Flutter IOS 通用链接打开应用程序，但无法导航到正确的页面

Question

我已经在 IOS 的 flutter 项目上设置了通用链接。

就像标题所示，当我单击与我的网站相关的链接时，我的应用程序确实会打开，但它不会导航到正确的页面。它只是打开应用程序。 我没有使用 uni_links 包，而是使用了指南的组合（包括官方文档）：

https://developer.apple.com/videos/play/wwdc2019/717/

https://nishbhasin.medium.com/apple-universal-link-setup-in-ios-131a508b45d1

https://www.kodeco.com/6080-universal-links-make-the-connection

我已将我的 apple-app-site-association 文件设置为如下所示：

{
    "applinks": {
        "details": [
            {
                "appIDs": [
                    "XXXXXXX.com.my.appBundle"
                ],
                "componenents": [
                    {
                        "/": "/*"
                    }
                ]
            }
        ]
    }
}

我已将其添加到我的 info.plist 文件中：

<key>FlutterDeepLinkingEnabled</key>
<true/>

我的 AppDelegate.swift 文件看起来像：

import UIKit
import Flutter
import Firebase

@UIApplicationMain
@objc class AppDelegate: FlutterAppDelegate {
  override func application(_ application: UIApplication, continue userActivity: NSUserActivity, 
    restorationHandler: @escaping ([UIUserActivityRestoring]?) -> Void) -> Bool {
    // This will allow us to check if we are coming from a universal link
    // and get the url with its components
    // The activity type (NSUserActivityTypeBrowsingWeb) is used
    // when continuing from a web browsing session to either
    // a web browser or a native app. Only activities of this
    // type can be continued from a web browser to a native app.
    guard userActivity.activityType == NSUserActivityTypeBrowsingWeb,
      let url = userActivity.webpageURL,
      let components = URLComponents(url: url, resolvingAgainstBaseURL: true) else {
        return false
    }
    // Now that we have the url and its components,
    // we can use this information to present
    // appropriate content in the app
    return true
  }

  override func application(
    _ application: UIApplication,
    didFinishLaunchingWithOptions launchOptions: [UIApplication.LaunchOptionsKey: Any]?
  ) -> Bool {
    FirebaseApp.configure()
    GeneratedPluginRegistrant.register(with: self)
    return super.application(application, didFinishLaunchingWithOptions: launchOptions)
  }
}

我的跑步者权利也已正确设置，例如：

<key>com.apple.developer.associated-domains</key>
<array>
    <string>applinks:www.example.com</string>
    <string>applinks:*.example.com</string>
</array>

问题是，如果我单击 www.example.com/mypath 的超链接，它不会到达 /mypath 处理的页面/路线，而只是打开应用程序。

我的路由是使用go_router完成的：^5.2.4

请问有人知道为什么会这样吗？我被这个挡住了我见过类似的问题，但没有一个答案对我有用。如有任何帮助，我们将不胜感激。

Answer 1

好吧，明白了。苹果官方文档要求在 AppDelegate.swift 文件中添加此函数的变体：

func application(_ application: UIApplication, continue userActivity: NSUserActivity, 
    restorationHandler: @escaping ([UIUserActivityRestoring]?) -> Void) -> Bool {
    // This will allow us to check if we are coming from a universal link
    // and get the url with its components
    // The activity type (NSUserActivityTypeBrowsingWeb) is used
    // when continuing from a web browsing session to either
    // a web browser or a native app. Only activities of this
    // type can be continued from a web browser to a native app.
    guard userActivity.activityType == NSUserActivityTypeBrowsingWeb,
      let url = userActivity.webpageURL,
      let components = URLComponents(url: url, resolvingAgainstBaseURL: true) else {
        return false
    }
    // Now that we have the url and its components,
    // we can use this information to present
    // appropriate content in the app
    return true
  }

似乎与处理通用链接的flutter框架冲突。取出该函数并将其放在我的 info.plist 中就可以了（其他一切都保持不变）：

<key>FlutterDeepLinkingEnabled</key>
<true/>

Flutter 文档尚未解决此问题（在发布此答案时），因此如果人们感兴趣，我可以写一篇关于必要步骤的小文章。

Answer 2

当您处理动态链接时，您将在以下函数的 userActivity 参数中获取通用链接和其他数据。

func application(_ application: UIApplication, continue userActivity: NSUserActivity, restorationHandler: @escaping ([UIUserActivityRestoring]?) -> Void) -> Bool {
    if let incomingURL = userActivity.webpageURL {
        debugPrint("incoming url is", incomingURL)
        let link = DynamicLinks.dynamicLinks().shouldHandleDynamicLink(fromCustomSchemeURL: incomingURL)
        print(link)
        let linkHandle = DynamicLinks.dynamicLinks().handleUniversalLink(incomingURL) { link, error in
            guard error == nil else {
                print("Error found.")
                return
            }
            if let dynamicLink = link {
                self.handleDynamicLinks(dynamicLink)
            }
        }
        if linkHandle {
            return true
        } else {
            return false
        }
    }
    return false
}

解析来自另一个函数的数据，或者您也可以在上面的代码中解析。就我而言，我解析了下面函数中的代码。

func handleDynamicLinks(_ dynamicLink: DynamicLink) {
    guard let link = dynamicLink.url else {
        return
    }
    
    if let landingVC = self.window?.rootViewController as? LandingViewController {
       // Do you your handling here with any controller you want to send or anything.
    }
          // example you are getting ID, you can parse it here

    if let idString = link.valueOf("id"), let id = Int.init(idString) {
        print(id)
    }
}

当您从链接中获取详细信息时，您可以简单地获取导航控制器或 VisibleController，然后可以推送到所需的流程。

Answer 3

只需将其添加到 info.plist 对我来说就足够了

<key>FlutterDeepLinkingEnabled</key>
<true/>