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
Related
I have a large flask application built inside a package called "MyApp" (exactly as shown here: http://flask.pocoo.org/docs/0.12/patterns/packages/)
According to the Flask documentation, the debug mode should enables the following features:
it activates the debugger
it activates the automatic reloader
it enables the debug mode on the Flask application.
At the beginning I've run my flask application with the following command and everything were worked fine:
export FLASK_APP=MyApp
export FLASK_DEBUG=1
flask run
Then I read about the correct way to setup a configuration system (including the debug mode).
So I created the following config.py file:
class Config(object):
DEBUG = False
...
class ProductionConfig(Config):
...
class DevelopmentConfig(Config):
DEVELOPMENT = True
DEBUG = True
...
CONFIGS = {
"development": DevelopmentConfig,
"production": ProductionConfig,
"default": DevelopmentConfig
}
And in my application __init__.py file, I wrote:
app = Flask(__name__)
config_name = os.getenv('FLASK_CONFIGURATION', 'default')
app.config.from_object(CONFIGS[config_name])
Now, to run the application I enter a new command:
export FLASK_APP=MyApp
export FLASK_CONFIGURATION=development
flask run
Unfortunately, this time the debug mode did not activated at all..
No debugger or automatic reloader has been activated.
The only thing that has been changed was that app.debug is now equals to True.
I don't get it.. It looks like the DEBUG = TRUE is not working correctly.
Do you have any idea why does it happen?
Running with the debugger is different than setting the DEBUG config. You have to do both. Running the server in debug mode sets the config automatically. Typically, you should rely on that rather than setting the config directly.
The "correct way to configure" you read about is a) just another way, not the "correct" way, and b) only sets the config, not the FLASK_DEBUG environment variable, which is what controls the debug mode for the server.
Setting the environment variable FLASK_DEBUG=1, or passing the --debug option to the flask command as of Flask 2.2, tells flask run to wrap the application with the debugger and reloader. (app.run(debug=True) does the same, but the flask run command is preferred). app.debug switches some internal behavior in the Flask app, such as passing through errors to the interactive debugger that development mode enabled.
If I run my Flask app through the command "flask run", not only the arguments in app.run are ignored (making debugger false and choosing the default port 5000), but changes made to the script too, e.g. changing url_prefix to "/views". The cmd prompt won't even respond when I save the script after making alterations. This way, changes are only effective after stopping the script then running again.
This doesn't happen when the app script is ran through VSCode commands (Ctrl+F5 or the play button): through those, the script is executed as written and changes are recognized right after saving a script.
Why is that so?
from flask import Flask
from views import views
app = Flask(__name__)
app.register_blueprint(views, url_prefix="/")
if __name__ == '__main__':
app.run(debug=True, port=8000)
C:\Users\*******\Documents\flask_quick_website>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
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
The flask executable works by importing the app object from your code. This is similar to how a WSGI server like gunicorn works (research what this is if you're unsure).
As for when you run the script with VSCode, or in-fact if you launch with the python executable by running python app.py or similar, then anything inside the if __name__ == '__main__': block is executed, in your case the app.run call.
Also note your app.run call is passed the debug=True argument. If you investigate the Flask source code, inside the run function, the use_reloader argument is set to the same value as debug so in this case the auto-reloader runs if debug is True.
So how to make the auto-reloader work with the flask command? Pass the --reload flag:
flask run --reload
You can also set the environment to development to achieve the same. On windows:
> set FLASK_ENV=development
> flask run
See Environment and Debug features for more on this.
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
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')
I have a large flask application built inside a package called "MyApp" (exactly as shown here: http://flask.pocoo.org/docs/0.12/patterns/packages/)
According to the Flask documentation, the debug mode should enables the following features:
it activates the debugger
it activates the automatic reloader
it enables the debug mode on the Flask application.
At the beginning I've run my flask application with the following command and everything were worked fine:
export FLASK_APP=MyApp
export FLASK_DEBUG=1
flask run
Then I read about the correct way to setup a configuration system (including the debug mode).
So I created the following config.py file:
class Config(object):
DEBUG = False
...
class ProductionConfig(Config):
...
class DevelopmentConfig(Config):
DEVELOPMENT = True
DEBUG = True
...
CONFIGS = {
"development": DevelopmentConfig,
"production": ProductionConfig,
"default": DevelopmentConfig
}
And in my application __init__.py file, I wrote:
app = Flask(__name__)
config_name = os.getenv('FLASK_CONFIGURATION', 'default')
app.config.from_object(CONFIGS[config_name])
Now, to run the application I enter a new command:
export FLASK_APP=MyApp
export FLASK_CONFIGURATION=development
flask run
Unfortunately, this time the debug mode did not activated at all..
No debugger or automatic reloader has been activated.
The only thing that has been changed was that app.debug is now equals to True.
I don't get it.. It looks like the DEBUG = TRUE is not working correctly.
Do you have any idea why does it happen?
Running with the debugger is different than setting the DEBUG config. You have to do both. Running the server in debug mode sets the config automatically. Typically, you should rely on that rather than setting the config directly.
The "correct way to configure" you read about is a) just another way, not the "correct" way, and b) only sets the config, not the FLASK_DEBUG environment variable, which is what controls the debug mode for the server.
Setting the environment variable FLASK_DEBUG=1, or passing the --debug option to the flask command as of Flask 2.2, tells flask run to wrap the application with the debugger and reloader. (app.run(debug=True) does the same, but the flask run command is preferred). app.debug switches some internal behavior in the Flask app, such as passing through errors to the interactive debugger that development mode enabled.