Can I get data from a label from one VC and pass the data to a textfield from another VC [duplicate] - swift

Question

I'm trying to move a user uploaded image from one UIImageView to another UIImageView on a different View Controller. I'm able to have the user upload an image and have it saved in the first UIImageView but after they press the "Post" button, the UIImageView on the next view controller remains blank.
Note: browsingImage is the name of the UIImageView on the second view controller (destination UIImageView)
Any help would be greatly appreciated!
#IBAction func cameraButton(sender: AnyObject) {
addNewPicture()
}
func addNewPicture() {
let picker = UIImagePickerController()
picker.allowsEditing = true
picker.delegate = self
presentViewController(picker, animated: true, completion: nil)
}
func imagePickerController(picker: UIImagePickerController, didFinishPickingImage image: UIImage!, editingInfo: [NSObject : AnyObject]!) {
postingImage.image = image
self.dismissViewControllerAnimated(true, completion: nil)
}
#IBAction func postButton(sender: AnyObject) {
performSegueWithIdentifier("toBrowsePage", sender: nil)
}
override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
if segue.identifier == "toBrowsePage" {
var itemToAdd = segue.destinationViewController as! ListPage
itemToAdd.postingImage.image = browsingImage.image
}
}

Answer 1

In prepare(for:) you can't access the #IBOutlets of the destination view controller because they haven't been set up yet. You should assign the image to a property of the destination view controller, and then move it into place in viewDidLoad():
In source view controller:
override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
if segue.identifier == "toBrowsePage" {
let dvc = segue.destination as! ListPage
dvc.newImage = postingImage.image
}
}
In destination view controller:
class ListPage: UIViewController {
#IBOutlet weak var browsingImage: UIImageView!
var newImage: UIImage!
override func viewDidLoad() {
super.viewDidLoad()
browsingImage.image = newImage
}
}

Answer 2

From your description,browsingImage is in destination viewController,so
in this line
itemToAdd.postingImage.image = browsingImage.image
You pass the destination imageview to source imageview

Answer 3

IBOutlets are not accessible in your current view controller. The outlets are not yet initialized since the your second view is not yet loaded.
Instead you can create a UIImage variable in your second viewController and assign the value to it in the prepareForSegue.
And in your second viewController's viewDidLoad method assign that value to your ImageView outlet.
override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
if segue.identifier == "toBrowsePage" {
let destination = segue.destinationViewController as! ListPage
destination.receivedImage = postingImage.image
}
}
In your next viewController declare a variable called
#IBOutlet weak var postingImage: UIImageView!
var receivedImage : UIImage?
In the viewDidLoad method
postingImage.image = receivedImage

Answer 4

At the time of the Segue happening your UIImageView is not initialised so you can not assign an image to it, the best practise would be to pass the UIImage and initialise your UIImageView in viewDidLoad.
so in your ListPage class make a property of type UIImage change your prepareForSegue line as
override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
if segue.identifier == "toBrowsePage" {
var itemToAdd = segue.destinationViewController as! ListPage
itemToAdd.image = browsingImage.image
}
}
in viewDidLoad or viewDidAppear of your destination viewController you will do something like this
browsingImage.image = image