ios - 将图像从 Collection View Controller 快速传递到另一个 View Controller

Question

我已经从 firebase 下载了一组用户照片并将其视为收藏 View (类似 Instagram)但我试图使用 segue 放大在另一个 View Controller 中单击的照片，但它不起作用。关于我的代码的任何想法:

class ProfileViewController: UIViewController{
    var photoThumbnail: UIImage! 
func collectionView(_ collectionView: UICollectionView, cellForItemAt indexPath: IndexPath) -> UICollectionViewCell {

        let cell = collectionView.dequeueReusableCell(withReuseIdentifier: "collection_cell", for: indexPath) as! ProfileCollectionViewCell
        cell.imageCollection.downloadImage2(from: currPosts[indexPath.row].photoUrl)
        cell.imageCollection.image = photoThumbnail
        return cell
    }

    func collectionView(_ collectionView: UICollectionView, didSelectItemAt indexPath: IndexPath) {
        performSegue(withIdentifier: "photoViewSegue", sender: nil)
    }

    override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
        var destViewController : EnlargePhotoViewController = segue.destination as! EnlargePhotoViewController
        destViewController.labelText = "test"  //**** this works
        destViewController.myImage = photoThumbnail //**** this doesnt work 
    }

}

Collection View 单元格是:

class ProfileCollectionViewCell: UICollectionViewCell {
    @IBOutlet weak var imageCollection: UIImageView!

}

最后是目标 View Controller :

class EnlargePhotoViewController: UIViewController {

        @IBOutlet weak var myLabel: UILabel!
        @IBOutlet weak var enlargedPhoto: UIImageView!

        var labelText = String()
        var myImage: UIImage!

        override func viewDidLoad() {
            super.viewDidLoad()

           myLabel.text = labelText
            enlargedPhoto.image = myImage
        }
    }

Answer 1

尝试改成这段代码:

func collectionView(_ collectionView: UICollectionView, didSelectItemAt indexPath: IndexPath) {
    let cell = collectionView.cellForItem(at: indexPath) as! ProfileCollectionViewCell

    photoThumbnail = cell.imageCollection.image
    performSegue(withIdentifier: "photoViewSegue", sender: nil)
}