "yourViewControllerID") as! YourViewController
        vc.url = url
        self.window?.rootViewController = vc

同样在YourViewController文件中，将您的url设置为：

var url = "" {
    didSet {
       //trigger your webView to start loading, you can also do it at viewDidAppear maybe.
       //example:
       let url = URL(string: url)
       let request = URLRequest(url: url)
       webView.loadRequest(request)
    }
}