building a 'interface' for .php upload api - php

Question

i have a .php file from a videohoster who says that this code is a comandline code which allows me to upload a video to their site and i need to create an interface for a random user that he can upload a video to the videohoster while staying on my website. i thought i simple form like below in html is enought but apparently it doesent work:
front-end, html:
<h2>This form allows you to upload a Video.</h2>
<form action="uploadapi.php" method="post" enctype="multipart/form-data"><br>
<p>Video Name: <input type="text" name="titel" size="50" /></p>
<p>Video Description:<br/><textarea name="text" rows="5" cols="50"> </textarea></p>
<p>Select File, allowed: .mpg </br><input type="file" name="file"></p>
<p><input type="submit" value="Upload"</p>
</form>
the uploadapi.php is supplied by the hoster and so i assume it is correct.
<?php
////////////////////////////////////////////////////////
// for php 5.6+ you need to make some changes in code
// method 1
// add the following line
// curl_setopt($ch, CURLOPT_SAFE_UPLOAD, 0);
//
// method 2
// change
// $post_fields['vfile'] = "#".$file;
// to
// $post_fields['vfile'] = CURLFile($file);
////////////////////////////////////////////////////////
$apiversion = "2.123.20150426";
//REQUIRED Registered Users - You can find your user token in API page.
$user_token = "xxx";
if(count($argv) < 2)
die("Usage: php $argv[0] [VIDEO TO UPLOAD] {SUB FILE}\n");
$file = $argv[1];
if(!file_exists("$file"))
die("ERROR: Can't find '$file'!\n");
$path_parts = pathinfo($file);
$ext = $path_parts['extension'];
$allowed = array("mov");
if (!in_array(strtolower($ext),$allowed))
die("ERROR: Video format not permitted. Formats allowed: .mov!\n");
if(isset($argv[2]))
{
$sub_file = $argv[2];
if(!file_exists("$sub_file"))
die("ERROR: Can't find '$file'!\n");
$path_parts = pathinfo($sub_file);
$ext = $path_parts['extension'];
$allowed = array("srt");
if (!in_array(strtolower($ext),$allowed))
die("ERROR: Subtitle format not permitted. Formats allowed: .srt!\n");
$post_fields['subfile'] = "#".$sub_file;
}
$converter = file_get_contents("http://.../getconv_uploadapi.php? upload_hash=".$user_token);
if($converter=="ERROR")
die("ERROR: Could not choose converter. Aborting... \n");
$post_fields['vfile'] = CURLFile($file);
$post_fields['upload'] = "1";
$post_fields ['token'] = 'xxx';
if(!empty($user_token))
$post_fields['upload_hash'] = $user_token;
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL,$converter);
curl_setopt($ch, CURLOPT_POST,1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $post_fields);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$result=curl_exec ($ch);
curl_close ($ch);
echo "$result\n";
?>
is my methodology correct (using html form to call the uploadapi.php on my server) or do i need in order to sumit my video to the videohoster via uploadapi.php other programming languages (ajax, javascript etc)?

Answer 1

Provided file is command-line file. It means it won't work on web as expected.
For example, on web you won't have $argv[1] or $argv[2].
You have two options:
you can rewrite uploadapi.php considering that source file is a file that came from your form
or upload your file and call uploadapi.php as a command-line script with arguments using shell_exec or exec, for example.

Answer 2

i thought that i define or fill the variable $argv[1] from the html form since: type="file" name="file"> = $file = $argv[1];?
... ok rewriting upload api is no alternative since it comes from the hoster.
so what i need to do is uploading the uploadapi.php on my server and then keeping my html form like it is? and insted of action=uploadapi.php i need a action=somethingsomething.php which has an exec command to execute uploadapi.php?:
<?php
function somethingsomething($uploadapi) {
exec($uploadapi . " > /dev/null &");
}
?>