ios - 从 swift 3 中的消息扩展打开应用程序

Question

我正在尝试通过单击按钮从消息扩展中打开我的母亲应用程序。我在扩展中使用了这段代码:

    @IBAction func open(_ sender: UIButton) {

    let url = URL(string: "swiftexamples://")

    self.extensionContext?.open(url!, completionHandler: {(succes) in })


}

当我的母应用程序在后台运行时一切正常，但是当我想打开已关闭的应用程序时，它崩溃了。没有崩溃日志，我有类似的情况iOS app crashes when first opened by URL Scheme .我很确定我必须向应用程序委托(delegate)添加一些东西。我找到了函数 application(_:open:options:)。问题是，我现在不知道如何实现“选项”部分。我写了这个:

    func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {
    // Override point for customization after application launch.
     let url = launchOptions?[UIApplicationLaunchOptionsKey.url] as? NSURL 
       let sourceApp = launchOptions?[UIApplicationLaunchOptionsKey.sourceApplication] as? String
       let annotation = launchOptions?[UIApplicationLaunchOptionsKey.annotation] as? AnyObject

        self.application(application, open: url, options:[sourceApp: String, annotation:AnyObject] )



    return true
}

我还听说过 Universal Links，它们与 url schemes 具有相似的能力。是否可以使用通用链接实现我想要的目标？

Answer 1

您可以试试这个解决方案。它对我有用:

func openUrl(url: URL?) {
    let selector = sel_registerName("openURL:")
    var responder = self as UIResponder?
    while let r = responder, !r.responds(to: selector) {
        responder = r.next
    }
    _ = responder?.perform(selector, with: url)
}

func canOpenUrl(url: URL?) -> Bool {
    let selector = sel_registerName("canOpenURL:")
    var responder = self as UIResponder?
    while let r = responder, !r.responds(to: selector) {
        responder = r.next
    }
    return (responder!.perform(selector, with: url) != nil)
}

如 https://stackoverflow.com/a/44694703/2064473 中所提议