Firebase 动态链接未打开应用程序 iOS 14

Question

I am attempting to create dynamic links to work as part of a referral system for my app. I believe I am successfully creating the links, however whenever I click on one from my notes app while running my app (which I am told should just open the app) I am directed to a Safari page then the App Store. This leads me to beleive the dynamic like for whatever reason isn't being handled at all by my app.

I am using a custom domain URI prefix, specifically https://mywebsite.com/invite and here is the complete code for creating my dynamic link

            guard let uid = Auth.auth().currentUser?.uid else {return}
            let link = URL(string: "https://mywebsite.com/invite/?invitedby=\(uid)")
            let referralLink = DynamicLinkComponents(link: link!, domainURIPrefix: "https://mywebsite.com/invite")
            
            referralLink?.iOSParameters = DynamicLinkIOSParameters(bundleID: "myBundleId")
            referralLink?.iOSParameters?.minimumAppVersion = "1.0"
            referralLink?.iOSParameters?.appStoreID = "123456789"
            
            referralLink?.shorten { (shortURL, warnings, error) in
                guard let referURL = shortURL else {
                    Service.showAlert(on: self, style: .alert, title: "Link Error", message: "We were unable to retreive your referral link at this time.")
                    return
                }
                let activityVC = UIActivityViewController(activityItems: [referURL], applicationActivities: nil)
                UIApplication.shared.windows.first?.rootViewController?.present(activityVC, animated: true, completion: nil)
            }

My code that I took from the Firebase documentation that allegedly is intended to handle the incoming dynamic link is as follows.

    func application(_ app: UIApplication, open url: URL, options: [UIApplication.OpenURLOptionsKey : Any]) -> Bool {
        return application(app, open: url, sourceApplication: options[UIApplication.OpenURLOptionsKey.sourceApplication] as? String, annotation: "")
    }
    
    func application(_ application: UIApplication, continue userActivity: NSUserActivity, restorationHandler: @escaping ([UIUserActivityRestoring]?) -> Void) -> Bool {
        let handled = DynamicLinks.dynamicLinks().handleUniversalLink(userActivity.webpageURL!) { (dynamicLink, error) in
            if (dynamicLink != nil) && !(error != nil) {
                self.handleDynamicLink(dynamicLink)
            }
        }
        return handled
    }

    func application(_ application: UIApplication, open url: URL, sourceApplication: String?, annotation: Any) -> Bool {
        if let dynamicLink = DynamicLinks.dynamicLinks().dynamicLink(fromCustomSchemeURL: url) {
            return handleDynamicLink(dynamicLink)
        }
        return false
    }

In addition to my code, I have ensured that in my info.plist I have listed "https://mywebsite.com/invite" under FirebaseDynamicLinksCustomDomains keys, I have added my apps bundle as a URL Type, and have added applinks:mywebsite.com to my apps Associated Domains.

I have checked multiple other posts and I'm not quite sure what is going wrong here. If someboyd could shed some light on any potential issues that would be great.

Answer 1

You are creating link wrong.

You need to set up:

let linkBuilder = DynamicLinkComponents(link: link, domainURIPrefix: "Firebase short link goes here")

let iOSParams = DynamicLinkIOSParameters(bundleID: "IOS app bundle id goes here")

iOSParams.customScheme = "url scheme" // for ios // forandroid firebase have different property