CODEWITHSUNDEEP
phplaravellaravel-4
panthro
panthro

Reputation: 24061

Getting an image server side using form data?

I send my image to php like so:

var formData = new FormData();

formData.append('file', file);
var xhr = new XMLHttpRequest();
    xhr.open('POST', '/upload', true);
    xhr.onload = function(e) { console.log(e) };
    xhr.upload.onprogress = function(e) {
        //loading bar
    };
    xhr.send(formData);

Then I get my file in php like so:

$data = Input::All();
var_dump($data['file']);

This outputs:

"object(Symfony\Component\HttpFoundation\File\UploadedFile)#9 (7) { ["test":"Symfony\Component\HttpFoundation\File\UploadedFile":private]=> bool(false) ["originalName":"Symfony\Component\HttpFoundation\File\UploadedFile":private]=> string(9) "space.jpg" ["mimeType":"Symfony\Component\HttpFoundation\File\UploadedFile":private]=> string(10) "image/jpeg" ["size":"Symfony\Component\HttpFoundation\File\UploadedFile":private]=> int(50974) ["error":"Symfony\Component\HttpFoundation\File\UploadedFile":private]=> int(0) ["pathName":"SplFileInfo":private]=> string(14) "/tmp/phpzMyVkq" ["fileName":"SplFileInfo":private]=> string(9) "phpzMyVkq"}

My question is, how can I get the file from the object for image processing?

Upvotes: 2

Views: 227

Answers (2)

clami219
clami219

Reputation: 3038

Use the move method to save the file from the tmp position where it is saved to where you need it:

$data['file']->move($destination_dir,$file_name);

Upvotes: 1

fire
fire

Reputation: 21531

Use the file method on input...

$file = Input::file('file');

Take a read of the documentation to see what methods are available... http://laravel.com/docs/requests#files

Upvotes: 2

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