1

在 IOS Swift 应用程序中收到推送通知后,我想根据通知中的内容做两件事:

  1. 要么导航到屏幕(深度链接),所以我必须从 rootviewcontroller 导航到几个屏幕。

  2. 导航到 rootviewcontroller,无论用户在应用程序中的哪个位置。

第二个我认为是第一个的先决条件。

我知道我需要在这两个函数中放置代码:

  • didReceiveRemoteNotification
  • didReceiveLocalNotification

错误消息:“UIViewcontroller?” 没有名为“navigationController”的成员

在文件 AppDelegate.swift 中:

func application(application: UIApplication, didReceiveRemoteNotification userInfo: [NSObject : AnyObject]) 
{
    println("didReceiveRemoteNotification")
    //Navigate to rootviewcontroller
    var rootViewController = self.window!.rootViewController
    let mainStoryboard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
    var setViewController = mainStoryboard.instantiateViewControllerWithIdentifier("CurrentShows")
        as ViewController

    //rootViewController.navigationController?
    //    .popToViewController(setViewController, animated: false)
}
4

3 回答 3

0

在您的方法中尝试此代码:

var rootViewController = self.window!.rootViewController
let mainStoryboard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)

var setViewController = mainStoryboard
    .instantiateViewControllerWithIdentifier("CurrentShows")
        as ViewController_CurrentShows

rootViewController
    .navigationController
    .popToViewController(setViewController, animated: false)
于 2015-02-23T11:56:00.387 回答
0

需要创建一个变量:var window: UIWindow

下面的代码在 appdelegate 中实现,在:func didBecomeActive()....

let storyboard = UIStoryboard(name: "Main", bundle: nil)
            let rootController = storyboard.instantiateViewControllerWithIdentifier("Login") as Login_ViewController

            self.window?.rootViewController?.dismissViewControllerAnimated(false, completion:
            {
                if self.window != nil
                {
                    self.window!.rootViewController = rootController
                }
            })
于 2015-03-11T20:32:19.697 回答
0

我编写了一个简单的类,只需传递类类型即可从一行代码中的任何位置导航到视图层次结构中的任何视图控制器,因此您将编写的代码也将与视图层次结构本身分离,例如:

Navigator.find(MyViewController.self)?.doSomethingSync()
Navigator.navigate(to: MyViewController.self)?.doSomethingSync()

..或者您也可以在主线程上异步执行方法:

Navigator.navigate(to: MyViewController.self) { (MyViewControllerContainer, MyViewControllerInstance) in
    MyViewControllerInstance?.doSomethingAsync()
}

这里是 GitHub 项目链接:https ://github.com/oblq/Navigator

于 2017-12-18T12:42:57.333 回答