Question

Suppose I have the following directory structure:

workspace/
  __init__.py
  ys_manage/
    __init__.py
    manage.py
  ys_utils/
    __init__.py
    project_dicts.py

Now, suppose I need access to project_dicts.py in manage.py. Also, my $PATH includes /home/rico/workspace/ys_manage.

I need to be able to run manage.py from any directory on my machine and still be able to access project_dicts.py.

My $PYTHONPATH only has /home/rico/workspace.

If I include the following in manage.py I can run the file from ~/workspace/ys_manage but not anywhere else.

import sys
sys.path.append('..')
from ys_utils import project_dicts

It appears that the '..' gives a relative path to where the directory where the file is run, not where the file is located. Is this correct?

I wanted to try and use ys_manage/__init__.py to import project_dicts.py so that it would be available in manage.py universally. Is this a good idea?

I’ve never used __init__.py for anything other than a “package creator”. That is, I’ve never used it for initialization purposes. Perhaps I’m doing it wrong.

Contents of ys_manage/__init__.py:

import sys
sys.path.append('..')
from ys_utils import project_dicts

Should I include something in manage.py to look for this import?

When I try and run manage.py I get the following error:

NameError: global name 'project_dicts' is not defined

As a secondary question, do I need to have workspace/__init__.py? I’d really rather not have it because ys_manage and ys_utils (and about a dozen other packages) are all under revision control and used by several other developers…workspace is not.

Answer 1

Try this instead of sys.path.append('..'):

sys.path.append(os.path.join(os.path.dirname(__file__), '..'))

(you’ll need to import both sys and os).