How to separate local and production settings in django?

I have .env.local and .env.prod files, and I've split settings.py into base, local, and production.

Running a Django app with production settings works without any problems (thanks to docker), but I'm not entirely sure how to easily run it with local settings.

I'm currently doing this with a sh script.

export $(cat .env.local | grep -v '^#' | xargs)
python manage.py runserver

but i dont think that its a best way to launch django with specific .env

i think it would be great to launch it with path to env file as a parametet. Like this

python manage.py runserver --env=./env.local

Anyway, what is the best way to launch django with specific environment (without docker)

how about django - How to use different .env files with python-decouple - Stack Overflow

  1. Switch environment via settings module

    Since you already have base.py, local.py, and production.py, you can run:

    # local
python manage.py runserver --settings=config.settings.local

# production
python manage.py runserver --settings=config.settings.production

    If you prefer an environment variable:

    DJANGO_SETTINGS_MODULE=config.settings.local python manage.py runserver

    You can wrap this in a small script / Makefile if you want a shorter command.

  2. Load the correct .env inside settings

    Instead of exporting everything before runserver, read the right .env inside your settings.
    For example with django-environ:

    # config/settings/base.py

import environ

from pathlib import Path

BASE_DIR = Path(_file_).resolve().parent.parent.parent

env = environ.Env()

# Use ENV_PATH if set, otherwise default to .env

env.read_env(env.str("ENV_PATH", BASE_DIR / ".env"))

SECRET_KEY = env("SECRET_KEY")

DEBUG = env.bool("DEBUG", False)

    Then run:

    # local
ENV_PATH=.env.local python manage.py runserver --settings=config.settings.local

# production-like
ENV_PATH=.env.prod python manage.py runserver --settings=config.settings.production
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