如何从PHPickerResult中获取PHAsset？iOS 14

Question

我之前发现了类似的问题，但我尝试了同样的方法后，它并没有像我预期的那样工作。我的目标是从PHPickerViewController中获取PHAsset。为此，我使用此源代码作为基础代码，并将其与这里获取的内容结合起来进行实验。

此外，在尝试时，我已经在info.plist文件中添加了“隐私-照片库用途说明”。

代码如下：

import UIKit
import PhotosUI

class ViewController: UIViewController {
    
    @IBOutlet weak var myImageView: UIImageView!
    
    override func viewDidLoad() {
        super.viewDidLoad()
    }
    
    @IBAction func buttonDidTap(_ sender: Any) {
        var configuration = PHPickerConfiguration()
        configuration.selectionLimit = 1
        configuration.filter = .any(of: [.images, .videos])
        
        let picker = PHPickerViewController(configuration: configuration)
        picker.delegate = self
        self.present(picker, animated: true, completion: nil)
    }
}

extension ViewController: PHPickerViewControllerDelegate {
    
    func picker(_ picker: PHPickerViewController, didFinishPicking results: [PHPickerResult]) {
        
        picker.dismiss(animated: true)
        let identifiers = results.compactMap(\.assetIdentifier)
        let fetchResult = PHAsset.fetchAssets(withLocalIdentifiers: identifiers, options: nil)
        fetchResult.enumerateObjects { (asset, index, stop) -> Void in
            PHImageManager.default().requestImage(for: asset,
                                                  targetSize: CGSize.init(width: 20, height: 20),
                                                  contentMode: PHImageContentMode.aspectFit,
                                                  options: nil) { (image: UIImage?, _: [AnyHashable : Any]?) in
                self.myImageView.image = image
            }
       }
    }
}

在尝试对代码进行调试之后，我发现 fetchResult.enumerateObjects 中的代码没有被调用，这就是为什么我无法获取图片的原因。

或许是我使用了错误的语法？有人可以帮我吗？

- Cado21

Answer 1

如果您稍后需要资产，则必须使用照片库初始化配置。因此，以下这行是错误的:

var configuration = PHPickerConfiguration()

你希望：

var configuration = PHPickerConfiguration(photoLibrary: PHPhotoLibrary.shared())

- matt

Answer 2

matt可能说得没错，但你还可以做以下一件事。这样，图像将直接显示在UIImageView中，与您所做的相同配置。

func picker(_ picker: PHPickerViewController, didFinishPicking results: [PHPickerResult]) {
    dismiss(animated: true)
    let itemProviders = results.map(\.itemProvider)
    for item in itemProviders {
        if item.canLoadObject(ofClass: UIImage.self) {
            item.loadObject(ofClass: UIImage.self) { (image, error) in
                DispatchQueue.main.async {
                    if let image = image as? UIImage {       
                        self.myImageView.image = image
                    }
                }
            }
        }
    }
}

- Abdul Samad Butt

Answer 3

你可以使用这个自定义类 ImagePickerManager:

import Foundation
import UIKit
import PhotosUI

class ImagePickerManager: NSObject, PHPickerViewControllerDelegate, UINavigationControllerDelegate {

var picker: PHPickerViewController?
var pickImageCallback : ((UIImage) -> ())?
var viewController: UIViewController?

override init(){
    super.init()
}

func pickSingleImage(_ viewController: UIViewController, _ callback: @escaping ((UIImage) -> ())) {
    pickImageCallback = callback
    self.viewController = viewController
    
    var configuration = PHPickerConfiguration()
    configuration.filter = .any(of: [.images, .livePhotos])
    configuration.selectionLimit = 1
    picker = PHPickerViewController(configuration: configuration)
    picker?.delegate = self
    viewController.present(picker!, animated: true, completion: nil)
}

func picker(_ picker: PHPickerViewController, didFinishPicking results: [PHPickerResult]) {
    picker.dismiss(animated: true) {
        let itemProviders = results.map(\.itemProvider)
        for item in itemProviders {
            if item.canLoadObject(ofClass: UIImage.self) {
                item.loadObject(ofClass: UIImage.self) { (image, error) in
                    DispatchQueue.main.async {
                        if let image = image as? UIImage {
                            self.pickImageCallback!(image)
                        }
                    }
                }
            }
        }
    }
  }
}

使用方法：

在视图控制器中初始化一个变量：

let imagePicker = ImagePickerManager()

在按钮操作中粘贴此行。

ImagePickerManager().pickSingleImage(self) { pickedImage in
        self.myImageView.image = pickedImage
    }

- Abdul Samad Butt