I am seeing a weird situation. I have put a search bar in the navigation bar and have linked a UISearchDisplayController with the search bar. Now, the search display control
In case anyone suffer from this issue.. Here comes my solution.
-(void) viewWillDisappear:(BOOL)animated {
[super viewWillDisappear:animated];
// check if searchDisplayController still active..
if ([searchDisplayController isActive]) {
[searchDisplayController setActive:NO];
}
}
-(void)viewDidLayoutSubviews{
[self.navigationController setNavigationBarHidden:NO animated:NO];
}
it will not hide navigation bar.
okay and just in case, if somebody faces this kind of situation. I implemented a work around for the above situation.
The problem was that the when I popped view controller B from the navigation stack, the searchDisplayController was still active in view controller A. Now, the searchDisplayController assumes that the search bar should always be below the navigation bar (AFAIK). Therefore, it did not displayed the navigation bar when view controller A was again displayed. To fix that, I wrote the following code in the viewWillLayoutSubviews function of view controller A.
-(void)viewWillLayoutSubviews
{
if(self.searchDisplayController.isActive)
{
[UIView animateWithDuration:0.001 delay:0.0 options:UIViewAnimationOptionTransitionCrossDissolve animations:^{
[self.navigationController setNavigationBarHidden:NO animated:NO];
}completion:nil];
}
[super viewWillLayoutSubviews];
}
The above provides a animation so that when the user pops view controller B, the view controller A shows its search bar activated (if the user had earlier tried to search for anything before going to view controller B). It is not a very smooth transition but it works :) ....
Note :- Do not use the above code in viewDidLoad
or viewDidAppear
functions as it might provide an undesirable animation.
My fix is working
override func viewDidLayoutSubviews() {
super.viewDidLayoutSubviews()
DispatchQueue.main.async {
self.navigationController?.setNavigationBarHidden(true, animated: false)
}
}