Set openAppWhenRun variable in AppIntent programatically.

As far as I can see openAppWhenRun is static, and is read and opens the app even before perform() is run (if openAppWhenRun is true that is). In other words, openAppWhenRun can not be dynamic. Your variable myNumber is always false at the time the App Intent is run so that is why your app never opens.

It would be interesting to see where you set, and changed, the myNumber variable.

You can't dynamically set openAppWhenRun. You need to create a new AppIntent with openAppWhenRun = true, and call it from you existing AppIntent using return .result(openIntent: NewAppletName()). The Dive into App Intents video has an example of this. at 17:19. 

Hello @Yodagama . I was curious how you implemented this.I have an HomeIntent whose openAppWhenRun value is false.Now i made another intent called OpenApp whose openAppWhenRun value is true.I returned this in perform function return .result(opensIntent: OpenApp() ).But still it does not solve the issue of imitating the dynamic value of showing the app in some cases and not showing in some as the other intent's open value remains always true.Any help regarding this is appreciated 😇

It appears there's a solution to a problem, but I've yet to find it. The One-Sec App seems to utilize precisely what is described here. Currently, I've implemented the suggested solution, which technically works by throwing an error, but it's not ideal as it could be irritating for users to encounter it repeatedly.

I've experimented with various methods, but nothing is working as intended: on one hand, it seems impossible to eliminate the error when attempting to use two different intents without throwing an error due to the mismatch in underlying types, and on the other hand, you can't catch the error to prevent the user from seeing it.

It's clear there must be a solution to this problem, but I've been unable to discover it. Any guidance or insights would be greatly appreciated.