On Thu, Apr 14, 2011 at 2:08 PM, Leandro Hermida
<soft...@leandrohermida.com> wrote:
> Hi,
>
> I have a tool with a <code file="my_script.py"/> tag and in that code file
> I'm trying to get the tool dirpath where that script and the tool XML
> exist.  I've tried:
>
> os.path.abspath(os.path.dirname(sys.argv[0]))
> os.path.abspath(os.path.dirname(__file__))
>
> And both don't work as expected. Is there a galaxy class I could import
> which will have the tool directory path?
>
> regards,
> Leandro

For standard Python tools in Galaxy, I'm using os.path.split(sys.argv[0])[0]
to get the path, which on reflection probably should be written as
os.path.dirname(sys.argv[0]) as you suggest.

What do __file__ and sys.argv[0] give you? The simplest way to debug
this is to add a print statement, since Galaxy will show the stdout.

Peter

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