我正在尝试通过单击按钮从消息扩展名打开我的母亲应用程序。我在扩展中使用了以下代码:

    @IBAction func open(_ sender: UIButton) {

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

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


}

当我的母亲应用程序运行并进入后台时,一切正常,但是当我要打开关闭的应用程序时,它崩溃。没有崩溃日志,我对此iOS app crashes when first opened by URL Scheme有类似的情况。
我很确定我必须向应用程序委托中添加一些内容。
我找到了函数application(_:open:options :)。问题是,我现在不怎么实现“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
}

我也听说过通用链接,它们具有与url方案相似的功能。是否可以使用通用链接来实现我想要的功能?

最佳答案

您可以尝试此解决方案。它为我工作:

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)
}


https://stackoverflow.com/a/44694703/2064473中所建议

关于ios - 通过Swift 3中的邮件扩展程序打开应用,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/42737059/

10-12 00:18
查看更多