Question

I have to send a file to an API. The API documentation provides an example of how to do this through a file upload script ($_FILES). But, I have the files on the same machine, so would like to skip this step and just read in the file directly to save time.

How can I read in the file (a video) so that it will work with this code snippet?

I understand that FILES is an array, but I could set the other parts of it (the filename) seperately, I just really need the data part of it to be read in the same format to work with this code (do I use fread? file get contents?)

<?php

# Include & Instantiate
require('../echove.php');
$bc = new Echove(
    'z9Jp-c3-KhWc4fqNf1JWz6SkLDlbO0m8UAwOjDBUSt0.',
    'z9Jp-c3-KhWdkasdf74kaisaDaIK7239skaoKWUAwOjDBUSt0..'
);

# Create new metadata
$metaData = array(
    'name' => $_POST['title'],
    'shortDescription' => $_POST['shortDescription']
);

# Rename the video file
$file = $_FILES['video'];
rename($file['tmp_name'], '/tmp/' . $file['name']);
$file_location = '/tmp/' . $file['name'];

# Send video to Brightcove
$id = $bc->createVideo($file_location, $metaData);

?>

Thanks in advance for any help!

Answer 1

Er – all you’d need to do is feed the location, not a file handle or anything, no?

$files = $_FILES['video'];
$filename = $files['name'];
$file_location = '/home/username/' . $filename; // maybe append the extension
$id = $bc->createVideo($file_location, $metaData);

Or more simply

$id = $bc->createVideo( '/foo/bar/baz.txt', $metaData );