16

我的以下代码在 swift 4 上运行良好，但在升级到 swift 4.2 后出现此错误，我浪费了 3 个小时来寻找问题所在，但失败了。请如果有人可以指导我如何解决这个问题。

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)
}
4

6 回答 6

27

你可以这样写...

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

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

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

编辑：

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]
}
于 2018-06-19T12:59:19.390 回答
7 
info[convertFromUIImagePickerControllerInfoKey(UIImagePickerController.InfoKey)]

没有任何意义。您正在指定整个枚举类型InfoKey而不是特定值，例如：

info[convertFromUIImagePickerControllerInfoKey(UIImagePickerController.InfoKey.originalImage)]

这可能也可以写成：

info[.originalImage] as! UIImage
于 2018-06-19T12:54:21.790 回答
3

您可以按如下方式访问属性。

var editedImage = (info[UIImagePickerControllerEditedImage] as? UIImage)!
var originalImage = (info[UIImagePickerControllerOriginalImage] as? UIImage)!
于 2019-08-09T07:14:21.800 回答
1 
func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any])
{
    // Local variable inserted by Swift 4.2 migrator.
    let info = convertFromUIImagePickerControllerInfoKeyDictionary(info)

    if picker.sourceType == .photoLibrary || picker.sourceType == .camera
    {
        let img: UIImage = info[convertFromUIImagePickerControllerInfoKey(UIImagePickerController.InfoKey.editedImage)] as! UIImage
        EditedImage=img
        WAProfile_UserImageView.image=EditedImage
        picker.dismiss(animated: true, completion: nil)
    }
}

// Helper function inserted by Swift 4.2 migrator.
fileprivate func convertFromUIImagePickerControllerInfoKeyDictionary(_ input: [UIImagePickerController.InfoKey: Any]) -> [String: Any] {
return Dictionary(uniqueKeysWithValues: input.map {key, value in (key.rawValue, value)})}

// Helper function inserted by Swift 4.2 migrator.
fileprivate func convertFromUIImagePickerControllerInfoKey(_ input: UIImagePickerController.InfoKey) -> String {
return input.rawValue}
于 2018-11-28T11:37:37.590 回答
1 
 func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any]) {
        picker.dismiss(animated: true, completion: nil)
        let image = info[UIImagePickerControllerOriginalImage] as! UIImage
        yourImgView.image = image
    }
于 2019-10-14T10:50:24.750 回答
0

虽然是一个旧线程，但迁移器添加了这个似乎可以很好地解决问题的功能：

// Function inserted by Swift 4.2 migrator.
fileprivate func convertFromUIImagePickerControllerInfoKeyDictionary(_ input: 
    [UIImagePickerController.InfoKey: Any]) -> [String: Any] {
    return Dictionary(uniqueKeysWithValues: input.map {key, value in (key.rawValue, value)})
}
于 2020-08-24T19:39:29.697 回答