Ios 尝试将多个图像上载到Firebase：SWIFT 4、Firebase 5.0.0中出现2个错误

我正在制作一个与Instagram的多图像post功能非常相似的项目，由于我的ImageUploadManager，我的capturePhotoController中有两个错误相互连接。错误是：

无法分配给“ImageUploadManager.type”类型的不可变表达式 类型“ImageUploadManager”没有成员“uploadImage”

此视图控制器中正在发生：*错误用注释标记

import UIKit

class takePicViewController: UIViewController, UIImagePickerControllerDelegate, UINavigationControllerDelegate {

    @IBOutlet weak var imageView: UIImageView!

    let imagePicker = UIImagePickerController()

    override func viewDidLoad() {
        super.viewDidLoad()

        imagePicker.delegate = self
        imagePicker.sourceType = .camera
        imagePicker.allowsEditing = false

    }

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

        if let userPickedImage = info[UIImagePickerControllerOriginalImage] as? UIImage {
 // Error       ImageUploadManager = ImageUploadManager()
 // Error        ImageUploadManager?.uploadImage(userPickedImage, progressBlock: { (percentage) in
                print(percentage)
            }, completionBlock: { (fileURL, errorMessage) in
                 print(fileURL)
                print(errorMessage)
            })

            PhotoArray.sharedInstance.photosArray.append(userPickedImage)  //append converted data in array

            imageView.image = userPickedImage

        }

        imagePicker.dismiss(animated: true, completion: nil)
    }

    @IBAction func cameraButtonPressed(_ sender: UIButton) {

        present(imagePicker, animated: true, completion: nil)

    }


}

以下是ImageUploadManager的代码：

import UIKit
import FirebaseStorage
import Firebase

struct Constants {

    struct Car {
        static let imagesFolder: String = "carImages"
    }

}

class ImageUploadManager: NSObject {

    func uploadImage(_ image: UIImage, /**/ progressBlock: @escaping (_ percentage: Double) -> Void, /**/ completionBlock: @escaping (_ url: String?, _ errorMessage: String?) -> Void) {

        let data = Data()

        let storage = Storage.storage()
        let storageRef = storage.reference()

        let imageName = "\(Date().timeIntervalSince1970).jpg"
        let imagesReference = storageRef.child(Constants.Car.imagesFolder).child(imageName)

        if let imageData = UIImageJPEGRepresentation(image, 0.8) {
            let metadata = storageRef.child("image/jpg")


            let uploadTask = metadata.putData(data, metadata: nil) { (metadata, error) in
                guard let metadata = metadata else {
                    // Uh-oh, an error occurred!
                    return
                }

                storageRef.downloadURL { (url, error) in
                    guard let downloadURL = url else {
                        // Uh-oh, an error occurred!
                        return
                    }
                }
            }

            uploadTask.observe(.progress, handler: { (snapshot) in
                guard let progress = snapshot.progress else {
                    return
                }

                let percentage = (Double(progress.completedUnitCount) / Double(progress.totalUnitCount)) * 100
                progressBlock(percentage)

            })
        } else {
            completionBlock(nil, "Image could not be converted to Data.")
        }

        }

}

是注释代码：

//错误ImageUploadManager=ImageUploadManager（）

您的

ImageUploadManager

声明？如果是这样的话，那是完全错误的。应声明如下：

if let userPickedImage = info[UIImagePickerControllerOriginalImage] as? UIImage {
    let imageUploadManager = ImageUploadManager()
    imageUploadManager.uploadImage(userPickedImage, prog.....

阅读

ImageUploadManager

时，它看起来像是一个无状态对象，您可以将函数声明为

类

函数

`class func uploadImage(_ image: UIImage`......`

然后，您可以将其称为：

if let userPickedImage = info[UIImagePickerControllerOriginalImage] as? UIImage {
    ImageUploadManager.uploadImage(userPickedImage, prog.....

---

这一行没有什么意义：

ImageUploadManager=ImageUploadManager（）

。左边应该是一个变量名，您将对象归因于类名。我使用了您的解决方案，现在出现了以下错误：无法对“ImageUploadManager”类型的非可选值使用可选链接，该值发生在“ImageUploadManager？”行中。uploadImage（userPickedImage，…”您提供的第一个解决方案代码的。您想在anwser中查看更新后的代码，还是希望从该评论中获得足够的信息？@StackOverflow更新了答案。我建议您阅读更多关于swift的信息。事实上，我对根据您的答案实现的代码进行了一些更改，并且似乎有效。谢谢！