php – 将图像文件变成body参数

我正在尝试将文件上传到服务器.我在后端代码中使用SlimFramework和 PHP,它采用REST形式.如果我通过邮递员上传图片上传功能非常好,但是我无法使用Swift和Alamofire库进行上传.

在我后端的路由,我得到这样的文件,从body参数：

$image = $request-> getUploadedFiles();

并上传我使用：

$图像 – >的moveto( “mypath中”);

在Swift我有路由器,它给了.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,但服务器并不告诉我什么.问题必须在Swift部分.所以我甚至不知道如何调试它.

我该如何解决问题？

我解决了这个问题.我这样重建：

>将图像转换为客户端上的Base64 String
>将其解码为后端图像
>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,只需使用base64_decode()将其转换回文件;并使用file_put_contents();功能将其移动到服务器.

php – 将图像文件变成body参数

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