I'm working on an app that displays a today extension with some information. When I tap on the today extension, it opens the app and navigates to a subview from the root to display the information. Normally the user would then click the back arrow to go back to the main view, but there is no way to tell if this is actually done. It is possible for the user to go back to the today extension and tap again. When this is done, the subview is opened once again with new information. If this is done a bunch of times, I end up with a bunch of instances of the subview and I have to click the back button on each of them to get back to the main view.
My question: Is it possible to check if the subview is already visible? I'd like to be able to just send updated information to it, instead of having to display an entirely new view.
I am currently handling this by keeping the instance of the UIViewController
at the top of my root. If it is not nil, then I just pass the information to it and redraw. If it is nil, then I call performSegue
and create a new one.
I just think that there must be a better way of handling this.
Edit: Thanks to the commenter below, I came up with this code that seems to do what I need.
if let quoteView = self.navigationController?.topViewController as? ShowQuoteVC {
quoteView.updateQuoteInformation(usingQuote: QuoteService.instance.getQuote(byQuoteNumber: quote))
}
else {
performSegue(withIdentifier: "showQuote", sender: quote)
}
This is different from the suggested post where the answer is:
if (self.navigationController.topViewController == self) {
//the view is currently displayed
}
In this case, it didn't work because I when I come in to the app from the Today Extension, it goes to the root view controller. I needed to check whether a subview is being displayed, and self.navigationController.topViewcontroller == self will never work because I am not checking to see if the top view controller is the root view controller. The suggestions in this post are more applicable to what I am trying to accomplish.