Symfony2 file upload get original file name

Question

I am unable to retrieve the originalName during a file upload.

print_r($request->file) returns the following:

Symfony\Component\HttpFoundation\FileBag Object
(
    [parameters:protected] => Array
        (
            [form_name] => Array
                (
                    [backgroundImageFile] => Symfony\Component\HttpFoundation\File\UploadedFile Object
                        (
                            [test:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 
                            [originalName:Symfony\Component\HttpFoundation\File\UploadedFile:private] => bg_image.jpg
                            [mimeType:Symfony\Component\HttpFoundation\File\UploadedFile:private] => image/jpeg
                            [size:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 65045
                            [error:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 0
                            [pathName:SplFileInfo:private] => /tmp/phpsU3X7a
                            [fileName:SplFileInfo:private] => phpsU3X7a
                        )

                )

        )

)


echo "<br /> Name= ".$this->getRequest()->files['backgroundImageFile']->getClientOriginalName();

Returns:

Fatal error: Call to a member function getClientOriginalName() on a non-object in ../path to the controller

So, how do I get the OriginalName?

Kal Zekdor · Accepted Answer · 2015-07-04 00:58:22Z

2

Try using the get() method on the FileBag, rather than array access.

E.g.,

$this->getRequest()->files->get("backgroundImageFile")->getClientOriginalName();

Alternatively, if it makes sense in your scenario, use a form upload, then access the uploaded file as form data.

answered Jul 4, 2015 at 0:58

Kal Zekdor

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3 Comments

kewpe20 Over a year ago

I still get the Fatal error: Call to a member function getClientOriginalName() on a non-object in ../path to the controller. I just need to get the Orginal File Name before the form is submitted

Kal Zekdor Over a year ago

@kewpe20 Before the form is submitted the file isn't uploaded. The upload happens on form submit, so before then there's nothing to get the name of.

Kal Zekdor Over a year ago

If you can put your code up somewhere for me to take a look, I'll see what's going on in your specific instance.

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