Get virtualenv's bin folder path from script

Question

I'm using virtualenvwrapper with a django project that has a management task that automatically writes some config files, so the user just has to

./manage.py generate_configuration > much_nice.conf 

And then move the file elsewhere. One of the generated config files is a task for supervisord that launches a celery worker. The problem I'm getting is that I don't know how to output the path of the celery executable that is within the bin folder of the virtualenv. Essentially, I'd like to have the output of the command

which celery 

One option is using sys.executable, get the folder (which seems to be the bin folder of the virtualenv) and that's it... but I'm not sure.

Doesn't virtualenv have any kind of method to get the path itself?

Answer 1

The path to the virtual env is in the environment variable VIRTUAL_ENV

echo $VIRTUAL_ENV 

Answer 2

The VIRTUAL_ENV environment variable is only available if the virtual environment is activated.

For instance:

$ python3 -m venv myapp $ source myapp/bin/activate (myapp) $ python  -c "import os; print(os.environ['VIRTUAL_ENV'])" /path/to/virtualenv/myapp 

If not activated, you have an exception:

(myapp) $ deactivate $ myapp/bin/python -c "import os; print(os.environ['VIRTUAL_ENV'])" Traceback (most recent call last):   File "<string>", line 1, in <module>   File "/usr/lib64/python3.4/os.py", line 635, in __getitem__     raise KeyError(key) from None KeyError: 'VIRTUAL_ENV' 

IMO, you should use sys.executable to get the path of your Python executable, and then build the path to celery:

import sys import os  celery_name = {'linux': 'celery', 'win32': 'celery.exe'}[sys.platform] celery_path = os.path.join(os.path.dirname(sys.executable), celery_name)