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? Continue reading...
