I'm investigating how to develop a decent web app with Python. Since I don't want some high-order structures to get in my way, my choice fell on the lightweight Flask framework. Time will tell if this was the right choice.
So, now I've set up an Apache server with mod_wsgi, and my test site is running fine. However, I'd like to speed up the development routine by making the site automatically reload upon any changes in py or template files I make. I see that any changes in site's .wsgi file causes reloading (even without WSGIScriptReloading On in the apache config file), but I still have to prod it manually (ie, insert extra linebreak, save). Is there some way how to cause reload when I edit some of the app's py files? Or, I am expected to use IDE that refreshes the .wsgi file for me?
Run the flask run CLI command with debug mode enabled, which will automatically enable the reloader. As of Flask 2.2, you can pass --app and --debug options on the command line.
$ flask --app main.py --debug run
--app can also be set to module:app or module:create_app instead of module.py. See the docs for a full explanation.
More options are available with:
$ flask run --help
Prior to Flask 2.2, you needed to set the FLASK_APP and FLASK_ENV=development environment variables.
$ export FLASK_APP=main.py
$ export FLASK_ENV=development
$ flask run
It is still possible to set FLASK_APP and FLASK_DEBUG=1 in Flask 2.2.
If you are talking about test/dev environments, then just use the debug option. It will auto-reload the flask app when a code change happens.
app.run(debug=True)
Or, from the shell:
$ export FLASK_DEBUG=1
$ flask run
http://flask.palletsprojects.com/quickstart/#debug-mode
In test/development environments
The werkzeug debugger already has an 'auto reload' function available that can be enabled by doing one of the following:
app.run(debug=True)
or
app.debug = True
You can also use a separate configuration file to manage all your setup if you need be. For example I use 'settings.py' with a 'DEBUG = True' option. Importing this file is easy too;
app.config.from_object('application.settings')
However this is not suitable for a production environment.
Production environment
Personally I chose Nginx + uWSGI over Apache + mod_wsgi for a few performance reasons but also the configuration options. The touch-reload option allows you to specify a file/folder that will cause the uWSGI application to reload your newly deployed flask app.
For example, your update script pulls your newest changes down and touches 'reload_me.txt' file. Your uWSGI ini script (which is kept up by Supervisord - obviously) has this line in it somewhere:
touch-reload = '/opt/virtual_environments/application/reload_me.txt'
I hope this helps!
If you're running using uwsgi look at the python auto reload option:
uwsgi --py-autoreload 1
Example uwsgi-dev-example.ini:
[uwsgi]
socket = 127.0.0.1:5000
master = true
virtualenv = /Users/xxxx/.virtualenvs/sites_env
chdir = /Users/xxx/site_root
module = site_module:register_debug_server()
callable = app
uid = myuser
chmod-socket = 660
log-date = true
workers = 1
py-autoreload = 1
site_root/__init__.py
def register_debug_server():
from werkzeug.debug import DebuggedApplication
app = Flask(__name__)
app.debug = True
app = DebuggedApplication(app, evalex=True)
return app
Then run:
uwsgi --ini uwsgi-dev-example.ini
Note: This example also enables the debugger.
I went this route to mimic production as close as possible with my nginx setup. Simply running the flask app with it's built in web server behind nginx it would result in a bad gateway error.
For Flask 1.0 until 2.2, the basic approach to hot re-loading is:
$ export FLASK_APP=my_application
$ export FLASK_ENV=development
$ flask run
you should use FLASK_ENV=development (not FLASK_DEBUG=1)
as a safety check, you can run flask run --debugger just to make sure it's turned on
the Flask CLI will now automatically read things like FLASK_APP and FLASK_ENV if you have an .env file in the project root and have python-dotenv installed
app.run(use_reloader=True)
we can use this, use_reloader so every time we reload the page our code changes will be updated.
I got a different idea:
First:
pip install python-dotenv
Install the python-dotenv module, which will read local preference for your project environment.
Second:
Add .flaskenv file in your project directory. Add following code:
FLASK_ENV=development
It's done!
With this config for your Flask project, when you run flask run and you will see this output in your terminal:
And when you edit your file, just save the change. You will see auto-reload is there for you:
With more explanation:
Of course you can manually hit export FLASK_ENV=development every time you need. But using different configuration file to handle the actual working environment seems like a better solution, so I strongly recommend this method I use.
Use this method:
app.run(debug=True)
It will auto-reload the flask app when a code change happens.
Sample code:
from flask import Flask
app = Flask(__name__)
#app.route("/")
def index():
return "Hello World"
if __name__ == '__main__':
app.run(debug=True)
Well, if you want save time not reloading the webpage everytime when changes happen, then you can try the keyboard shortcut Ctrl + R to reload the page quickly.
From the terminal you can simply say
export FLASK_APP=app_name.py
export FLASK_ENV=development
flask run
or in your file
if __name__ == "__main__":
app.run(debug=True)
Enable the reloader in flask 2.2:
flask run --reload
Flask applications can optionally be executed in debug mode. In this mode, two very convenient modules of the development server called the reloader and the debugger are enabled by default.
When the reloader is enabled, Flask watches all the source code files of your project and automatically restarts the server when any of the files are modified.
By default, debug mode is disabled. To enable it, set a FLASK_DEBUG=1 environment variable before invoking flask run:
(venv) $ export FLASK_APP=hello.py for Windows use > set FLASK_APP=hello.py
(venv) $ export FLASK_DEBUG=1 for Windows use > set FLASK_DEBUG=1
(venv) $ flask run
* Serving Flask app "hello"
* Forcing debug mode on
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
* Restarting with stat
* Debugger is active!
* Debugger PIN: 273-181-528
Having a server running with the reloader enabled is extremely useful during development, because every time you modify and save a source file, the server automatically restarts and picks up the change.
To achieve this in PyCharm set 'Environment Variables' section to:
PYTHONUNBUFFERED=1;
FLASK_DEBUG=1
For Flask 'run / debug configurations'.
To help with fast automatic change in browser:
pip install livereload
from livereload import Server
if __name__ == '__main__':
server = Server(app.wsgi_app)
server.serve()
Next, Start your server again:
eg. your .py file is app.py
python app.py
I believe a better solution is to set the app configuration. For me, I built the tool and then pushed it to a development server where I had to set up a WSGI pipeline to manage the flask web app. I had some data being updated to a template and I wanted it to refresh every X minutes (WSGI deployment for the Flask site through APACHE2 on UBUNTU 18). In your app.py or whatever your main app is, add app.config.update dictionary below and mark TEMPLATES_AUTO_RELOAD=True, you will find that any templates that are automatically updated on the server will be reflected in the browser. There is some great documentation on the Flask site for configuration handling found here.
app = Flask(__name__)
app.config.update(
TEMPLATES_AUTO_RELOAD=True
)
I'm trying to deploy a flask app on heroku. I've gotten to the point where the app builds and deploys, but when I try to go to the URL, the app times out with the following error.
Error R10 (Boot timeout) -> Web process failed to bind to $PORT within 60 seconds of launch
I think the problem is with my procfile. It has one line.
web: python add_entry3.py
Other people have procfiles that look like this:
web: gunicorn app:app
This is just a toy app and I don't care about performance so I don't think I need to use gunicorn for the web server. Should I be putting a colon and command after my app's file name (add_entry3.py)?
Most likely your flask app isn't answering on the port and interface the Heroku expects. By default, Flask only listens on 127.0.0.1, and I think on port 5000. Heroku passes your app a PORT environment variable and you'd need to tell Flask to listen on all interfaces.
But there are reasons other than performance you want to avoid Flask's default debug server for production code. It's got memory leaks, there are security implications, and really ... just don't do it. Add gunicorn to your requirements.txt and use that.
But if you must use the Flask test/debug server, change your app.run() call to something like this:
app.run(host='0.0.0.0', port=int(os.environ.get("PORT", 5000)))
I'm running a simple bottle application with gunicorn as a webserver, the application is working fine. My code:
from bottle import route, run, template
#route('/hello/<name>')
def index(name):
return template('<b>Hello {{name}}</b>!', name=name)
bottle.run(server='gunicorn', workers="3")
The Problem
Now I would like to create my own gunicorn config file and use it with bottle. I want to add a lot of extra functionality to the gunicorn workers (like SSL for example) and using a config file is a great way to do this.
I've tried this:
bottle.run(server='gunicorn', config="settings.py.ini")
AND
bottle.run(server='gunicorn', -c="settings.py.ini")
I know that in the CLI this the settings file can be set as an extra option like so:
-c CONFIG, --config CONFIG
gunicorn --config="settings.py.ini"
Anyone knows how to achieve the same thing when using the bottle gunicorn controller?
Solved it by taking a different aproach.
I'm using the code from this question: Bottle with Gunicorn
This makes it possible to use gunicorn from the CLI and load the bottle app in gunicorn. Now I can use the CLI to use gunicorn with config file, it works with bottle. Bottle is now basicly an Gunicorn module.
based on this structure: http://flask.pocoo.org/docs/patterns/packages/
I also tried this post: Deploying Flask app to Heroku
I am having trouble getting this to work on heroku. I usually get the PORT does not set within 60 seconds error. I have read other SO posts and just can't figure out if my project structure is wrong or my procfile. I tried other ports than 5000 as well.
Here is my current project structure:
/myapplication
Procfile
runserver.py
/applicationfolder
__init__.py
views.py
Here is my Procfile
web: python runserver.py $PORT
Here is my runserver.py
from applicationfolder import app
app.run()
if __name__ == '__main__':
import os
port = int(os.environ.get('PORT', 5000))
app.run(host='0.0.0.0', port=port)
Here is my init.py
import os
from flask import Flask
from flask import render_template, jsonify, request
app = Flask(__name__)
app.config.from_object('config')
import applicationfolder.views
From there views.py runs.
This works locally with foreman start and python runserver.py, but does not work with heroku. I have tried many things with PORT but port doesn't seem to set even with a different PORT than 5000. I think it has something to do with my project structure.
The app.run() was in there twice, which as you noted is what's screwing things up. The app.run() invokes a simply pure-python development server so that you can easily run and/or debug your script.
By invoking it at the module level (right under your import in runserver.py), you were effectively trying to start the development server as the python code was loaded, and then when it went to run it when invoked from the Procfile, the development server was already in flight, having been starting with it's defaults (latest version of Flask is pulling relevant defaults from the SERVER_NAME environment variable). By having it in both places, you were trying to invoke that method twice.
You basically want either the straight up module load (in which case, kill off the code under "if name ...", or you use the code when invoking under main, in which case don't start the service at module load time.
I found this 0 dependency python websocket server from SO: https://gist.github.com/jkp/3136208
I am using gunicorn for my flask app and I wanted to run this websocket server using gunicorn also. In the last few lines of the code it runs the server with:
if __name__ == "__main__":
server = SocketServer.TCPServer(
("localhost", 9999), WebSocketsHandler)
server.serve_forever()
I cannot figure out how to get this websocketserver.py running in gunicorn. This is because one would think you would want gunicorn to run server_forever() as well as the SocketServer.TCPServer(....
Is this possible?
GUnicorn expects a WSGI application (PEP 333) not just a function. Your app has to accept an environ variable and a start_response callback and return an iterator of data (roughly speaking). All the machinery encapsuled by SocketServer.StreamRequestHandler is on gunicorn side. I imagine this is a lot of work to modify this gist to become a WSGI application (But that'll be fun!).
OR, maybe this library will get the job done for you: https://github.com/CMGS/gunicorn-websocket
If you use Flask-Sockets extension, you have a websocket implementation for gunicorn directly in the extension which make it possible to start with the following command line :
gunicorn -k flask_sockets.worker app:app
Though I don't know if that's what you want to do.