UIStoryboard: What's the Correct Way to Get the Active Storyboard?

Question

I am currently furiously digging through all the docs, and haven't quite found what I'm looking for. I suspect it is a real d'oh! answer.

I simply need to find the active storyboard in the main bundle, and want to know the best way to do this.

This is so that I can use the [UIStoryboard storyboardWithName:@"XXX" bundle:mainBundle] to extract the running storyboard.

I know how to kludge it by switching on the idiom, but I feel that this is a...kludge.

What's a correct way of doing this?

That's the main storyboard, not the "active" storyboard, whatever that means. How many storyboards are in your app? — jrturton, Mar 24 '12 at 18:35
2. One for iPhone, and one for iPad. This works fine. Only one is active at a time. — Chris Marshall, Mar 24 '12 at 18:38
Oh right, I get you. You should add your code as an answer, then. — jrturton, Mar 24 '12 at 18:43
I'm too new. I'll add it later, when SO lets me, so I edited the original. — Chris Marshall, Mar 24 '12 at 19:02
An easier way, if you are trying to get this from an existing view controller that was loaded from the storyboard (very common) is simply: `self.storyboard` — lnafziger, Mar 25 '12 at 03:24
Your answer pushed me in the right direction :) the code I'm using now is `UIStoryboard *sb = [[self.window rootViewController] storyboard];` — Ja͢ck, May 31 '12 at 07:10

score 41 · Answer 1 · edited Jan 18 '19 at 23:30

In case you want to get the active storyboard for a viewController, there's a storyboard property. This is how I solved it, instead of making a new instance:

LoginViewController *vc = [navController.storyboard instantiateViewControllerWithIdentifier:@"firstLaunch"];
[navController presentModalViewController:vc animated:YES];

In Swift you'd call:

let loginViewController = navigationController?.storyboard?.instantiateViewController(withIdentifier: "firstLaunch") as! LoginViewController
navigationController?.present(loginViewController, animated: true, completion: nil)

You could also be a lot safer by using guards against the navigation controller and the storyboard. I've used as! so as to guarantee that you're getting a LoginController.

+1 for the .storyboard property. This property is not only available for a navigation controller but also for UITableViewController, UITabBarController, UICollectionViewController, etc. — Leon, Dec 22 '15 at 11:36

score 27 · Accepted Answer · edited May 23 '17 at 12:18

27

OK. As my comment above indicates, I found the answer to the (badly phrased question):

I wanted to be able to get the main (not active) storyboard, as I'm not using multiple storyboards per incarnation. I'm using the standard model of 1 storyboard for iPhone, and 1 for iPad. I just wanted the cleanest way to get the storyboard, so that I could use it to generate a view controller.

I found the answer in this post on Stack Overflow, and implemented it with the following code:

UIStoryboard *st = [UIStoryboard storyboardWithName:[[NSBundle mainBundle].infoDictionary objectForKey:@"UIMainStoryboardFile"] bundle:[NSBundle mainBundle]];

edited May 23 '17 at 12:18

Community

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answered Mar 25 '12 at 03:17

Chris Marshall

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calling `[UIStoryboard storyboardWithName:...` creates a new instance of the storyboard. This won't give you access to the same instances that are actually loaded. – Matt Connolly Apr 06 '12 at 01:49
2

Thanks. Yeah, but that's OK. I want a new instance. The already loaded one is off doing something else. – Chris Marshall Apr 06 '12 at 11:42
Definitely the best answer for my problem. Thanks! – mpemburn Feb 23 '13 at 16:57
1

Or may be adding this to the AppDelegate as a Class method, so you can reuse the storyboard and avoid creating unnecessary instances. – Frederic Yesid Peña Sánchez Oct 18 '13 at 17:11
Yeah...not a big deal to do that. But why bother, if it is only gonna be used once? Was that REALLY worth a downvote? – Chris Marshall Oct 06 '14 at 15:08
@FredericYesidPeñaSánchez Good point, I've done this to avoid creating additional instances. Create storyboard once in ``application:DidFinishLaunching... `` or hold onto a ``vc.storyboard`` that was already created by system – Protongun Dec 14 '16 at 16:13

nburk · Answer 3 · 2016-07-20T18:25:31.857

In Swift, you'd use the following syntax:

let storyboard = UIStoryboard(name: "Main", bundle: nil)

Note that passing nil to bundle will make the call refer to your main bundle automatically.

If you're in a view controller that you have on the Storyboard and want to instantiate the Storyboard from there directly, you can just do:

let storyboard: UIStoryboard? = self.storyboard // call this inside a VC that is on the Storyboard

Note that in the last case, self.storyboard will return an optional Storyboard (Storyboard?), so if you'd like to use it unwrap it like so:

if let storyboard = self.storyboard {
  // access storyboard here
}

score 1 · Answer 4 · answered Oct 16 '22 at 21:33

The better approach would be to get existing storyboard instance from window's rootViewController if not available then create a new instance.

let storyboard = self.window?.rootViewController?.storyboard ?? UIStoryboard.init(name: "Main", bundle: nil)

Also using globally accessible helper functions like these can be a good idea provided that you pass the already active storyboard as parameter.

class Helper {
    static func getLoginVC(storyboard: UIStoryboard) -> LoginVC {
         return storyboard.instantiateViewController(withIdentifier: String(describing: LoginVC.self)) as! LoginVC
    }
}

you can pass the active storyboard instance from a controller like this -

let loginVC = Helper.getLoginVC(storyboard: self.storyboard)

In case you are trying to access storyboard from appDelegate or sceneDelegate you can use -

let storyboard = self.window?.rootViewController?.storyboard ?? UIStoryboard.init(name: "Main", bundle: nil)
let loginVC = Helper.getLoginVC(storyboard: storyboard)

let navigationController = UINavigationController.init(rootViewController: loginVC)
self.window?.rootViewController = navigationController
self.window?.makeKeyAndVisible()

score 0 · Answer 5 · answered Jun 21 '16 at 09:18

I have just copy pasted the code form above updated question so that everyone can see it as an answer.

UIStoryboard *st = [UIStoryboard storyboardWithName:[[NSBundle mainBundle].infoDictionary objectForKey:@"UIMainStoryboardFile"] bundle:[NSBundle mainBundle]];

UIStoryboard: What's the Correct Way to Get the Active Storyboard?

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