How to call different storyboards through Swift for iOS?

Question

I created an application with three different Storyboards for each iOS device family. Now I do not know how to choose the right storyboard when starting the application? I check the height of the screen to recognize various devices:

func application(application: UIApplication!, didFinishLaunchingWithOptions launchOptions: NSDictionary!) -> Bool {
    // Check Device Family
    var bounds: CGRect = UIScreen.mainScreen().bounds
    var screenHeight: NSNumber = bounds.size.height
    var deviceFamily: String
    if screenHeight == 480 {
        deviceFamily = "iPhoneOriginal"
        // Load Storyboard with name: iPhone4
    } else if screenHeight == 568 {
        deviceFamily = "iPhone5Higher"
        // Load Storyboard with name: iPhone5
    } else {
        deviceFamily = "iPad"
        // Load Storyboard with name: iPad
    }

    return true
}

Can someone give me a working solution in Swift? I found solutions for ObjC.

Thank.

Answer 1

Think you want to open a show? If so, this code will complete the task:

var mainView: UIStoryboard!
mainView = UIStoryboard(name: "vcLogin", bundle: nil)
let viewcontroller : UIViewController = mainView.instantiateViewControllerWithIdentifier("iPhone5") as UIViewController
self.window!.rootViewController = viewcontroller

It will open the view controller with identifier: yourViewControllerId

viewcontroller. , : StoryBoard.

, :

if screenHeight == 480 {
  deviceFamily = "iPhoneOriginal"
  // Load Storyboard with name: iPhone4
  var mainView: UIStoryboard!
  mainView = UIStoryboard(name: "vcLogin", bundle: nil)
  let viewcontroller : UIViewController = mainView.instantiateViewControllerWithIdentifier("iPhone4") as UIViewController
  self.window!.rootViewController = viewcontroller
}