ios - Swift 3 '[UIApplicationLaunchOptionsKey : Any]?' 不可转换为 '[String : NSString]'

Question

我有一个 TVOS 应用程序，它已从 Swift 2 转换为 Swift 3，但出现以下错误。我不确定如何让它静音。

'[UIApplicationLaunchOptionsKey:任何]？不能转换为“[String : NSString]”

它出现在这段代码中

appControllerContext.launchOptions["BASEURL"] = AppDelegate.TVBaseURL

        if let launchOptions = launchOptions as? [String: AnyObject] {
            for (kind, value) in launchOptions {
                appControllerContext.launchOptions[kind] = value
            }
        }

添加:

/*
Copyright (C) 2015 Hani Hamrouni. All Rights Reserved.


*/

import UIKit
import TVMLKit

@UIApplicationMain
class AppDelegate: UIResponder, UIApplicationDelegate, TVApplicationControllerDelegate {
    // MARK: Properties
    
    var window: UIWindow?
    
    var appController: TVApplicationController?
    
    //change the link to your host url
    
    static let TVBaseURL = "http://google.com"
    
    static let TVBootURL = "\(AppDelegate.TVBaseURL)js/application.js"

    // MARK: UIApplication Overrides
    
    func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {
        // Override point for customization after application launch.
        window = UIWindow(frame: UIScreen.main.bounds)
        
        /*
            Create the TVApplicationControllerContext for this application
            and set the properties that will be passed to the `App.onLaunch` function
            in JavaScript.
        */
        let appControllerContext = TVApplicationControllerContext()
        
        /*
            The JavaScript URL is used to create the JavaScript context for your
            TVMLKit application. Although it is possible to separate your JavaScript
            into separate files, to help reduce the launch time of your application
            we recommend creating minified and compressed version of this resource.
            This will allow for the resource to be retrieved and UI presented to
            the user quickly.
        */
        if let javaScriptURL = URL(string: AppDelegate.TVBootURL) {
            appControllerContext.javaScriptApplicationURL = javaScriptURL
        }
        
        appControllerContext.launchOptions["BASEURL"] = AppDelegate.TVBaseURL
        
        if let launchOptions = launchOptions {
            for (kind, value) in launchOptions {
                appControllerContext.launchOptions[kind.rawValue] = value as AnyObject
            }
        }
        

        appController = TVApplicationController(context: appControllerContext, window: window, delegate: self)
        
        return true
    }
    
    // MARK: TVApplicationControllerDelegate
    
    func appController(_ appController: TVApplicationController, didFinishLaunching options: [String: Any]?) {
        print("\(#function) invoked with options: \(options)")
    }
    
    func appController(_ appController: TVApplicationController, didFail error: Error) {
        print("\(#function) invoked with error: \(error)")
        
        let title = "Error Launching Application"
        //error message
        let message = error.localizedDescription
        let alertController = UIAlertController(title: title, message: message, preferredStyle:.alert )
        
        self.appController?.navigationController.present(alertController, animated: true, completion: { () -> Void in
            // ...
        })
    }
    
    func appController(_ appController: TVApplicationController, didStop options: [String: Any]?) {
        print("\(#function) invoked with options: \(options)")
    }
}

Answer 1

你最好按原样使用 [UIApplicationLaunchOptionsKey : Any]。

这是怎么回事？

    if let launchOptions = launchOptions {
        for (kind, value) in launchOptions {
            appControllerContext.launchOptions[kind.rawValue] = value
        }
    }

---

已更新

似乎 TVApplicationControllerContext 的属性 launchOptions 的类型是 [String: Any]，所以你不需要用 作为 AnyObject。