Back and forth, removeFromSuperView
Im a bit confused...
I have a mainview, which by the click of a button shows a subview. What I want to do next is to be able to switch back and forth between the Main and the Sub, but when I do this, the return button FROM the subView makes the TO SubView button disabled...I cant seem to be able to access the Subview a second time. Nothing happens.
What I have is a UIButton
, and an IBAction
, which I have written the "[self.view removeFromSuperView];
" in.
Please do not get me wrong here, I do get back to the mainView. The problem is that I cant go back to the subview afterwards.
I've read something in the documentation about retaining the subView...But I did not quite understand how to do it.
开发者_开发百科Could anyone please shed some light over this?
Thanks!
this is just a guess, since you don't provide much code:
- you might first need to retain the view. in case it's only retained by its current superview
[self.view retain];
remember to free it manually when it's no longer needed.
[self.view release];
Why dont you try out this?
to open ur subview:
Ursubview *Ursubview =[[Ursubview alloc] initWithNibName:nil bundle:nil];
Ursubview.modalPresentationStyle = UIModalPresentationFullScreen;
[self presentModalViewController:Ursubview animated:YES];
to go back:
[self dismissModalViewControllerAnimated:YES];
Why not just hide it instead of removing it?
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