在我的iOS 9应用中添加"快速操作"

Question

我想将iOS 9的快速操作添加到我的应用程序中.

我把这段代码放在我的app委托中:

import UIKit
enum ShortcutType: String {
    case NewScan = "QuickAction.NewScan"
    case Settings = "QuickAction.Settings"
}
@UIApplicationMain
class AppDelegate: UIResponder, UIApplicationDelegate {

    var window: UIWindow?
    static let applicationShortcutUserInfoIconKey = "applicationShortcutUserInfoIconKey"

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

        UIViewController.prepareInterstitialAds()

        if(UIApplication.instancesRespondToSelector(Selector("registerUserNotificationSettings:"))) {
            UIApplication.sharedApplication().registerUserNotificationSettings(UIUserNotificationSettings(forTypes: [.Alert, .Badge, .Sound], categories: nil))
        }

        // QUICK ACTIONS
            var launchedFromShortCut = false

            if #available(iOS 9.0, *) {
                if let shortcutItem = launchOptions?[UIApplicationLaunchOptionsShortcutItemKey] as? UIApplicationShortcutItem {
                    launchedFromShortCut = true
                    handleShortCutItem(shortcutItem)
                }
            } else {
                return true
            }
            return !launchedFromShortCut

    }

    /**************** QUICK ACTIONS ****************/
    @available(iOS 9.0, *)
    func application(application: UIApplication, performActionForShortcutItem shortcutItem: UIApplicationShortcutItem, completionHandler: Bool -> Void) {
            let handledShortCutItem = handleShortCutItem(shortcutItem)
            completionHandler(handledShortCutItem)
    }
    @available(iOS 9.0, *)
    func handleShortCutItem(shortcutItem: UIApplicationShortcutItem) -> Bool {
        var handled = false
        if let shortcutType = ShortcutType.init(rawValue: shortcutItem.type) {
            let rootNavigationViewController = window!.rootViewController as? UINavigationController
            let rootViewController = rootNavigationViewController?.viewControllers.first as UIViewController?
            rootNavigationViewController?.popToRootViewControllerAnimated(false)
            switch shortcutType {
                case .NewScan:

                    rootViewController?.performSegueWithIdentifier("goToCamera", sender: nil)
                    handled = true

                case.Settings:
                    rootViewController?.performSegueWithIdentifier("goToSettings", sender: nil)
                    handled = true
            }
        }
        return handled
    }
}

现在我可以强行触摸我的应用程序图标>将显示快速操作>我选择快速操作"新扫描">应用程序将打开并显示我已离开的最后一个视图.

但是segue不会被执行.

这是我的故事板的一部分:

说明:

答:导航控制器和初始控制器

B:ViewController,经过检查后,这将成为导航控制器C的一个segue

C:导航控制器

D:表视图控制器

E:ViewController

如果我选择快速操作的新扫描 - 我想显示ViewController E.

Answer 1

根据文档中的示例代码,您似乎正在正确地执行操作.但是,您的实现中有很多可选的链接handleShortCutItem:.您是否使用调试器来验证这些表达式都没有nil值？此外,从我所看到的(虽然图像模糊),该故事板中的第一个导航控制器的根视图控制器没有到E的segue.所以我不确定你打算如何到达那里.

我建议你在handleShortCutItem:实现中设置一个断点,首先验证你正在使用的值是不是nil,代码实际上是在执行.完成此操作后,您可以使用故事板来实例化所需的视图控件,只需创建一个数组就可以将视图控制器层次结构放在导航控制器中,并将导航控制器的viewControllers属性设置为此数组.同样,很难准确地说出你想要的图像,但也许是这样的:

func handleShortCutItem(shortcutItem: UIApplicationShortcutItem) -> Bool {
    guard let shortcutType = ShortcutType.init(rawValue: shortcutItem.type) else {
        return false
    }

    guard let rootNavigationController = window?.rootViewController as? UINavigationController else {
        return false
    }

    guard let rootViewController = rootNavigationController?.viewControllers.first else {
        return false
    }

    guard let storyboard = rootNavigationController.storyboard else {
        return false
    }

    var viewControllers = [rootViewController]
    switch shortcutType {
    case .NewScan:
        // Instantiate the necessary view controllers for this case
        viewControllers += [storyboard.instantiateViewControllerWithIdentifier("<#Identifier for some view controller#>")]
        ...
        viewControllers += [storyboard.instantiateViewControllerWithIdentifier("<#Identifier for some other view controller#>")]

    case.Settings:
        // Instantiate the necessary view controllers for this case
        viewControllers += [storyboard.instantiateViewControllerWithIdentifier("<#Identifier for some view controller#>")]
        ...
        viewControllers += [storyboard.instantiateViewControllerWithIdentifier("<#Identifier for some other view controller#>")]
    }

    // Set the new view controllers array
    rootNavigationController.setViewControllers(viewControllers, animated: false)

    return true
}

注意:由于您使用Swift2标记了此问题,因此我冒昧地调整代码以使用保护语句.