Debugging with echo

[russia5]

I have a php problem and I read that I can debugg my script calling echo for all of he commands. I am not sure what I am suppose to see.  Here is an example of what I did and I got know results. (in bold at the end of the script and the variable is about three lines down.

[code]<?php
$command = $imagemagik_path.'convert -resize ';

// resize for thumbnail 1 size
$current_command = $command.$thumb_size1.' "'.$old.'" '.$thumb1;
exec($current_command, $retarray, $retval);
// echo $current_command;
if ($retval != 0) // error
{
echo 'error upon resizing for thumbnail 1 : '.$old.' to '.$thumb1;
}

// resize for thumbnail 2 size
$current_command = $command.$thumb_size2.' "'.$old.'" '.$thumb2;
exec($current_command, $retarray, $retval);
// echo $current_command;
if ($retval != 0) // error
{
// echo 'error upon resizing for thumbnail 2';
}

[b]echo '$current_command';[/b]?>

?>[/code]

Also, can anyone explain what the $command.$thumb_size1  means?  I know $command is the path to imagemagick.  But what would $thumb_size1 mean and does it need declared someplace within the directory it is included in?