Swift didReceiveRemoteNotification - 导航到 rootviewcontroller,无论用户位于应用程序的何处(现在有错误消息)
Swift didReceiveRemoteNotification - Navigate to rootviewcontroller no matter where in app user is located (Now With Error Message)
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:
要么导航到屏幕(深层链接),这样我将不得不从 rootviewcontroller 导航到多个屏幕。
导航到 rootviewcontroller,无论用户位于应用程序的哪个位置。
我认为第二个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)
}
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)
需要创建一个变量:
var window: UIWindow
下面的代码是在 appdelegate 中实现的,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
}
})
我写了一个简单的 class 只需传递 class 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将也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 link: https://github.com/oblq/Navigator
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:
要么导航到屏幕(深层链接),这样我将不得不从 rootviewcontroller 导航到多个屏幕。
导航到 rootviewcontroller,无论用户位于应用程序的哪个位置。
我认为第二个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)
}
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)
需要创建一个变量: var window: UIWindow
下面的代码是在 appdelegate 中实现的,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
}
})
我写了一个简单的 class 只需传递 class 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将也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 link: https://github.com/oblq/Navigator