An application that uses Flask framework, can start the built-in werkzeug development server simply by calling app.run() :
from flask import Flask
app = Flask(__name__)
app.run()
Consider an empty Django project "myproject" created as follows: django-admin.py startproject myproject. Is there a non-hackish way to start Django's development server for myproject as it's done in Flask? Thank you.
You can use management.call_command() to emulate what manage.py does:
from django.core import management
management.call_command('runserver')
See https://docs.djangoproject.com/en/dev/ref/django-admin/
Related
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
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')
I am trying to deploy my flask app. Usually I would have an app.py and put all code in it.
app.py
templates/
|
|--index.html
for my really small projects. But then I have a slightly larger app and follow the larger app guide by flask.
So I have this:
setup.py
app/
__init__.py
views.py
models.py
forms.py
templates/
| ---index.html
I now have all my routes and views in views.py and running the app in __init__.py:
from flask import Flask
app = Flask(__name__)
import app.views # Name in setup.py
if __name__ == "__main__":
app.run()
(This is just an example)
So now I follow the guide by running it with pip install -e . and running with:
>set FLASK_APP=app(name I set in setup.py) flask run and it works. Except I do not know how to run it with one command. Since there is no one file to run I can not use gunicorn or anything like that. I am not sure how to go about executing this app. How would I run pip install . on the cloud server heroku?
My problem is because I have to import the app from __init__.py and views using import blog.[insert import] (models, views etc.) Any help is appreciated. Thank you.
EDIT: I do not want to use blueprints though. That might be too much. My app is medium, not small but not large either
You absolutely can use Gunicorn to run this project. Gunicorn is not limited to a single file, it imports Python modules just the same as flask run can. Gunicorn just needs to know the module to import, an the WSGI object to call within that module.
When you use FLASK_APP, all that flask run does is look for module.app, module.application or instances of the Flask() class. It also supports a create_app() or make_app() app factory, but you are not using such a factory.
Gunicorn won't search, if you only give it a module, it'll expect the name application to be the WSGI callable. In your case, you are using app so all you have to do is explicitly tell it what name to use:
gunicorn app:app
The part before the : is the module to import (app in your case), the part after the colon is the callable object (also named app in your module).
If you have set FLASK_APP as a Heroku config var and want to re-use that, you can reference that on the command line for gunicorn:
gunicorn $FLASK_APP:app
As for heroku, it can handle requirement.txt or setup.py
c.f. https://devcenter.heroku.com/articles/python-pip#local-file-backed-distributions
If your Python application contains a setup.py file but excludes a requirements.txt file, python setup.py develop will be used to install your package and resolve your dependencies.
If you already have a requirements file, but would like to utilize this feature, you can add the following to your requirements file:
-e .
And about run command, i think you cat put Procfile like
web: FLASK_APP=app flask run
or
web: FLASK_APP=app python -m flask run
I am new to flask and GAE. I am trying to deploy a simple flask app to GAE. I am using https://github.com/kamalgill/flask-appengine-template/ as the template for deploying.
When I run
dev_appserver.py src/
I get the following error -
Debugged import:
- 'application' found in '/home/murtaza/workspace/flask/sim-sim/src/application/__init__.py'.
- 'application.settings' not found.
Below is the code from the application.settings file -
"""
Initialize Flask app
"""
from flask import Flask
app = Flask('application')
app.config.from_object('application.settings')
import urls
What is the utility of application.settings, is it a GAE or flask config file, or a custom file for the above template that can be ignored ?
Any other templates / approaches for deploying flask on GAE? Or sample flask project on GAE?
This is a flask config. Take a look here: http://flask.pocoo.org/docs/config/#configuring-from-files
I haven't tried this, but it looks like a decent start: https://github.com/kamalgill/flask-appengine-template/
There are also a bunch of other boilerplate templates if you google for it.