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?

Shorttempered

Answer 1

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

echo $VIRTUAL_ENV

Caesarea

That's it, thanks. I managed to get the value in python using os.environ['VIRTUAL_ENV']. – Hilleary

Note that VIRTUAL_ENV is set by the virtualenv's activate script, and it's possible to use the virtualenv python without activating the virtualenv. See: https://mcmap.net/q/14761/-how-to-determine-if-python-is-running-inside-a-virtualenv – Aleydis

Also, conda do not use this variable. This will beak for users using CONDA or any other ways to manage their python distributions. I don't think this is a robust approach. Using sys.executable as proposed by @laurent-laporte seems more reliable. – Worshipful

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)

Goolsby

Answer 3

How about referencing sys.prefix? It always outputs a result regardless of a virtualenv is activated or not, and also it's more convenient than getting grand parent position of sys.executable.

$ python -c 'import sys;print(sys.prefix)'
/usr
$ . venv/bin/activate
(venv) $ python -c 'import sys;print(sys.prefix)'
path/to/venv

Assortment

Answer 4

You can use fabric to do such things from python

>>> from fabric.api import local
>>> local('which celery')

Languorous

This will probably not work. A sub-shell started is not going to have the virtual env on $PATH so which won't work. – Hawkie