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SLACK BOLT PYTHON - Although the app should be installed into this workspace, the AuthorizeResult (returned value from authorize) for it was not found

I want to build a simple Slack bolt python project so I followed this document.
but when I use python_dotenv and then run my main file (app.py) I face this error:
As `installation_store` or `authorize` has been used, `token` (or SLACK_BOT_TOKEN env variable) will be ignored.
Although the app should be installed into this workspace, the AuthorizeResult (returned value from authorize) for it was not found.
NOTE: by deleting this line in the main file (app.py):
load_dotenv()
and use the export method for defining tokens, everything works correctly.
this is my main file:
import os
from dotenv import load_dotenv
from slack_bolt import App
from slack_bolt.adapter.socket_mode import SocketModeHandler
import logging
load_dotenv() # by deleting this file, the error will be gone, but I want using the dotenv pip
app = App(token=os.environ.get("SLACK_BOT_TOKEN"))
logger = logging.getLogger(__name__)
#app.message("hello")
def message_hello(message, say):
# say() sends a message to the channel where the event was triggered
say(f"Hey there <#{message['user']}>!")
# Start your app
if __name__ == "__main__":
SocketModeHandler(app, os.environ["SLACK_APP_TOKEN"]).start()

I don't know exactly why, but I removed extra keys in the .env file, and everything went fine.
in .env file:
before:
SLACK_APP_TOKEN=xapp-fake-token
SLACK_BOT_TOKEN=xoxb-fake-token
USER_TOKEN=xoxp-fake-token
SLACK_CLIENT_ID=aa.bb
SLACK_CLIENT_SECRET=YOUR_CLIENT_SECRET
SLACK_API_TOKEN_APP_LEGACY=xoxb-fake-token
SLACK_SIGNING_SECRET=YOUT_SIGNING_SECRET
after:
SLACK_APP_TOKEN=xapp-fake-token
SLACK_BOT_TOKEN=xoxb-fake-token

Celery for Testing Configuration in Django

I have a file celery_test.py which should load the configuration settings for testing(config.test) in Django but instead, it loads configuration of Development(config.development)
from __future__ import absolute_import
import os
from django.conf import settings
from celery import Celery
os.environ.setdefault(
"DJANGO_SETTINGS_MODULE", "config.testing"
)
test_app = Celery('test')
test_app.config_from_object('django.conf:settings')
test_app.autodiscover_tasks()
print("Testing celery")
print(settings)
#test_app.task(bind=True)
def debug_task(self):
print('Request: {0!r}'.format(self.request))
When I am printing settings it printing config.development instead of config.testing. Can anyone help me here how to load config.testing?

Seeing that you use django I would suggest not actually overriding the settings file like this. Instead use the django.test.override_settings decorator to override the specific settings you want to change. This way you're always testing the most current settings file, so you don't have to keep both up-to-date.
More information is also in the documentation: https://docs.djangoproject.com/en/3.0/topics/testing/tools/#django.test.override_settings.
I can't really provide an example, since you don't provide any unittest example. But I hope the linked documentation is clear enough.

Assuming your are using pytest, you could put this in your pytest.ini file:
[pytest]
DJANGO_SETTINGS_MODULE = celery_test.py
Or use:
DJANGO_SETTINGS_MODULE=celery_test.py pytest
for a one-liner.

Python Config Parser environment variables

I am trying to read some environment variables into my ConfigParser file.
I was referring to this answer but I am getting
"InterpolationDepthError: Value interpolation too deeply recursive" error.
section: [amazon]
option : amazon_access_key
rawval : %(AMAZON_ACCESS_KEY)s
Here is the portion of my config file:
[amazon]
amazon_access_key=%(AMAZON_ACCESS_KEY)s
amazon_secret_key=%(AMAZON_SECRET_KEY)s
And this is what I am writing to call the file:
from ConfigParser import SafeConfigParser
import os
config = SafeConfigParser(os.environ)
config.read('config.txt')
When I am calling these variables directly in the "amazon" section, I get the above stated error.
When I call these variables in the "default" section, like:
[default]
aws_access_key=%(AMAZON_ACCESS_KEY)s
aws_secret_key=%(AMAZON_SECRET_KEY)s
[amazon]
amazon_access_key=%(aws_access_key)s
amazon_secret_key=%(aws_secret_key)s
I get the following error:
ConfigParser.InterpolationMissingOptionError: Bad value substitution:
section: [amazon]
option : amazon_access_key
key : aws_access_key
rawval : %(aws_access_key)s
What am I missing here?
Also, how can I have separate config files for local and production deployments? Currently, all the configurations are same for local and production environments.

from ConfigParser import SafeConfigParser
import ConfigParser
import os
class CaseConfigParser(SafeConfigParser):
def optionxform(self, optionstr):
return optionstr
config = CaseConfigParser(os.environ)
config.read('config.ini')
print config.get('amazon', 'amazon_access_key')
Explanation:
The problem here is with optionxform, which turns all options to lower case by default. eventually, it will have key and value equal, similar to following.
[amazon]
amazon_access_key=%(amazon_access_key)s
Another solution can be - change names of keys in config.txt NOT equal to env vars.
[amazon]
aak=%(AMAZON_ACCESS_KEY)s
and then
config = SafeConfigParser(os.environ)
config.read('config.ini')
print config.get('amazon', 'aak')

You should remember that the default section should be capitalised.
[DEFAULT]
Making this change should solve the problem.

How to get config params in another module in flask python?

In my main.py have the below code:
app.config.from_object('config.DevelopmentConfig')
In another module I used import main and then used main.app.config['KEY'] to get a parameter, but Python interpreter says that it couldn't load the module in main.py because of the import part. How can I access config parameters in another module in Flask?

Your structure is not really clear but by what I can get, import your configuration object and just pass it to app.config.from_object():
from flask import Flask
from <path_to_config_module>.config import DevelopmentConfig
app = Flask('Project')
app.config.from_object(DevelopmentConfig)
if __name__ == "__main__":
application.run(host="0.0.0.0")
if your your config module is in the same directory where your application module is, you can just use :
from .config import DevelopmentConfig

The solution was to put app initialization in another file (e.g: myapp_init_file.py) in the root:
from flask import Flask
app = Flask(__name__)
# Change this on production environment to: config.ProductionConfig
app.config.from_object('config.DevelopmentConfig')
Now to access config parameters I just need to import this module in different files:
from myapp_init_file import app
Now I have access to my config parameters as below:
app.config['url']
The problem was that I had an import loop an could not run my python app. With this solution everything works like a charm. ;-)

Import variables into a class

I have a file that contains all the variables my application needs:
config.py
AWS_KEY=1234
AWS_SECRET=5678
modules/db.py
from project.libraries import boto as boto
conn = boto.connect_dynamodb(
aws_access_key_id=AWS_KEY
aws_secret_access_key=AWS_SECRET)
How can I get the variables from config.py into modules/db.py?
(Or, is this the wrong approach?)
Thanks.

modules/db.py
import config
AWS_KEY = config.AWS_KEY
Alternatively,
from config import AWS_KEY, AWS_SECRET
Or, if you want to keep the config prefix,
import config
print config.AWS_KEY
Forgive me if you've already done so, but have you read about python modules?