I have an issue when uploading iOS app with watchOS, when I upload the iOS app it's showing that there is no WatchOS app attached to it. I tested the app in the simulator and it is working fine, here is the targets for my app it is showing there are watchOS targets attached to it
every time I upload the iOS build it is showing there is no Apple Watch app attached. can anyone help me out to figure out what is the problem for this one
I could think this is it. Watch app for iOS is installed with the iOS app, stand-alone watch app is installed standalone via the watch app store
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Situation:
I have a Xcode project that included an iPhone app and a Watch App.
Goal:
Only release the Watch App
Issue at I Tunes Connect:
I get the message that I should upload product images for iPhone.
But I don't want to launch the IPHone app in it.
The Watch app works without iPhone fine.
Question:
Do I have the wrong project chosen in I Tunes Connect or must I delete targets in Xcode?
Snippet From Xcode
The iOS bundles had been inside the watch bundles.
We needed to separate them and it worked.
Is there a way to download an app container for a watchOS app running on a real device?
PS: for an iOS app, the one can use Xcode Devices and Simulators window, but there only iOS apps are listed and there is no way to check the apps installed on Apple Watch.
We have a app that was submitted as a universal app with support for all devices iPad and iPhone but we no longer want it to show in the store as a iPad app we want it to just be a iPhone only app.
This app was built with Appcelerator so i have changed the app tiapp.xml details to only support iPhone no longer universal, but it will not submitted to the store.
I can't upload the existing app because the family group in the bundle Id. it won't accept removing the iPad.
How can we remove the app from being listed as a iPad app really important as the app was not built with iPad in mind it would be fine if it was just a scaled up version of the app.
Any help please?
Developers who wish to issue updates, but remove device support, have
three choices:
Fix their app so that it can work on the devices they originally set
out to support. Target a newer version of iOS that requires a newer
device. Remove their app from the store, and upload the new app with a
different bundle ID.
I'm trying to test my apple watch glance for my iOS app using the simulator, however I'm getting some very strange behaviour. When I launch the iOS app in the simulator it works without problem and the app is fully functional. However , when I then try to launch the glance app for apple watch afterwards it shows up, but the iOS app no longer works (if I try to launch it from the simulator it immediately crashes) and the data that's in the app sharing group doesn't show up on the watch (but if I relaunch the iOS app simulator from Xcode it shows up there).
I've checked both apps and they are configured for the same sharing groups and that is definitely the case.
I just don't understand why running the iOS app and the apple watch app in the simulator won't work / behave correctly together?
Credit to #WhiteTiger for this answer:
if you're using watchos 2 appgroup the system does not work on the device, you should use wcsession as the documentation indicates.
Or look at updating to use the WatchConnectivity approach instead
I am not able to attach watch app process into debugger while running from iPhone app.
I am following below steps.
Run iPhone app from Xcode in iPhone.
Make app in background.
Open watch app in apple watch.
Now I am doing Debug -> Attach to process. in that list apple watch process is not showing..
Any help?
I have only played with the Simulator (watch not arrived), but you better launch the watch app using Xcode and then open the iPhone app manually.
Xcode will only attach to one of them, so may need to debug one at a time.
You should run the WatchKit app target from Xcode, then launch the iPhone app manually or from Xcode. Note that you can attach to the iPhone app while attached to the WatchKit app to debug both at the same time.