Swift 4.2 Cannot convert value of type '(UIImagePickerController.InfoKey).Type' to expected argument type 'UIImagePickerController.InfoKey'

Question

My below code was working just fine with swift 4 but after upgrading to swift 4.2 I am getting this error, I had wasted my 3 hours searching what's the issue but failed. Please if anyone can guide me how to fix this.

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {

    if postType == 2 {
        let image = info[convertFromUIImagePickerControllerInfoKey(UIImagePickerController.InfoKey)] as! UIImage
        mediaType.image = image
    } else {
        videoURL = info[convertFromUIImagePickerControllerInfoKey(UIImagePickerController.InfoKey.mediaURL)] as? URL
        do {
            let asset = AVURLAsset(url: videoURL!, options: nil)
            let imgGenerator = AVAssetImageGenerator(asset: asset)
            imgGenerator.appliesPreferredTrackTransform = true
            let cgImage = try imgGenerator.copyCGImage(at: CMTime.init(value: 0, timescale: 1), actualTime: nil)
            let thumbnail = UIImage(cgImage: cgImage)
            self.mediaType.image = thumbnail
        } catch {
            print("*** Error generating thumbnail: \(error)")
        }
    }
    picker.dismiss(animated: true, completion: nil)
}

Answer 1

You can write like...

if let image = info[.originalImage] as? UIImage {

    print("image found")
    //do something with an image

} else {
     print("Not able to get an image")
}

EDIT:

func imagePickerController(_ picker: UIImagePickerController,
 didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {

    //for original image
    let originalImage = info[.originalImage]

    //for edited image
    let editedImage = info[.editedImage]

    if #available(iOS 11.0, *) {
        //gives URL of image picked by user
        let imageURL = info[.imageURL]
    }

    //gives URL of media picked by user
    let mediaURL = info[.mediaURL]
}