manually creating an Symfony UploadedFile - php

Question

I'm facing following problem and can't seem to figure this one out.
I wrote an API endpoint accepting a POST with binary data (header: content-type:image/jpeg).
I know i can read out the raw string with file_get_content('php://input') or Laravel's $request->getContent().
PHP also has a function createimagefromstring($string) which also seems to read the string in correctly.
What i'd like to do is create an UploadedFile from this raw data , so that I can handle it with already written functions.
Is this possible?
Thank you in advance

Answer 1

I think I found it... Still curious if there are improvements that can be made..
$imgRaw = imagecreatefromstring( $request->getContent() );
if ($imgRaw !== false) {
imagejpeg($imgRaw, storage_path().'/tmp/tmp.jpg',100);
imagedestroy($imgRaw);
$file = new UploadedFile( storage_path().'/tmp/tmp.jpg', 'tmp.jpg', 'image/jpeg',null,null,true);
// DO STUFF WITH THE UploadedFile
}

Answer 2

You can try to use base64 encoding. Symfony have some nice stuff for this.
Your code will be smthng like this:
$base64Content = $request->request->get('base64Content'); // this is your post data
$yourFile = new UploadedBase64EncodedFile(new Base64EncodedFile($base64Content)); // this is an instance of UploadedFile
Hope it helps!

Answer 3

As per Laravel 8
Just follow the constructor:
* #param string $path The full temporary path to the file
* #param string $originalName The original file name of the uploaded file
* #param string|null $mimeType The type of the file as provided by PHP; null defaults to application/octet-stream
* #param int|null $error The error constant of the upload (one of PHP's UPLOAD_ERR_XXX constants); null defaults to UPLOAD_ERR_OK
* #param bool $test Whether the test mode is active
$file = new UploadedFile(
$pathIncludingFilename,
$fileName,
'image/jpeg',
null,
false
);

Answer 4

There is no need to manually create it, Symfony parses received $_FILES array for you. Http Request object has a FileBag property called $files with a get() method which returns an UploadedFile instance.
/** #var UploadedFile $file */
$file = $request->files->get('user-pictures-upload')[0];
$cmd = new UploadPictureCmd($file, $this->getUser()->getId());

Answer 5

Here is the example of generating images files using fzaninotto/faker in Symfony 4 Fixtures
class FileFixtures extends Fixture
{
private $faker;
private $parameterBag;
public function __construct(ParameterBagInterface $parameterBag)
{
$this->faker = Factory::create();
$this->parameterBag = $parameterBag;
}
public function load(ObjectManager $manager)
{
$tempFixturesPath = $this->parameterBag->get('kernel.project_dir') . '/tmp';
if (!file_exists($tempFixturesPath)) {
mkdir($tempFixturesPath);
}
$fileName = $this->faker->image($tempFixturesPath, 640, 480, 'cats', false, true);
$file = new UploadedFile($tempFixturesPath . '/' . $fileName, $fileName, 'image/jpeg', null, null, true);
//do something with $file
}
}

Answer 6

If it counts for anything, this is how I did it in Laravel 5.4. In my case, I wanted to be able to easily resize an image and be able to do something like this.
request()->file('image')->resize(250, 250)->store('path/to/storage');
This is what I did to the UploadedFile class.
Illuminate\Http\UploadedFile.php ~this file ships with the Laravel framework
public function resize($w, $h) {
$image = Intervention::make($this)->fit($w, $h)->save($this->getPathname());
$filesize = filesize($this->getPathname());
return new static(
$this->getPathname(),
$this->getClientOriginalName(),
$this->getClientMimeType(),
$filesize,
null,
false
);
}
Using Intervention, I resized the image that is stored in the /tmp/ folder when files are uploaded and then I saved it in the same place. Now all I do after that is create an instance of UploadedFile so that I can keep using Laravel's methods on request()->file('image'). Hope this helps.