Get Uploaded File's Original Name

Question

$request = $event->getRequest();
print_r($request->files);die;

gives me

Symfony\Component\HttpFoundation\FileBag Object
(
    [parameters:protected] => Array
        (
            [files] => Array
                (
                    [0] => Symfony\Component\HttpFoundation\File\UploadedFile Object
                        (
                            [test:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 
                            [originalName:Symfony\Component\HttpFoundation\File\UploadedFile:private] => Chrysanthemum.jpg
                            [mimeType:Symfony\Component\HttpFoundation\File\UploadedFile:private] => image/jpeg
                            [size:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 879394
                            [error:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 0
                            [pathName:SplFileInfo:private] => /tmp/phpmUl54W
                            [fileName:SplFileInfo:private] => phpmUl54W
                        )

                )

        )

)

I'm trying to get at the value for 'originalname' i.e "Chrysanthemum.jpg" without resorting to a loop, but I can't seem to find the right syntax

Using the 1UP file uploader, but I dont think that's important

Answer 1

When you upload files you get UploadFile (API link) objects (basically the wrappers of array).

$this->getRequest()->files[0]->getClientOriginalName();

Can't try this now but you might need to do this instead:

$this->getRequest()->files['name_of_file_field_in_post']->getClientOriginalName();

where you would replace name_of_file_field_in_post with your form field's name.

Answer 2

2015 Update:

$request->files->get('your-file-name')->getClientOriginalName();

your-file-name for me was just file.

This will probably help too: https://github.com/1up-lab/OneupUploaderBundle/issues/21