On 22 Feb, 17:13, Brandon Mintern <[EMAIL PROTECTED]> wrote:
>
> toplevel_dir
> +-main script
> +-wrapper_dir
> +-some_wrapper
> +-utility_dir
> +-some_external_utility
[...]
> And then in some_wrapper, I would have code like:
>
> import os
>
> def use_external_utility():
> f = os.popen('utility_dir/some_external_utility')
> lines = f.readlines()
> f.close()
> return lines
And you really want to refer to utility_dir relative to some_wrapper.
What you can try is to split the __file__ attribute of some_wrapper -
it's a standard attribute on imported modules - in order to refer to
the module's parent directory (which should correspond to
wrapper_dir):
parent_dir, filename = os.path.split(__file__)
Then you can join the parent directory to the path of the command:
cmd = os.path.join(parent_dir, "utility_dir", "some_external_utility")
The __file__ attribute of modules is documented here:
http://docs.python.org/ref/types.html#l2h-109
Paul
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