从 Storyboard 转换为子类

Question

Imagine i have a BaseViewController. Then i have 2 scenarios, New and Edit, where both shares the same UI and the most of logic. So i created class NewViewController and EditViewController, subclassing BaseViewController. The problem comes when i try to instantiate "BaseViewController" from the storyboard cause i want to specify which implementation is.

if isEdit { 
    storyboard.instantiateViewControllerWithIdentifier("baseVCIdentifier") as! EditViewController 
} else {
   storyboard.instantiateViewControllerWithIdentifier("baseVCIdentifier") as! NewViewController 
}

Then i get an error: Could not cast value of type 'Test.BaseViewController' (0x10ee5e0f0) to 'Test.EditViewController' (0x10ee5f000).

I dont wanto to have both ViewController on the storyboard since i dont want to redo the same UI 2 times.

Answer 1

You can do this using instantiateViewController(identifier:creator:) .

I assume you have the view controller in a storyboard, with identifier template . The class assigned to the view controller in the storyboard should be the superclass:

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

let viewController = storyboard.instantiateViewController(identifier: "template") { coder in 
    // this passes us with a coder for the storyboard, we can now init the preferred subclass.

    if useSubclass {
        return SpecialViewController(coder: coder)
    } else {
        return BaseViewController(coder: coder)
    }
}

Here is the documentation

Answer 2

You can't do that. Instead of sub classing create 'interaction manager' classes, or state manager classes. The base view controller would then be provided with a manager instance as part of the segue and it would forward all UI interaction to the manager for processing. You then have a single VC in the storyboard as is required and you can supply a new or edit manager. The managers can also have specific instance variables which the view controller doesn't care about.