iOS 9 how to invoke “Back to” feature programmatic

问题:

Is there an api to invoke "Back to" action programmatically in order to return to the app which called (openurl) mine?

回答1:

It is not possible for you to redirect back to the app

The reason is

The only possible way for launching another app programmatically(Exposed by Apple) is via URLScheme. That means the current app should know the URL scheme of the to-be-launched app. In your case the launched app does not not know the URL scheme of the launched app. So there is no way to go back to the previous app.