From: tsuraan <tsuraan@...>
Date: 2007-04-07T03:51:25+09:00
Subject: Re: [Painfully OT] A python question...

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>
> My interpretation of the above is that, to be importable, a zip file
> must be explicitly named in sys.path, so the mere fact that a zip file
> lies somewhere in a directory which is part of the sys.path does not
> make it importable.


I just realized that I left out the tricky part of what I'm doing.  Within
a/__init__.py, I have the following:

import zipimport
import os
here = os.path.join(os.getcwd(), __path__[0])
zip = os.path.join(here, 'b.zip')
mod = os.path.split(zip)[-1][:-4]
globals()[mod] = zipimport.zipimporter(zip).load_module(mod)

So, this explicitly uses zipimport to manually load the b.zip module into
a's namespace.  I sort of forgot that was important :)  Anyhow, b does exist
within a as a module, but it has no contents, and there doesn't seem to be
any way to get them, unless anyone knows a cool trick.  Thanks for the help
though :)

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