I am using python-decouple 3.4 for setting up environment variables for my django application. My .env file is in the same directory as that of manage.py. Except for SECRET_KEY (in settings.py), loading other environment variables in either settings.py or views.py directly fails stating that they have not been defined. The other environment variables which give error will be used in views.py.
Here is my .env file:-
SECRET_KEY=<django_app_secret_key>
file_path=<path_to_the file>
If I try to define them in settings.py like:-
from decouple import config
FILE_PATH = config('file_path')
and then use them in views.py,
from django.conf.settings import FILE_PATH
print(FILE_PATH)
then also I get the same error. How can I define environment variable for my views.py specifically?
[Edit: This is the error which I get:-
raise UndefinedValueError('{} not found. Declare it as envvar or define a default value.'.format(option))
decouple.UndefinedValueError: file_path not found. Declare it as envvar or define a default value.
whether I used this
from decouple import config
FILE_PATH = config('file_path')
in settings.py directly or views.py directly or first in settings.py and then in views.py like the example shown above]
I reproduced the same code you explained and it is working for me. The .env file should be in the root folder where manage.py exists. Make sure you are referencing the same settings file:
python manage.py runserver --settings=yourproj.settings.production
In the .env file:
file_path='/my/path'
In the settings.py file:
from decouple import config
FILE_PATH = config('file_path')
Also in the views.py file import the should be like this:
from django.conf import settings
print(settings.FILE_PATH)
In the .env file, the values assigned to variables were not enclosed in qoutes and that was why it was giving the error that it was unable to find file_path variable.
The .env file should be like this:-
SECRET_KEY='<django_app_secret_key>'
file_path='<path_to_the file>'
Anyways thanks to #Iain Shelvington and #Prakash S for your help.
Related
I need to change the output/working directory of the hydra config framework in such a way that it lies outside of my project directory. According to my understanding and the doc, config.yaml would need to look like this:
exp_nr: 0.0.0.0
condition: something
hydra:
run:
dir: /absolute/path/to/folder/${exp_nr}/${condition}/
In my code, I then tried to access and set the path like this:
import os
import hydra
from omegaconf import DictConfig
#hydra.main(config_path="../../config", config_name="config", version_base="1.3")
def main(cfg: DictConfig):
print(cfg)
cwd = os.getcwd()
print(f"The current working directory is {cwd}")
owd = hydra.utils.get_original_cwd()
print(f"The Hydra original working directory is {owd}")
work_dir = cfg.hydra.run.dir
print(f"The work directory should be {work_dir}")
But I get the following output and error:
{'exp_nr': '0.0.0.0', 'condition': 'something'}
The current working directory is /project/path/subdir/subsubdir
The Hydra original working directory is /project/path/subdir/subsubdir
Error executing job with overrides: ['exp_nr=1.0.0.0', 'condition=somethingelse']
Traceback (most recent call last):
File "/project/path/subdir/subsubdir/model.py", line 13, in main
work_dir = cfg.hydra.run.dir
omegaconf.errors.ConfigAttributeError: Key 'hydra' is not in struct
full_key: hydra
object_type=dict
I see that hydra.run.dir doesn't appear in the cfg dict printed first but how can I access the path through the config if os.getcwd() isn't set already? Or what did I do wrong?
The path is correct as I already saved files to the folder before integrating hydra and if the process isn't killed due to the error the folder also gets created but hydra doesn't save any files to it, not even the log file with the parameters it should save by default. I also tried to set the path relative to the standard output path or having an extra config parameter work_dir: ${hydra.run.dir} (returns an Interpolation error).
You can access the Hydra config via the HydraConfig singleton documented here.
from hydra.core.hydra_config import HydraConfig
#hydra.main()
def my_app(cfg: DictConfig) -> None:
print(HydraConfig.get().job.name)
I need help on how to call specific variables depending on the environment. Currently, the project's root contains .env file that contains all environment variables being used by the project, but I need to load it according to environment
local -> where development happens
staging -> where we test our project
production -> where we deploy our final project
So in my project root, the goal is to use the file that contains specific values for variables depending on the environment where the [project runs. We have 3 files for the environment variables
local.env
INTEGRATION_URL=http://sandbox.api.com
staging.env
INTEGRATION_URL=http://sandbox2.api.com
production.env
INTEGRATION_URL=http://production.api.com
So inside client.py
from dotenv import load_dotenv
load_dotenv(os.getenv('local.env'))
class Client(BaseClient):
def _init_(self):
self.endpoint = os.environ.get('INTEGRATION_URL') // it should be `http://sandbox.api.com` if I run the project locally
But with this implementation, it still loads the default environment variables defined in the .env file. I need to remove the.env file and just used the 3 env files
Any help?
I solved it by removing all variables in .env, and putting a new variable that would specify which file should be called
.env
ENV_PATH=local.env // or prod.env // or staging.env
Then in client.py
from dotenv import load_dotenv
load_dotenv(os.getenv('ENV_PATH'), 'prod.env')// default to prod if not set
class Client(BaseClient):
def _init_(self):
self.endpoint = os.environ.get('INTEGRATION_URL')
Some variables had to be moved from the main server.py in to a configuration file called config.py.
Its contents are something like this:
class Config(object):
#All the settings and variables go here
In server.py, in invoke this file with:
app = Flask(__name__)
app.config.from_object("config.py")
However, i get this error:
werkzeug.utils.ImportStringError: import_string() failed for 'config.py'. Possible reasons are:
- missing __init__.py in a package;
- package or module path not included in sys.path;
- duplicated package or module name taking precedence in sys.path;
- missing module, class, function or variable;
Debugged import:
- 'config' found in '/mnt/io_files/config.py'.
- 'config.py' not found.
Original exception:
ImportError: module 'config' has no attribute 'py'
I don't know what to do in order to properly invoke this file. In the same directory, i do have a __init__.py file.
This is what worked eventually (i don't know what was going on previously):
#server.py
app = Flask(__name__)
app.config.from_object("config")
#config.py
class Config(object):
#String variables go here
I'm a total newbie and I'm trying to do this project this is my first time, and it's almost done. I tried every method mentioned in this SO thread to move secret key from settings. In every method i got some kind of error, even from this official django doc mathod. I couldn't find where I'm making mistake.
When the secret key is inside the settings.py, everything is working super smooth. But I need to push my code in git, so i have to hide it from settings.py.
Right now im adding the details when i tried using django-environ, to keep secret key outside of settings.py.
im putting the contents inside the root project folder.
im using miniconda: 4.10.1. here is my requirement.txt.
# platform: linux-64
_libgcc_mutex=0.1=main
_openmp_mutex=4.5=1_gnu
appdirs=1.4.4=py_0
asgiref=3.3.4=pyhd3eb1b0_0
attrs=21.2.0=pyhd3eb1b0_0
black=19.10b0=py_0
ca-certificates=2021.5.30=ha878542_0
certifi=2021.5.30=py39hf3d152e_0
click=8.0.1=pyhd3eb1b0_0
django=3.2.4=pyhd3eb1b0_0
django-environ=0.4.5=py_1
importlib-metadata=3.10.0=py39h06a4308_0
krb5=1.17.1=h173b8e3_0
ld_impl_linux-64=2.35.1=h7274673_9
libedit=3.1.20210216=h27cfd23_1
libffi=3.3=he6710b0_2
libgcc-ng=9.3.0=h5101ec6_17
libgomp=9.3.0=h5101ec6_17
libpq=12.2=h20c2e04_0
libstdcxx-ng=9.3.0=hd4cf53a_17
mypy_extensions=0.4.1=py39h06a4308_0
ncurses=6.2=he6710b0_1
openssl=1.1.1k=h7f98852_0
pathspec=0.7.0=py_0
pip=21.1.2=py39h06a4308_0
psycopg2=2.8.6=py39h3c74f83_1
python=3.9.5=h12debd9_4
python_abi=3.9=1_cp39
pytz=2021.1=pyhd3eb1b0_0
readline=8.1=h27cfd23_0
regex=2021.4.4=py39h27cfd23_0
setuptools=52.0.0=py39h06a4308_0
six=1.16.0=pyh6c4a22f_0
sqlite=3.35.4=hdfb4753_0
sqlparse=0.4.1=py_0
tk=8.6.10=hbc83047_0
toml=0.10.2=pyhd3eb1b0_0
typed-ast=1.4.2=py39h27cfd23_1
typing_extensions=3.7.4.3=pyha847dfd_0
tzdata=2020f=h52ac0ba_0
wheel=0.36.2=pyhd3eb1b0_0
xz=5.2.5=h7b6447c_0
zipp=3.4.1=pyhd3eb1b0_0
zlib=1.2.11=h7b6447c_3
settings.py
import os
import environ
from pathlib import Path
env = environ.Env(
# set casting, default value
DEBUG=(bool, False)
)
# reading .env file
environ.Env.read_env()
# Build paths inside the project like this: BASE_DIR / 'subdir'.
BASE_DIR = Path(__file__).resolve().parent.parent
# Quick-start development settings - unsuitable for production
# See https://docs.djangoproject.com/en/3.2/howto/deployment/checklist/
# SECURITY WARNING: keep the secret key used in production secret!
SECRET_KEY = env('SECRET_KEY')
# False if not in os.environ
DEBUG = env('DEBUG')
im not adding the rest of settings. i dont think its important. if need please mention I.ll update.
i placed .env file in root of the project where manage.py and db.sqlite3 are placed
.env
#env file
DEBUG=on
#copied the entire line from settings.py
SECRET_KEY ='xxxx django secret key here xxxx'
while running "python manage.py runserver", i got this error.
im not sure what im missing. i got some kind of error, when i tried each method and errors are not same. sorry that i cannot explain every method and error here.
there are several questions asked in this form. but most are not answered and some are not accurately explains my situation. please mention if anything else is needed or for more clarification.
First check that you have installed django-environ and maybe you have a typing mistake in your requirements.txt it should be django-environ=0.4.5 instead of django-environ=0.4.5=py_1
you can pass the path of your .env inside read_env(env_file="relative_path_of_your_env_file")
it read a .env file into os.environ.
If not given a path to a dotenv path, does filthy magic stack backtracking
to find manage.py and then find the dotenv.
go through this code https://github.com/joke2k/django-environ/blob/master/environ/environ.py#L614
From the structure of the file tree, its clear that .env file is placed in the root folder of the project. When checking the error message, its visible that whoever is searching for .env file is checking at the same place as settings.py.
So, the short answer is if you are using django-environ to keep secret-key outside, place .env file together with settings.py in the same directory.
For a bit more elaborated content, you can refer to this link. I felt it is suitable for newbies.
I have a serverless code in python. I am using serverless-python-requirements:^4.3.0 to deploy this into AWS lambda.
My code imports another python file in same directory as itself, which is throwing an error.
serverless.yml:
functions:
hello:
handler: functions/pleasework.handle_event
memorySize: 128
tags:
Name: HelloWorld
Environment: Ops
package:
include:
- functions/pleasework
- functions/__init__.py
- functions/config
(venv) ➜ functions git:(master) ✗ ls
__init__.py boto_client_provider.py config.py handler.py sns_publish.py
__pycache__ cloudtrail_handler.py glue_handler.py pleasework.py
As you can see, pleasework.py and config are in same folder, but when I do import config in pleasework I get an error:
{
"errorMessage": "Unable to import module 'functions/pleasework': No module named 'config'",
"errorType": "Runtime.ImportModuleError"
}
I am struggling with this for few days and think I am missing something basic.
import boto3
import config
def handle_event(event, context):
print('lol: ')
ok, so i found out my isssue. Way i was importing the file was wrong
Instead of
import config
I should be doing
import functions.config
#Pranay Sharma's answer worked for me.
An alternate way is creating and setting PYTHONPATH environment variable to the directory where your handler function and config exist.
To set environment variables in the Lambda console
Open the Functions page of the Lambda console.
Choose a function.
Under Environment variables, choose Edit.
Choose Add environment variable.
Enter a key and value.
In our case Key is "PYTHONPATH" and value is "functions"