当我从 Scheme Url（Swift 项目）启动应用程序时未调用 openURL

Question

当我通过myapp://param1=abc 打开我的应用程序时，不会调用 open url 函数。

正如 Apple 文档所述，我添加了函数 didFinishLaunchingWithOptions 和 willFinishLaunchingWithOptions 使其成为 true，但它仍然没有被调用。应用程序完美打开，但没有调用函数 open url 我无法获取 param1

这是我的AppDelegate 文件：

class AppDelegate: UIResponder, UIApplicationDelegate  {

    func application(_ application: UIApplication,
                     didFinishLaunchingWithOptions launchOptions: [UIApplication.LaunchOptionsKey : Any]? = nil) -> Bool {
        return true
    }

    func application(_ application: UIApplication, willFinishLaunchingWithOptions launchOptions: [UIApplication.LaunchOptionsKey : Any]? = nil) -> Bool {
        return true
    }

    func application(_ application: UIApplication, open url: URL, options: [UIApplication.OpenURLOptionsKey : Any] = [:]) -> Bool {
        // Here doesn't come! :(
        return true
    }


    // MARK: UISceneSession Lifecycle


    func application(_ application: UIApplication, configurationForConnecting connectingSceneSession: UISceneSession, options: UIScene.ConnectionOptions) -> UISceneConfiguration {
        // Called when a new scene session is being created.
        // Use this method to select a configuration to create the new scene with.
        return UISceneConfiguration(name: "Default Configuration", sessionRole: connectingSceneSession.role)
    }

    func application(_ application: UIApplication, didDiscardSceneSessions sceneSessions: Set<UISceneSession>) {
        // Called when the user discards a scene session.
        // If any sessions were discarded while the application was not running, this will be called shortly after application:didFinishLaunchingWithOptions.
        // Use this method to release any resources that were specific to the discarded scenes, as they will not return.
    }
}

知道会发生什么吗？

谢谢！

Answer 1

AppDelegate中openUrl方法的签名更新为，

func application(_ app: UIApplication, open url: URL, options: [UIApplication.OpenURLOptionsKey : Any] = [:]) -> Bool {
    return true
}

您需要在项目设置的信息选项卡下的URL类型中添加您的urlScheme，即

---

对于 Swift-5 和 iOS-13：

在SceneDelegate.swift文件中，需要实现下面的方法来处理urlSchemes，即

func scene(_ scene: UIScene, openURLContexts URLContexts: Set<UIOpenURLContext>) {
    print(URLContexts)
}

Answer 2

application(_ application: UIApplication, open url 不会在应用启动时调用。您需要在[UIApplication.LaunchOptionsKey.url] 的帮助下访问launchOptions: [UIApplication.LaunchOptionsKey: Any] 和您的网址

这里是可能对你有帮助的 sn-p

   func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplication.LaunchOptionsKey: Any]?) -> Bool {

        let url = launchOptions![UIApplication.LaunchOptionsKey.url]
        ...

        return true
    }

Answer 3

从 iOS 14 开始，您可以改用深层链接：

struct ShapifyApp: App {
    var body: some Scene {
        WindowGroup {
            ContentView()
                .onOpenURL { url in
                    //Whatever you need to do goes here
                }
        }
    }
}