Howto print full path of open files

I am using windows xp sp2. I am kicking ruby scripts off remotely and want to be able to see which scripts are currently running.

For instance, I have a script called c:\instantrails\www\healthchecks\tmp\core.rb and another in the same directory called dsnsrv.rb and I want to be able to see which one in particular is running.

I have downloaded handle.exe from the sysinternals suite and it works perfectly except for the fact that it doesn't print the full path name, only the first few subdirectories. For instance, if I am running core.rb and then run "handle -p ruby instantrails" it will return only return "c:\instantrails\www\healthchecks\" as opposed to "c:\instantrails\www\healthchecks\tmp\core.rb", which makes it kind of difficult to see if core.rb or dsnsrv.rb is the actual running script.

Anyone know how I can get handle.exe to print the full path name or know of another program that can do exactly what I am looking for? I tried openfiles.exe and it does the same thing.

Thanks in advance.

Here's a short WSH script that will list any process that contains the word "ruby" in the command line. Name it with a .vbs extension and call it using the following command:

cscript.exe myscript.vbs