This question already has answers here:
Warning message while running Flask
(12 answers)
Closed 2 years ago.
from flask import Flask
app = Flask(_name_)
#app.route('/')
def index():
return '<h1>Hello, world</h1>'
I get this error
* Serving Flask app "application.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
This is just a warning as flask server is not meant for production.
You can use
export FLASK_ENV=development
before flask run
It depends if you are running it from command line or adding python code to run it.
If you want to run it with Python, add this to the bottom of your code.
if __name__ == "__main__":
app.run(host='0.0.0.0', port=80)
(you can change the port number, to access it on a browser, type: 127.0.0.1:80)
If you are running it from command line, you run
flask run
Note this appears every time:
Serving Flask app "application.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
All it tells you is that this server is only used for development or production. Hosting it for commercial/public use would require additional files and web hosting.
You can allow debugging to be on with
if __name__ == "__main__":
app.run(host='0.0.0.0', port=80, DEBUG=True)
OR
export FLASK_ENV=development (linux/macos)
set FLASK_ENV=development (Windows)
Hope this helps (follow me on GitHub #haydenso)
Related
I'm trying to run a flask application in debug mode (or at least a mode where it will reload after changing the files).
I'm aware of export FLASK_ENV=development, however I am working on a university online development environment, and I lose the environment variables every time the site reloads, which while not the end of the world, is slightly annoying, and I'd rather avoid having to keep typing it (lazy I know).
If I include the following, and run using python3 main.py, debug mode is activated, however when using flask run, debug remains off.
if __name__ == "__main__":
app.run(debug=True)
However, as I understand it, using the flask run command is the preferred way to launch the app, not using python app.py.
I've found ideas such as including the following, however none of these have activated debug mode, so I'm wondering whether it is even possible:
app.config['ENV'] = 'development'
app.config['DEBUG'] = True
app.config['TESTING'] = True
I've simplified my code to the following to see if it was an error in my original piece, but it doesn't seem to be:
from flask import Flask
app = Flask(__name__)
app.config['ENV'] = 'development'
app.config['DEBUG'] = True
app.config['TESTING'] = True
#app.route('/')
def home():
return '<h1>debugging!</h1>'
if __name__ == "__main__":
app.run(debug=True)
In short, there does not seem to be a way of using flask run how I want without assigning environment variables, however using a .flaskenv file will allow environment variables to be loaded at run time.
The .flaskenv file for example could include the following ENVs among others to be loaded:
FLASK_APP=main:app
FLASK_ENV=development
FLASK_DEBUG=1
Note - this does require python-dotenv to be installed to use.
All credit to #cizario who answered here with some more detail:
https://stackoverflow.com/a/64623193/12368419
Well, when it's Flask 2.2 or later, you can just do
flask --debug run
* Debug mode: on
WARNING: This is a development server. Do not use it in a production deployment. Use a production WSGI server instead.
* Running on http://127.0.0.1:5000
Press CTRL+C to quit
* Restarting with stat
* Debugger is active!
* Debugger PIN: 560-342-853
This question already has answers here:
How to host Python 3.7 flask application on Windows Server?
(1 answer)
Are a WSGI server and HTTP server required to serve a Flask app?
(3 answers)
Serving Flask app with waitress on windows
(8 answers)
Closed 1 year ago.
I want to run a local server which executes as specific script when a GET request to the IP is made,
because of that I need to run the server in the background (while a other script is runing too) without printing any stuff.
Here is my Code:
from flask import Flask
app = Flask(__name__)
#app.route('/', methods = ['GET'])
def index():
return "Hello How are you?"
if __name__ == '__main__':
app.run(port = 5000)
I tried it with Threading but I always get stuck at this Output
When running this code I get this Output:
* 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
* Running on http://IP:5000/ (Press CTRL+C to quit)```
You could use pythonw to run the application in the background:
pythonw -m flask run > log.txt 2>&1
make sure the name of the Python file is app.py.
It's hard to remember when, but at one point the auto-reload function of Flask started to not work anymore in my project.
This is the output upon starting my app :
FLASK_APP = back/python/app/app.py:app
FLASK_ENV = development
FLASK_DEBUG = 1
In folder C:/path/to/project
ssh://[VirtualMachineIP]:22/root/env/bin/python3.7 -u -m flask run -h 0.0.0.0 -p 1234
* Serving Flask app 'back/python/app/app.py:app' (lazy loading)
* Environment: development
* Debug mode: on
* Running on all addresses.
WARNING: This is a development server. Do not use it in a production deployment.
* Running on http://[VirtualMachineIP]:1234/ (Press CTRL+C to quit)
* Restarting with stat
* Debugger is active!
* Debugger PIN: 106-048-128
The development environment and Debug mode are both on. Thus, upon saving changes in a file (while the app is deployed) I get the usual message :
* Detected change in '/path/to/changed/file.py', reloading
Signaling that the app is reloading with the new code. Except it doesn't reload anything, and the message doesn't appear on any further changes until I'm forced to restart the app.
PyCharms runs on Windows and communicates via ssh to my Virtual Machine, where the code is executed. I have installed the following modules:
flask
flask-socketio
eventlet
flask-cors
Any help is welcomed. Thanks :)
The FLASK_DEBUG environment variable is badly supported, it may not behave as expected if set in code. (Quoted from the source of flask).
It suggest to use flask run in debug mode.
eg: $ flask --app hello --debug run
If it still not work, you can force to use reloader like this:
if __name__ == '__main__':
app.run(host=config.HOST, port=config.PORT, debug=True)
Take care, the app.run() must be wrapped with if __name__ == '__main__'.
doc: https://flask.palletsprojects.com/en/2.2.x/config/#DEBUG
I installed the Flask plugin in PyCharm Community Edition and I just have this simple code in my flask app:
from flask import Flask
app = Flask(__name__)
#app.route('/')
def index():
return '<h1>Hello!</h1>'
if __name__ == "__main__":
app.run(debug=True)
And I get this message:
WARNING: This is a development server. Do not use it in a production deployment.
Use a production WSGI server instead
* Restarting with stat
* Debugger is active!
* Debugger PIN: 123-456-789
* Running on http://127.0.0.1:5000/
Why am I getting this error when I run Flask?
A previous version of the message read "Do not use the development server in a production environment."
For deploying an application to production, one option is to use Waitress, a production WSGI server.
Here is an example of using waitress in the code.
from flask import Flask
app = Flask(__name__)
#app.route("/")
def index():
return "<h1>Hello!</h1>"
if __name__ == "__main__":
from waitress import serve
serve(app, host="0.0.0.0", port=8080)
Running the application:
$ python hello.py
Waitress also provides a command line utility waitress-serve. To use that, you can modify the code to the following:
from flask import Flask
app = Flask(__name__)
#app.route("/")
def index():
return "<h1>Hello!</h1>"
def create_app():
return app
Then we can use waitress-serve as the following:
waitress-serve --port=8080 --call hello:create_app
And BTW, 8080 is the default port.
To validate the deployment, open a separate window:
% curl localhost:8080
<h1>Hello!</h1>%
Or directly in your browser http://localhost:8080/.
Other alternatives to deploy your app include Gunicorn and uWSGI. For more details, please refer to the flask deploy doc.
As of Flask 2.2, the development server always shows this warning, it is not possible to disable it. The development server is not intended for use in production. It is not designed to be particularly efficient, stable, or secure. Use a production WSGI server instead. See the deployment docs from Flask for more information.
That warning is just a warning though, it's not an error preventing your app from running. If your app isn't working, there's something else wrong with your code.
That warning applies to the development server, not Flask itself. The Flask framework is appropriate for any type of application and deployment.
To avoid these messsages, inside the CLI (Command Line Interface), run these commands.
export FLASK_APP=app.py
export FLASK_ENV=development
export FLASK_DEBUG=0
flask run
If for some people (like me earlier) the above answers don't work, I think the following answer would work (for Mac users I think)
Enter the following commands to do flask run
$ export FLASK_APP=hello.py
$ export FLASK_ENV=development
$ flask run
Alternatively you can do the following (I haven't tried this but one resource online talks about it)
$ export FLASK_APP=hello.py
$ python -m flask run
source: For more
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
This worked for me on windows:
$env:FLASK_APP="flask_project.py"
$env:FLASK_ENV="development"
flask run
flask_project.py is on the same path as my virtual environment.
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')