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.
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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.
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
I'm starting a flask application like this:
from idapi.wsgi import app
app.run(port=8080, debug=True)
Where app is an object of a subclass of flask.Flask. Within the application, I'm only ever calling app.logger.<level> (or self.logger.<level> within the class itself). I'm never importing the Python logging module itself.
My understanding was that when startring a flask app with debug=True, I should see see messages from app.logger.debug(...) and app.logger.info(...), but I'm not.
I can solve this problem by explicitly configuring the root logger:
import logging
from idapi.wsgi import app
logging.basicConfig(level='DEBUG')
app.run(port=8080, debug=True)
Or I can use app.logger.warning(...) instead. What could be going on here...am I incorrect in my understanding of how logging is configured by Flask when running in debug mode?
I'm not sure, but in the documentation of Flask I've seen that you need to set variable FLASK_ENV to development if you want to see debug messages:
To enable all development features (including debug mode) you can
export the FLASK_ENV environment variable and set it to development
before running the server:
$ export FLASK_ENV=development
$ flask run
How are you meant to debug errors in Flask? Print to the console? Flash messages to the page? Or is there a more powerful option available to figure out what's happening when something goes wrong?
Running the app in debug mode will show an interactive traceback and console in the browser when there is an error. As of Flask 2.2, to run in debug mode, pass the --app and --debug options to the flask command.
$ flask --app example --debug run
Prior to Flask 2.2, this was controlled by the FLASK_ENV=development environment variable instead. You can still use FLASK_APP and FLASK_DEBUG=1 instead of the options above.
For Linux, Mac, Linux Subsystem for Windows, Git Bash on Windows, etc.:
$ export FLASK_APP=example
$ export FLASK_DEBUG=1
$ flask run
For Windows CMD, use set instead of export:
set FLASK_DEBUG=1
For PowerShell, use $env:
$env:FLASK_DEBUG = "1"
If you're using the app.run() method instead of the flask run command, pass debug=True to enable debug mode.
Tracebacks are also printed to the terminal running the server, regardless of development mode.
If you're using PyCharm, VS Code, etc., you can take advantage of its debugger to step through the code with breakpoints. The run configuration can point to a script calling app.run(debug=True, use_reloader=False), or point it at the venv/bin/flask script and use it as you would from the command line. You can leave the reloader disabled, but a reload will kill the debugging context and you will have to catch a breakpoint again.
You can also use pdb, pudb, or another terminal debugger by calling set_trace in the view where you want to start debugging.
Be sure not to use too-broad except blocks. Surrounding all your code with a catch-all try... except... will silence the error you want to debug. It's unnecessary in general, since Flask will already handle exceptions by showing the debugger or a 500 error and printing the traceback to the console.
You can use app.run(debug=True) for the Werkzeug Debugger edit as mentioned below, and I should have known.
From the 1.1.x documentation, you can enable debug mode by exporting an environment variable to your shell prompt:
export FLASK_APP=/daemon/api/views.py # path to app
export FLASK_DEBUG=1
python -m flask run --host=0.0.0.0
One can also use the Flask Debug Toolbar extension to get more detailed information embedded in rendered pages.
from flask import Flask
from flask_debugtoolbar import DebugToolbarExtension
import logging
app = Flask(__name__)
app.debug = True
app.secret_key = 'development key'
toolbar = DebugToolbarExtension(app)
#app.route('/')
def index():
logging.warning("See this message in Flask Debug Toolbar!")
return "<html><body></body></html>"
Start the application as follows:
FLASK_APP=main.py FLASK_DEBUG=1 flask run
If you're using Visual Studio Code, replace
app.run(debug=True)
with
app.run()
It appears when turning on the internal debugger disables the VS Code debugger.
If you want to debug your flask app then just go to the folder where flask app is. Don't forget to activate your virtual environment and paste the lines in the console change "mainfilename" to flask main file.
export FLASK_APP="mainfilename.py"
export FLASK_DEBUG=1
python -m flask run --host=0.0.0.0
After you enable your debugger for flask app almost every error will be printed on the console or on the browser window.
If you want to figure out what's happening, you can use simple print statements or you can also use console.log() for javascript code.
To activate debug mode in flask you simply type set FLASK_DEBUG=1 on your CMD for windows, or export FLASK_DEBUG=1 on Linux terminal then restart your app and you are good to go!!
Install python-dotenv in your virtual environment.
Create a .flaskenv in your project root. By project root, I mean the folder which has your app.py file
Inside this file write the following:
FLASK_APP=myapp
FLASK_ENV=development
Now issue the following command:
flask run
When running as python app.py instead of the flask command, you can pass debug=True to app.run.
if __name__ == "__main__":
app.run(debug=True)
$ python app.py
with virtual env activate
export FLASK_DEBUG=true
you can configure
export FLASK_APP=app.py # run.py
export FLASK_ENV = "development"
to start
flask run
the result
* Environment: development
* Debug mode: on
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
* Restarting with stat
* Debugger is active!
* Debugger PIN: xxx-xxx-xxx
and if you change
export FLASK_DEBUG=false
* Environment: development
* Debug mode: off
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
For Windows users:
Open Powershell and cd into your project directory.
Use these commandos in Powershell, all the other stuff won't work in Powershell.
$env:FLASK_APP = "app"
$env:FLASK_ENV = "development"
If you have PyCharm Professional, you can create a Flask server run configuration and enable the FLASK_DEBUG checkbox. Go to Run > Edit Configurations, select or create a Flask server configuration, and enable the FLASK_DEBUG checkbox. Click OK, then click the run button.
You can install python-dotenv with
pip install python-dotenv then create a .flask_env or a .env file
The contents of the file can be:
FLASK_APP=myapp
FLASK_DEBUG=True
Use loggers and print statements in the Development Environment, you can go for sentry in case of production environments.
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')