Develop Reference

Flask: not seeing .env outside of root directory

As per flask documentation, FLASK_ENV environmental variable determines whether flask runs in development or production mode.
Hence I have a .env file like so:
FLASK_ENV="development"
and my app.py looks like this:
load_dotenv(find_dotenv())
app = Flask(__name__)
config = DevConfig() if os.environ.get('FLASK_ENV') == 'development' else ProdConfig()
app.config.from_object(config)
Now here's the problem: if I move .env into another folder (in my case config), flask stops seeing it. More specifically (and weirdly):
The env variable is set ok (I can print it from different parts of the app)
The config loads ok (dev config loads indeed)
But flask app itself says:
Loading .env environment variables…
* Serving Flask app "app.py"
* Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
Use a production WSGI server instead.
* Debug mode: off
How is it possible that env variable is set, but flask still thinks it's running in prod? Again this only happens when I move .env away from the root folder.

Error: Could not locate a Flask application in VSCode

I am trying to learn Flask using VScode.
The tutorial that I am following is: Python Flask Tutorial: Full-Featured Web App Part 1 - Getting Started.
I did the following things:
Created a new virtualenv in a folder using: virtualenv venv
activated it as: venv\Scripts\activate (I am on Windows 10)
After that, I created a new directory named Flask_Blog using mkdir Flask_Blog and in it, I created a new flaskblog.py file containing the following code:
from flask import Flask
app = Flask(__name__)
#app.route('/')
def hello():
return 'Hello'
Then, in the terminal of VScode, I changed my working directory in order to be in the Flask_Blog directory using cd Flask_Blog.
Now, when I am doing set FLASK_APP=flaskblog.py followed by flask run, I am getting the following error:
(venv) PS C:\Users\kashy\OneDrive\Desktop\Flask\Flask_Blog> flask run
* Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
Use a production WSGI server instead.
* Debug mode: off
Usage: flask run [OPTIONS]
Error: Could not locate a Flask application. You did not provide the "FLASK_APP" environment variable, and a "wsgi.py" or "app.py" module was not found in the current directory.
But
When I do the same in the cmd prompt, the code runs and I get to see the output.
I am completely new to this. Can anyone please tell me what is the mistake I am doing in VSCode and why is it working in the cmd?

Issue raised in VsCode
Under Powershell, you have to set the FLASK_APP environment variable as follows:
$env:FLASK_APP = "webapp"
Then you should be able to run "python -m flask run" inside the hello_app folder. In other words, PowerShell manages environment variables differently, so the standard command-line "set FLASK_APP=webapp" won't work.

Try Set FLASK_APP = Full path of the folder/filename.py.
This worked for me

This worked for me on the VSCode:
$env:FLASK_APP= 'C:\Python\ex003\main:app'

Set Flask environment to development mode as default?

Every time I start up my flask app the environment variable is set to production. I want to have it set to development mode by default. Otherwise every time I start my app i have to run ..
export FLASK_ENV=development
How can I set environment's default value as development in every startup?
EDIT: I am using flask in a virtual environment on a raspberry pi.

You can edit your main flask app file and add these lines:
if __name__ == '__main__':
app.run(debug=True)
Using this method you have to run your flask app with Python interpreter like this => python app.py
Best Practice:
Install python-dotenv package inside your working environment =>pip install python-dotenv
Create a file named .env, put your environment variables in it, for your case it's FLASK_ENV=development
Then add this code to your config.py or some file that will get loaded before Flask main App
from dotenv import load_dotenv
dotenv_path = join(dirname(__file__), '.env') # Path to .env file
load_dotenv(dotenv_path)
Note that: If you are using flask command to run your application, you don't need to do the third step, flask will find .env files in the project directory by itself.
Using this method, it will only set Environment variable for the project that you have added this code to.

On Linux distro, like "Raspberry pi o.s", specify the environment on the terminal with the code below.
Unless you specify the environment, flask will assume production.
export FLASK_ENV=development
flask run

Like the first answer and instead of adding the variable to a .env file which can be forgotten, do this instead.
This way, if you try to run the file in production, you'll get an assertion error to remind you to actually use a dedicated web server (which "imports" the app). If you run locally, not only will you be reminded to use a .env file, but in the case no environment file is needed, the flask env is set to development to avoid any production conflicts.
import os
app = Flask(__name__)
IS_DEV = app.env == 'development' # FLASK_ENV env. variable
# code
if __name__ == '__main__':
# guaranteed to not be run on a production server
assert os.path.exists('.env') # for other environment variables...
os.environ['FLASK_ENV'] = 'development' # HARD CODE since default is production
app.run(debug=True)

Warning message while running Flask

While I am running Flask code from my command line, a warning is appearing:
Serving Flask app "hello_flask" (lazy loading)
* Environment: production
WARNING: Do not use the development server in a production environment.
Use a production WSGI server instead.
What does this mean?

As stated in the Flask documentation:
While lightweight and easy to use, Flask’s built-in server is not suitable for production as it doesn’t scale well and by default serves only one request at a time.
Given that a web application is expected to handle multiple concurrent requests from multiple users, Flask is warning you that the development server will not do this (by default). It recommends using a Web Server Gateway Interface (WSGI) server (numerous possibilities are listed in the deployment docs with further instructions for each) that will function as your web/application server and call Flask as it serves requests.

Try gevent:
from flask import Flask
from gevent.pywsgi import WSGIServer
app = Flask(__name__)
#app.route('/api', methods=['GET'])
def index():
return "Hello, World!"
if __name__ == '__main__':
# Debug/Development
# app.run(debug=True, host="0.0.0.0", port="5000")
# Production
http_server = WSGIServer(('', 5000), app)
http_server.serve_forever()
Note: Install gevent using pip install gevent

As of Flask 1.x, the default environment is set to production.
To use the development environment, create a file called .flaskenv and save it in the top-level (root) of your project directory. Set the FLASK_ENV=development in the .flaskenv file. You can also save the FLASK_APP=myapp.py.
Example:
myproject/.flaskenv:
FLASK_APP=myapp.py
FLASK_ENV=development
Then you just execute this on the command line:
flask run
That should take care of the warning.

To remove the "Do not use the development server in a production environment." warning, run:
export FLASK_ENV=development
before flask run.

I was typing flask run and then saw this message after that I solve this issue with these:
1- Add this text in your myproject/.flaskenv :
FLASK_APP=myapp.py
FLASK_ENV=development
also you should type "pip3 install python-dotenv" for using this file .flaskenv
2-in your project folder type in terminal your flask command which one you use :
flask-3 run

First, try to the following :
set FLASK_ENV=development
then run your app.

I have been using flask for quite some time now, and today, suddenly this warning turned up. I found this.
As mentioned here, as of flask version 1.0 the environment in which a flask app runs is by default set to production. If you run your app in an older flask version, you won't be seeing this warning.
New in version 1.0.
Changelog
The environment in which the Flask app runs is set by the FLASK_ENV environment variable. If not set it defaults to production. The other recognized environment is development. Flask and extensions may choose to enable behaviors based on the environment.

in configurations or config you can add this code :
ENV = ""
same as if you try to add debug set to true like this
DEBUG = True
for more detail you can check this http://flask.pocoo.org/docs/1.0/config/#ENV

It means the programe is run on production mode even in developing environment.so to avoid that warning, you need to define this is development environment.for that,Type and run below command in project directory on terminal(linux).
export FLASK_ENV=development
if you are windows user then run,
set FLASK_ENV=development

To disable the message I use:
app.env = "development"
You have to put this in the Python-Script before you run the app with:
app.run(host="localhost")

If you encounter NoAppException and you see lazy loading the following seemed to fix the issue:
cd <project directory>
export FLASK_APP=.
export FLASK_ENV=development
export FLASK_DEBUG=1

You can begin your main script like this :
import os
if __name__ == '__main__':
os.environ.setdefault('FLASK_ENV', 'development')

flask.cli.NoAppException: The file/path provided (new_app.py) does not appear to exist

I keep getting this error flask.cli.NoAppException: The file/path provided (new_app.py) does not appear to exist. Please verify the path is correct. If app is not on PYTHONPATH, ensure the extension is .py it goes away after I restart the Flask server.
I am running flask run in the correct directory where my app is. This just started happening after working for 2 weeks. I've read that it could be due to an import error, but I am not finding any modules that are not installed on my virutalenv.
from flask import Flask
app = Flask(__name__)
app.debug=True

Most likely you haven't set the FLASK_APP environment variable.
To run the application you can either use the flask command or
python’s -m switch with Flask. Before you can do that you need to tell
your terminal the application to work with by exporting the FLASK_APP
environment variable:
$ export FLASK_APP=hello.py
$ flask run * Running on http://127.0.0.1:5000/
If you are on Windows you need to use set
instead of export.
Alternatively you can use python -m flask:
$ export FLASK_APP=hello.py
$ python -m flask run * Running on http://127.0.0.1:5000/
EDIT
If you have FLASK_APP set then try adding this to new_app.py
app.run(debug=True, port=8800)
Or if you're on Windows:
if __name__ == '__main__':
app.run(debug=True, port=8800)
And then just execute the app with python new_app.py.