Creating a segue programmatically

Question

I have a common UIViewController that all my UIViewsControllers extend to reuse some common operations.

I want to set up a segue on this \"

Answer 1

I reverse-engineered and made an open source (re)implementation of UIStoryboard's segues: https://github.com/acoomans/Segway

With that library, you can define segues programmatically (without any storyboard).

Hope it may help.

Answer 2

By definition a segue can't really exist independently of a storyboard. It's even there in the name of the class: UIStoryboardSegue. You don't create segues programmatically - it is the storyboard runtime that creates them for you. You can normally call performSegueWithIdentifier: in your view controller's code, but this relies on having a segue already set up in the storyboard to reference.

What I think you are asking though is how you can create a method in your common view controller (base class) that will transition to a new view controller, and will be inherited by all derived classes. You could do this by creating a method like this one to your base class view controller:

- (IBAction)pushMyNewViewController
{
    MyNewViewController *myNewVC = [[MyNewViewController alloc] init];

    // do any setup you need for myNewVC

    [self presentModalViewController:myNewVC animated:YES];
}

and then in your derived class, call that method when the appropriate button is clicked or table row is selected or whatever.

Answer 3

well , you can create and also can subclass the UIStoryBoardSegue . subclassing is mostly used for giving custom transition animation.

you can see video of wwdc 2011 introducing StoryBoard. its available in youtube also.

http://developer.apple.com/library/ios/#documentation/UIKit/Reference/UIStoryboardSegue_Class/Reference/Reference.html#//apple_ref/occ/cl/UIStoryboardSegue

Answer 4

Storyboard Segues are not to be created outside of the storyboard. You will need to wire it up, despite the drawbacks.

UIStoryboardSegue Reference clearly states:

You do not create segue objects directly. Instead, the storyboard runtime creates them when it must perform a segue between two view controllers. You can still initiate a segue programmatically using the performSegueWithIdentifier:sender: method of UIViewController if you want. You might do so to initiate a segue from a source that was added programmatically and therefore not available in Interface Builder.

You can still programmatically tell the storyboard to present a view controller using a segue using presentModalViewController: or pushViewController:animated: calls, but you'll need a storyboard instance.

You can call UIStoryboards class method to get a named storyboard with bundle nil for the main bundle.

storyboardWithName:bundle:

Answer 5

Guess this is answered and accepted, but I just would like to add a few more details to it.

What I did to solve a problem where I would present a login-view as first screen and then wanted to segue to the application if login were correct. I created the segue from the login-view controller to the root view controller and gave it an identifier like "myidentifier".

Then after checking all login code if the login were correct I'd call

[self performSegueWithIdentifier: @"myidentifier" sender: self];

My biggest misunderstanding were that I tried to put the segue on a button and kind of interrupt the segue once it were found.

Answer 6

Here is the code sample for Creating a segue programmatically:

class ViewController: UIViewController {
    ...
    // 1. Define the Segue
    private var commonSegue: UIStoryboardSegue!
    ...
    override func viewDidLoad() {
        ...
        // 2. Initialize the Segue
        self.commonSegue = UIStoryboardSegue(identifier: "CommonSegue", source: ..., destination: ...) {
            self.commonSegue.source.showDetailViewController(self.commonSegue.destination, sender: self)
        }
        ...
    }
    ...
    override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
        // 4. Prepare to perform the Segue
        if self.commonSegue == segue {
            ...
        }
        ...
    }
    ...
    func actionFunction() {
        // 3. Perform the Segue
        self.prepare(for: self.commonSegue, sender: self)
        self.commonSegue.perform()
    }
    ...
}