I noticed a few PassKit wallet passes can open their originating app, right from the pass. The iOS app opens when the pass icon is clicked, for example in Delta Airlines passes.
I can't seem to find any documentation or examples on this. How can one add this to a PKPass?
You can use the appLaunchURL key in the passes’ pass.json.
https://developer.apple.com/library/archive/documentation/UserExperience/Reference/PassKit_Bundle/Chapters/TopLevel.html#//apple_ref/doc/uid/TP40012026-CH2-SW1
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I'm currently working on an iOS app, Swift 4, that automatically launches on click of a link to block users from accidentally navigating to that link.
How would I set this up?
For example, YouTube opens up youtube.com links.
Sorry if this is a vague question, if more info is required feel free to ask ^^
If you create a custom URL protocol like myapp://somepath (where myapp is the protocol) then you just register that custom protocol, and when the user clicks such a link it’ll automatically open your app. You can’t intercept a general purpose URL protocol like HTTP or YouTube.
(At least not without OS support. That's how YouTube and the app store are able to open HTTP links.)
I haven't done anything like this so far but it seems like you should take a look at Apple URL Scheme and Universal Links:
https://developer.apple.com/library/content/featuredarticles/iPhoneURLScheme_Reference/Introduction/Introduction.html
https://coderwall.com/p/mtjaeq/ios-custom-url-scheme
https://developer.apple.com/library/content/documentation/General/Conceptual/AppSearch/UniversalLinks.html
https://www.raywenderlich.com/128948/universal-links-make-connection?utm_source=raywenderlich.com%20Weekly&utm_campaign=95f29cf4fc-raywenderlich_com_Weekly5_31_2016&utm_medium=email&utm_term=0_83b6edc87f-95f29cf4fc-415182745
I am implementing firebase dynamic links in my iOS app and I can already parse the link, redirect to AppStore etc. Now I want to distinguish the first run of the app, when user installs it from the dynamic link - I want to skip the intro and show him the content that is expected to be shown.
Is there some parameter, that I could catch in application(_:didFinishLaunchingWithOptions:) so I could say that it was launched thru the dynamic link?
The method application(_:continueUserActivity:userActivity:restorationHandler:) is called later, so the intro is already launched.
This case is difficult to test, because you have to have your app published on the AppStore.
You actually don't need to have the app published in the App Store for this to work — clicking a link, closing the App Store, and then installing an app build through Xcode (or any other beta distribution platform like TestFlight or Fabric) has exactly the same effect.
According to the Firebase docs, the method that is called for the first install is openURL (no, this makes no sense to me either). The continueUserActivity method is for Universal Links, and is only used if the app is already installed when a link is opened.
I am not aware of any way to detect when the app is opening for the first time after install from a 'deferred' link, but you could simply route directly to the shared content (skipping the intro) whenever a deep link is present. If a deep link is NOT present, show the regular intro.
Alternative Option
You could check out Branch.io (full disclosure: I'm on the Branch team). Amongst other things, Branch is a great, free drop-in replacement for Firebase Dynamic Links with a ton of additional functionality. Here is an example of all the parameters Branch returns immediately in didFinishLaunchingWithOptions:
{
"branch_view_enabled" = 0;
"browser_fingerprint_id" = "<null>";
data = "{
\"+is_first_session\":false,
\"+clicked_branch_link\":true,
\"+match_guaranteed\":true,
\"$canonical_identifier\":\"room/OrangeOak\",
\"$exp_date\":0,
\"$identity_id\":\"308073965526600507\",
\"$og_title\":\"Orange Oak\",
\"$one_time_use\":false,
\"$publicly_indexable\":1,
\"room_name\":\"Orange Oak\", // this is a custom param, of which you may have an unlimited number
\"~channel\":\"pasteboard\",
\"~creation_source\":3,
\"~feature\":\"sharing\",
\"~id\":\"319180030632948530\",
\"+click_timestamp\":1477336707,
\"~referring_link\":\"https://branchmaps.app.link/qTLPNAJ0Jx\"
}";
"device_fingerprint_id" = 308073965409112574;
"identity_id" = 308073965526600507;
link = "https://branchmaps.app.link/?%24identity_id=308073965526600507";
"session_id" = 319180164046538734;
}
You can read more about these parameters on the Branch documentation here.
Hmm... as far as I'm aware, there's not really anything you can catch in the application:(_:didFinishLaunchingWithOptions) phase that would let you know the app was being opened by a dynamic link. You're going to have to wait until the continueUserActivity call, as you mentioned.
That said, FIRDynamicLinks.dynamicLinks()?.handleUniversalLink returns a boolean value nearly instantly, so you should be able to take advantage of that to short-circuit your into animation without it being a bad user experience. The callback itself might not happen until several milliseconds later, depending on if it's a shortened dynamic link (which requires a network call) or an expanded one (which doesn't).
How to Open an iOS App using Firebase Dynamic Links and Pass or get Parameters To an App Via Custom URL Scheme in iOS(swift)?
for eg :- https://q3tyj.app.goo.gl/abcd
My URL Scheme is ‘q3tyj.app.goo.gl’ in iOS app in Url Types.
If I type q3tyj.app.goo.gl in safari, I am able to open the application. But if I type q3tyj.app.goo.gl with some extra parameter like https://q3tyj.app.goo.gl/abcd in safari, , I am not able to open the application.
please also explain me how to get “link” parameter (which associated with dynamic link) from dynamic link in iOS app ( Swift ) .
I followed steps which were mentioned in Firebase.google.com for iOS ( Swift ).but its not working.
Thanks,
Nirav Virpara
It's hard to know exactly where your error is, but a few things to help you debug:
If you go to https://q3tyj.app.goo.gl/apple-app-site-association, you should see some JSON that points to your app. If you don't see this, make sure you've entered your team ID and your App Store ID in the project settings in the Firebase console
Make sure you're using your Bundle ID, not the shortlink domain, as your custom URL scheme in Xcode
Make sure you've enabled Associated Domains in the Capabilities tab of your Xcode project, and your domain looks like applinks:q3tyj.app.goo.gl
Universal Links (and, therefore, Dynamic Links) generally don't work if you type them directly into the Safari address bar. Instead, try typing the URL into an app like Notes and then clicking on them from there.
Good luck!
If you type https://q3tyj.app.goo.gl/apple-app-site-association this link in your browser and check, you will end up seeing a json, meaning your apple-app-site-association is correctly configured on your google website. However, you need to make sure that teamid in your google console or appid is correct.
I implemented Universal Links in our iOS 9 app and they work by calling a method in AppDelegate.swift, in which I get an NSUserActvity with an URL attached to it.
Is there a way to get the (HTTP-) referer? I need to know on which website the user has tapped the link that opened the app.
You should be able to use NSUserActivity > referrerURL instance property. See https://developer.apple.com/documentation/foundation/nsuseractivity/2875762-referrerurl.
No, there is no way to get the referrer.
An Apple employee made that pretty clear on the developer forums: https://forums.developer.apple.com/thread/65423
They say that the only way to get some kind of referrer would be to append it to the URL. Maybe take a look Google's campaign tracking URLs (utm_...).
I found another interesting solution if you need to support iOS 9 and 10 - https://appmetrica.yandex.com/blog/referrer-based-tracking-dlya-ios-tochnaya-atributsiya-dlya-lyubogo-istochnika-trafika.
Since iOS 11 I guess can be used NSUserActivity > referrerURL.
In short:
SFSafariViewController uses the same cookies as Safari app.
You can open invisible SFSafariViewController inside your app.
If your site contains some cookie - use Universal Link to pass it back to the app.
Though it requires to implement something like tracking link on your site.
I am having trouble figuring out how to switch to Safari from a native app in iOS 7+. I've used UIApplication.sharedApplication.openURL(), but that opens a new tab. I would like to return the user to the current page he/she was viewing before without opening a new tab. I found this SO post, but it is a few years old, so I was hoping things have changed since then.
Here is the workflow I am envisioning:
User taps on a link on an HTML page on Safari to open/install my app
User performs an action on my app
After the user is done performing the action, my app opens Safari automatically, and the user is back on the page where he/she left off
Google has somehow done this with their Google Maps app. If you search for an address on google.com on Safari, you can tap on the map that appears in the search results, and it will open the Maps app. At the top of the Maps app will be a "Return to Safari" bar that you can tap. Once you tap it, you are returned to Safari without loading another tab. I can't seem to find anything regarding how Google did this. If I can replicate that behavior in my app, that would work just fine.
Any help would be greatly appreciated!
There is a way to accomplish what you want using standard iOS APIs. No need to use external components.
You control your webpage and your app, so you know the exact URL that has the link to your app.
These are the steps:
1) In your app, define a custom URL scheme. In this case let's assume you use the scheme myawesomeapp://. You can do this in your Xcode project by going to the Info section of your target. See below
2) In your web page you need to handle the two scenarios: app installed / not installed. It is just a matter of detecting if an app responds to the scheme myawesomeapp://.
To detect from your webpage if your app is not installed please refer to this post
I will explain the case where your app is already installed.
Let's say the webpage that contains the link is:
http://www.mywebsite.com/mypage.html#mytag
The link you provide in your webpage should pass some parameters to the app and one of these should be the URL that you want the app to return. Following with the example the link could be:
myawesomeapp://?action=my_action_1&sourceurl=http%3A%2F%2Fwww.mywebsite.com%2Fmypage.html%23mytag
Note that the URL you pass as a parameter inside the scheme has to be URL encoded or it won't work properly.
3) In your app delegate you need to implement the method:
- (BOOL)application:(UIApplication *)application openURL:(NSURL *)url sourceApplication:(NSString *)sourceApplication annotation:(id)annotation
In this method, parse the URL, decode the query items and pass the sourceURL to the view controller responsible of handling the action prior to calling it. In this case I set a public property in the ViewController that will store the URL.
#property (nonatomic, strong) NSURL *sourceURL;
4) In the view controller when the user finishes the interaction, you just call:
[[UIApplication sharedApplication] openURL:self.sourceURL];
Because self.sourceURL contains the URL of your webpage, Safari will be launched with the URL. However, because that page is already opened, iOS detects this and re-opens that page.
I have a sample project in my Github page that implements all this.
And to finalize, after you install the sample project in your iPhone, open this stack overflow post from mobile Safari and open my awesome app
Once the app is opened, click on the button and you will return to this stack overflow post.
The behaviour you described is exactly what FB's AppLinks is designed for, and you'll get the same behaviour with all iOS apps that support it (which is quite a lot) out of box!
By the way Google Maps uses the same component: you can see it if you open Google Maps from let's say Fantastical.app!