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Hi guys. I'm new here and I am also new to php, and I need help about something.
So I have this basic piece of code in my html fileĀ 

	<input type="file" id="plg_upload" name="filename">
	<input type="submit">

And I found this code for FTP uploads

	// FTP access parameters
	$host = 'hostexample.com';
	$usr = 'user';
	$pwd = 'pass';
	 
	// file to move:
	$local_file = './example.txt';
	$ftp_path = '/data/example.txt';
	 
	// connect to FTP server (port 21)
	$conn_id = ftp_connect($host, 21) or die ("Cannot connect to host");
	 
	// send access parameters
	ftp_login($conn_id, $usr, $pwd) or die("Cannot login");
	 
	// turn on passive mode transfers (some servers need this)
	// ftp_pasv ($conn_id, true);
	 
	// perform file upload
	$upload = ftp_put($conn_id, $ftp_path, $local_file, FTP_ASCII);
	 
	// check upload status:
	print (!$upload) ? 'Cannot upload' : 'Upload complete';
	print "\n";
	 
	/*
	** Chmod the file (just as example)
	*/
	 
	// If you are using PHP4 then you need to use this code:
	// (because the "ftp_chmod" command is just available in PHP5+)
	if (!function_exists('ftp_chmod')) {
	   function ftp_chmod($ftp_stream, $mode, $filename){
			return ftp_site($ftp_stream, sprintf('CHMOD %o %s', $mode, $filename));
	   }
	}
	 
	// try to chmod the new file to 666 (writeable)
	if (ftp_chmod($conn_id, 0666, $ftp_path) !== false) {
		print $ftp_path . " chmoded successfully to 666\n";
	} else {
		print "could not chmod $file\n";
	}
	 
	// close the FTP stream
	ftp_close($conn_id);

I am using provided code in my .php file that is POST called.
How can I pass the information about file I want to upload from html page to this .php file? I know how to pass strings, field results, etc. But never worked with the files.

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https://forums.phpfreaks.com/topic/313779-ftp-remote-upload/
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