将图像文件转换为body参数

Question

我正在尝试将文件上传到服务器.我在我的后端代码中使用SlimFramework和PHP,它是RESTful形式.如果我通过Postman上传它,图像上传效果很好但是我无法使用Swift和Alamofire库来使用它.

在后端的路由中,我从body参数得到这样的文件:

$image = $request->getUploadedFiles();

并上传我使用:

$image->moveTo("myPath");

在Swift我有路由器,它提供.post方法和路由URL,我尝试这样:

func uploadImage(user_id: Int, image: Data, completion: @escaping (_ error: Error?, _ success: Bool)->Void) {
        let parameters = [
            "user_id": user_id,
            "newFile": image
        ] as [String : Any]

        Alamofire.request(Router.imageUplaod(parameters: parameters))
            .validate(statusCode: 200..<300)
            .responseJSON { response in

                switch response.result{

                case .failure(let error):
                    completion(error, false)

                    print(error)

                case .success(let value):
                    //Registered sucesfully! let json = JSON(value)
                    completion(nil, true)

                }//Switch case
        }//Alamofire
    }

我称之为:

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any]) {
        let chosenProfileImage = info[UIImagePickerControllerEditedImage] as! UIImage
        let imageData = UIImageJPEGRepresentation(chosenProfileImage, 0.2)
        profileImageView.image = chosenProfileImage

        dismiss(animated: true) { 
            API.uploadImage(user_id: User.sharedInstance.user_id, image: imageData!) { (error, success) in

                if success{
                    print("IMAGE UPLOADDED")
                }
            }
        }
    }

这是我的后端路线:

 $this->post('/image/upload', function($request, $response, $args){
    global $database;
    $data = $request->getParsedBody();
    $user_id = $data['user_id'];

    $files = $request->getUploadedFiles();
    if (empty($files['newfile'])) {
        throw new Exception('Expected a newfile');
    }

    $newfile = $files['newfile'];

    if ($newfile->getError() === UPLOAD_ERR_OK) {
    $uploadFileName = "profileImage".$user_id.".jpg";
    $newfile->moveTo("/path/to/image/$uploadFileName");
    }

    $database->query("UPDATE `users` SET `profile_image_url` = :image WHERE user_id = :user_id");
    $database->bind(':image', $uploadFileName);
    $database->bind(':user_id', $user_id);
    $database->execute();


    $Jsonarray['status'] = "true";
    $Jsonarray['message'] = "Image Uploaded Successfully";
    return $response->withStatus(200)->withHeader('Content-Type', 'application/json')->write(json_encode($Jsonarray));

    });

奇怪的是,Alamofire输出状态代码500,但服务器并没有告诉我任何关于它的信息.问题必须出在Swift部分.所以我甚至不知道如何调试它.

我该如何解决这个问题？

Answer 1

我解决了这个问题.我这样重建了它:

1. 在客户端上将图像转换为Base64字符串
2. 将其解码为后端的图像
3. 上传它

所以在Swift我转换它像这样:

let chosenProfileImage = info[UIImagePickerControllerEditedImage] as! UIImage
let imageData = UIImageJPEGRepresentation(chosenProfileImage, 0.2)
profileImageView.image = chosenProfileImage

let base64String = imageData?.base64EncodedString(options: .lineLength64Characters)

后端几乎相同,只是将其转换回文件使用base64_decode();并使用file_put_contents();函数将其移动到服务器.