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')