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Django: Execute code only for `manage.py runserver`, not for `migrate`, `help` etc

We are using Django as backend for a website that provides various things, among others using a Neural Network using Tensorflow to answer to certain requests.
For that, we created an AppConfig and added loading of this app config to the INSTALLED_APPS in Django's settings.py. This AppConfig then loads the Neural Network as soon as it is initialized:
settings.py:
INSTALLED_APPS = [
...
'bert_app.apps.BertAppConfig',
]
.../bert_apps/app.py:
class BertAppConfig(AppConfig):
name = 'bert_app'
if 'bert_app.apps.BertAppConfig' in settings.INSTALLED_APPS:
predictor = BertPredictor() #loads the ANN.
Now while that works and does what it should, the ANN is now loaded for every single command run through manage.py. While we of course want it to be executed if you call manage.py runserver, we don't want it to be run for manage.py migrate, or manage.py help and all other commands.
I am generally not sure if this is the proper way how to load an ANN for a Django-Backend in general, so does anybody have any tips how to do this properly? I can imagine that loading the model on startup is not quite best practice, and I am very open to suggestions on how to do that properly instead.
However, there is also some other code besides the actual model-loading that also takes a few seconds and that is definitely supposed to be executed as soon as the server starts up (so on manage.py runserver), but also not on manage.py help (as it takes a few seconds as well), so is there some quick fix for how to tell Django to execute it only on runserver and not for its other commands?

I had a similar problem, solved it with checking argv.
class SomeAppConfig(AppConfig):
def ready(self, *args, **kwargs):
is_manage_py = any(arg.casefold().endswith("manage.py") for arg in sys.argv)
is_runserver = any(arg.casefold() == "runserver" for arg in sys.argv)
if (is_manage_py and is_runserver) or (not is_manage_py):
init_your_thing_here()
Now a bit closer to the if not is_manage_py part: in production you run your web server with uwsgi/uvicorn/..., which is still a web server, except it's not run with manage.py. Most likely, it's the only thing that you will ever run without manage.py
Use AppConfig.ready() - it's intended for it:
Subclasses can override this method to perform initialization tasks such as registering signals. It is called as soon as the registry is fully populated. - [django documentation]
To get your AppConfig back, use:
from django.apps import apps
apps.get_app_config(app_name)
# apps.get_app_configs() # all

This is another way, in your manage.py will have something probably look like this
def main():
os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'slambook.settings')
try:
from django.core.management import execute_from_command_line
except ImportError as exc:
raise ImportError(
"Couldn't import Django. Are you sure it's installed and "
"available on your PYTHONPATH environment variable? Did you "
"forget to activate a virtual environment?"
) from exc
execute_from_command_line(sys.argv)
# check if has runserver
if `runserver` in sys.argv:
#execute your custom function
if __name__ == '__main__':
main()
you can check sys.argv if it have runserver, if so then execute your script or function

Error "no such file or directory" when using os.chdir in Flask

I can boot the app up with Flask, it loads and displays localhost viewing info, then immediately quits with the error:
can't open file 'server.py': [Errno 2] No such file or directory
This is normally what you get when there is no file of that name in the folder, but there definitely is in this case as it successfully loads it initially
EDIT: Adding code for server.py:
from flask import Flask, jsonify
from services.controller import Controller
app = Flask(__name__)
path = '/Users/bhouwens/some/path'
ctrl = Controller(path)
#app.route('/')
def home():
stats = {'path' : path}
return jsonify(stats)
#app.route('/task-runners')
def task_runners():
return jsonify({'task_runners': ctrl.task_runners})
if __name__ == '__main__':
app.run(debug=True)

As mentioned, os.chdir from within a Flask app doesn't seem possible without breaking the server.
In my case, I'm trying to read a file in another directory using another module, where the path to the directory is passed into the module as a parameter in this line:
ctrl = Controller(path)
I've discovered that the open function that Python makes available can open files in other directories, so I got around this issue by using
with open(path + '/file_to_open.txt', 'r') as file:
from the Controller module.
Hopefully this helps someone else who runs into the same problem.

Like you already discovered, you can't do chdir.
import os
from flask import Flask
app = Flask(__name__)
os.chdir('..')
app.run(debug=True)
When you execute this file, this is the output:
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
* Restarting with stat
/usr/bin/python: can't open file 'example.py': [Errno 2] No such file or directory

I don't know exactly what causes this issue, but if you check app.root_path after an os.chdir(), you would find it messed up, changing app.config['APPLICATION_ROOT'] doesn't solve this either, you need to change app.root_path attribute manually with the new directory path, you can do it while initializing the flask app like so app = Flask.flask(__name__, root_path=BASE_DIR), but you can also change it at any time while flask is running.
Without doing so, Flask does this:
if your app was at ~/one/two/three/app.py and you try to run the app from ~/one , doing an os.chdir() to os.path.dirname(os.path.abspath(__file__)) inside the app would result the app.root_path to be equal to '~/one/two/three/two/three', it basically doubles the portion of the path left between the location where you called the app and the app file itself, and if you were running the app from ~/one/two/three/four/five and doing os.chdir() to os.path.dirname(os.path.abspath(__file__)) inside the app this would result app.root_path to be equal to '~/one' ,i.e truncates the number of directories it's far from the app file starting from the app.py directory and going backwards.
os.path.dirname(os.path.abspath(__file__)) equals '~/one/two/three'
Weird bug :(

How can I make a Python package which serves a website with Flask to work with Apache?

I have written a Python package hwrt (see installation instructions if you want to try it) which serves a website when executed with
$ hwrt serve
2014-12-04 20:27:07,182 INFO * Running on http://127.0.0.1:5000/
2014-12-04 20:27:07,183 INFO * Restarting with reloader
I would like to let it run on http://www.pythonanywhere.com, but when I start it there I get
19:19 ~ $ hwrt serve
2014-12-04 19:19:59,282 INFO * Running on http://127.0.0.1:5000/
Traceback (most recent call last):
File "/home/MartinThoma/.local/bin/hwrt", line 108, in <module>
main(args)
File "/home/MartinThoma/.local/bin/hwrt", line 102, in main
serve.main()
File "/home/MartinThoma/.local/lib/python2.7/site-packages/hwrt/serve.py", line 95, in main
app.run()
File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 739, in run
run_simple(host, port, self, **options)
File "/usr/local/lib/python2.7/dist-packages/werkzeug/serving.py", line 613, in run_simple
test_socket.bind((hostname, port))
File "/usr/lib/python2.7/socket.py", line 224, in meth
return getattr(self._sock,name)(*args)
socket.error: [Errno 98] Address already in use
I only found this in the documentation:
Flask
never use app.run(), it will break your webapp. Just import the
app into your wsgi file...
By searching for wsgi file, I found mod_wsgi (Apache). However, I don't understand how I can adjust my current minimalistic Flask application to work with that. Currently, the script behind hwrt serve is:
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""Start a webserver which can record the data and work as a classifier."""
import pkg_resources
from flask import Flask, request, render_template
from flask_bootstrap import Bootstrap
import os
import json
# hwrt modules
import hwrt
import hwrt.utils as utils
def show_results(results, n=10):
"""Show the TOP n results of a classification."""
import nntoolkit
classification = nntoolkit.evaluate.show_results(results, n)
return "<pre>" + classification.replace("\n", "<br/>") + "</pre>"
# configuration
DEBUG = True
template_path = pkg_resources.resource_filename('hwrt', 'templates/')
# create our little application :)
app = Flask(__name__, template_folder=template_path)
Bootstrap(app)
app.config.from_object(__name__)
#app.route('/', methods=['POST', 'GET'])
def show_entries():
heartbeat = request.args.get('heartbeat', '')
return heartbeat
#app.route('/interactive', methods=['POST', 'GET'])
def interactive():
if request.method == 'POST':
raw_data_json = request.form['drawnJSON']
# TODO: Check recording
# TODO: Submit recorded json to database
# Classify
model_path = pkg_resources.resource_filename('hwrt', 'misc/')
model = os.path.join(model_path, "model.tar")
print(model)
results = utils.evaluate_model_single_recording(model, raw_data_json)
# Show classification page
page = show_results(results, n=10)
page += 'back'
return page
else:
# Page where the user can enter a recording
return render_template('canvas.html')
def get_json_result(results, n=10):
s = []
for res in results[:min(len(results), n)]:
s.append({res['semantics']: res['probability']})
return json.dumps(s)
#app.route('/worker', methods=['POST', 'GET'])
def worker():
# Test with
# wget --post-data 'classify=%5B%5B%7B%22x%22%3A334%2C%22y%22%3A407%2C%22time%22%3A1417704378719%7D%5D%5D' http://127.0.0.1:5000/worker
if request.method == 'POST':
raw_data_json = request.form['classify']
# TODO: Check recording
# TODO: Submit recorded json to database
# Classify
model_path = pkg_resources.resource_filename('hwrt', 'misc/')
model = os.path.join(model_path, "model.tar")
results = utils.evaluate_model_single_recording(model, raw_data_json)
return get_json_result(results, n=10)
else:
# Page where the user can enter a recording
return "Classification Worker (Version %s)" % hwrt.__version__
def get_parser():
"""Return the parser object for this script."""
from argparse import ArgumentParser, ArgumentDefaultsHelpFormatter
parser = ArgumentParser(description=__doc__,
formatter_class=ArgumentDefaultsHelpFormatter)
return parser
def main():
app.run()
if __name__ == '__main__':
main()

Ok, a not so non-sequitur answer to your question is around what mod_wsgi does to interface with your app. A typical flask app would look something like this:
from flask import Flask
app = Flask(__name__)
app.route("/")
def hello():
return "Holy moly that tunnel was bright.. said Bit to NIC"
if __name__ == "__main__":
app.run()
Unfortunately, Apache has no way to know what to do with this (though the app would run happily on its own). In order to get the app and Apache to play nice together we're going to use something called mod_wsgi. What Mod_WSGI does that's important to us, is that it provides a known interface (a file type called wsgi) that's going to wrap our application and initialize it so that we can serve it through Apache.
I'm going to assume you are using a python virtual environment, but if you aren't you can omit the step that deals with this in the instructions below. If you're curious why virtual environments are so great, feel free read about the python ecosystem.
Also - you can include an extra flag (assuming you are running wsgi as a daemon) to automatically reload the daemon whenever you touch or alter your wsgi file. This is quite useful during development and debugging so I'll include is below.
Anyway, let's get started. I'll break this down to steps below.
Configuring Apache for mod_wsgi
Enable mod_wsgi in Apache:
sudo apt-get install libapache2-mod-wsgi
Edit your /etc/apache2/sites-available/<yoursite>.conf.
<VirtualHost interface:port>
WSGIDaemonProcess yourapp user=someUser processes=2 threads=15
WSGIProcessGroup yourapp
# In this case / refers to whatever relative URL path hosts flask
WSGIScriptAlias / /absolute/path/to/yourapp.wsgi
<Directory /path/to/your/main/py/file/ >
# Use good judgement here when server hardening, this assumes dev env
Order allow,deny
Allow from all
Require all granted
#The below enables 'auto-reload' of WSGI
WSGIScriptReloading On
</Directory>
# If you want to serve static files as well and bypass flask in those cases
Alias /relative/url/to/static/content/
<Directory /absolute/path/to/static/root/directory/>
Order allow,deny
Allow from all
</Directory>
</VirtualHost>
Create your yourapp.wsgi file and put it in the appropriate place: Be wary of file permissions!
#!/usr/bin/python
import sys
import logging
# Activate virtual environment.
# If you are not using venv, skip this.
# But you really should be using it!
activate_this = "/path/to/venv/bin/activate_this.py"
execfile(activate_this, dict(__file__=activate_this))
# Handle logging
logging.basicConfig(stream=sys.stderr)
sys.path.insert(0, "/path/to/your/main/py/file/")
from YourMainPyFileName import app as application
application.secret_key = "your_secret_key"
Reload Apache and troubleshoot problems. I set this up probably every few weeks for a different project or idea I have and... I usually have to fix one thing or another when doing it from scratch. Don't despair though! Flask has great documentation on this.
Once you've done all this you should be at a place where flask runs all on its own. The sample flask app above is the actual code I use to verify everything works whenever I set this up.
This was left here in case it's some use, but is not really directly related to the question...
The answer here is to use x-send-file. This takes advantage of letting Apache do what it's good at (serving static content), while at the same time first letting flask (or other python framework) do it's work first. I do this often to let flask handle my auth layers in single page web apps and have so far been happy with the results.
Doing so requires two things:
First - Enable xsendfile on Apache2 sudo apt-get install libapache2-mod-xsendfile.
Second - Alter your apache2 configuration so allow x-send-file headers:
Alter your conf file in /etc/apache2/sites-available/<yoursite>.conf and add...
XSendFile On
XSendFilePath /path/to/static/directory
This can be entered top level within the <Virtualhost></Virtualhost> tags.
Don't forget to restart Apache sudo service apache2 restart.
Finally - Configure your flask app to use x-send-file in your app.py file:
app.user_x_sendfile = True
Note: Must be done after app initialization. Consequently can also be passed as an initialization parameter.
Flask has documentation on this (excerpt below):
use_x_sendfile
Enable this if you want to use the X-Sendfile feature. Keep in mind that the server has to support this. This only affects files sent with the send_file() method.
New in version 0.2.
This attribute can also be configured from the config with the USE_X_SENDFILE configuration key. Defaults to False.

I ran into a similar issue #moose was having. Getting connection refused and couldnt even telnet localhost 5000.
Turns out theres a ports.conf file i had to add Listen 5000
Happy days.

Writing doctests for pyramid web app which depend on settings in ini file

I would like to write doctests for my pyramid web app, using the webtest module. I tried it like this:
from my_webapp import main
from webtest import TestApp
app = TestApp(main({}))
result = app.get('/')
This raises a KeyError (because some.url is not known) when my code reaches this line:
url = request.registry.settings['some.url']
The value of some.url is specified in the paster ini file of my application. Is there a simple way to use my development.ini when running my test code? I did not yet fully understand how/when the ini file is loaded during pyramid start up, so it's hard to figure out where to load it while testing.

main is invoked with the contents of your ini file. A simple way to load your app from an ini is:
from pyramid.paster import get_app
app = get_app('testing.ini#main')
test_app = TestApp(app)
This expects "testing.ini" to be in the current working directory, so you may need to tweak that. If you'd like it to be relative to a spot in your tree you can use:
import os.path
import some_module
here = os.path.dirname(some_module.__file__)
app = get_app(os.path.join(here, 'testing.ini'))

Easy application logging/debugging with nginx, uwsgi, flask?

I'm not looking to turn on the dangerous debugging console, but my application is getting a 500 error and doesn't seem to be writing any output for me to investigate more deeply.
I saw this exchange on the mailing list, which led me to this page on logging errors.
However, I still find this very confusing and have a couple of questions:
(1) In which file should the stuff below go?
ADMINS = ['yourname#example.com']
if not app.debug:
import logging
from logging.handlers import SMTPHandler
mail_handler = SMTPHandler('127.0.0.1',
'server-error#example.com',
ADMINS, 'YourApplication Failed')
mail_handler.setLevel(logging.ERROR)
app.logger.addHandler(mail_handler)
...assuming the "getting bigger" file pattern for larger applications? __init__.py? config.py? run.py?
(2) I am overwhelmed by options there, and can't tell which I should use. Which loggers should I turn on, with what settings, to replicate the local python server debug I get to stdout when I run run.py? I find that default, local output stream very useful, more so than the interactive debugger in the page. Does anyone have a pattern they could share on setting up something replicating this with an nginx deployment, outputting to a log?
(3) Is there anything I need to change, not at the flask level, but in nginx, say in my /etc/nginx/sites-available/appname file, to enable logging?
UPDATE
Specifically, I'm looking for information like I get when python runs locally as to why, say, a package isn't working, or where my syntax error might be, or what variable doesn't exist:
$ python run.py
Traceback (most recent call last):
File "run.py", line 1, in <module>
from myappname import app
File "/home/me/myappname/myappname/__init__.py", line 27, in <module>
file_handler.setLevel(logging.debug)
File "/usr/lib/python2.7/logging/__init__.py", line 710, in setLevel
self.level = _checkLevel(level)
File "/usr/lib/python2.7/logging/__init__.py", line 190, in _checkLevel
raise TypeError("Level not an integer or a valid string: %r" % level)
When I run flask on a server, I never see this. I just get a uWSGI error in the browser, and have no idea which code was problematic. I would just like something like the above to be written to a file.
I notice also that setting the following logging didn't really write much to file, even when I turn the log way up to the DEBUG level:
from logging import FileHandler
file_handler = FileHandler('mylog.log')
file_handler.setLevel(logging.DEBUG)
app.logger.addHandler(file_handler)
mylog.log is blank, even when my application errors out.
I'll also add that I've tried to set debug = True in the following ways, in __init__.py:
app = Flask(__name__)
app.debug = True
app.config['DEBUG'] = True
from werkzeug.debug import DebuggedApplication
app.wsgi_app = DebuggedApplication(app.wsgi_app, True)
app.config.from_object('config')
app.config.update(DEBUG=True)
app.config['DEBUG'] = True
if __name__ == '__main__':
app.run(debug=True)
While in my config.py file, I have...
debug = True
Debug = True
DEBUG = True
Yet, no debugging happens, and without logging or debugging, this is rather hard to track down. Errors simply terminate the application with the un-useful browser message:
uWSGI Error
Python application not found

Set config['PROPAGATE_EXCEPTIONS'] to True when running app in production and you want tracebacks to be logged into log files. (I haven't tried with SMTP handler, though..)

The part where you create handlers, add to loggers etc. should be in the if __name__ == '__main__' clause, i.e. your main entry point. I assume that would be run.py.
I'm not sure I can answer this - it depends on what you want. I'd advise looking at the logging tutorial to see the various options available.
I don't believe you need to change anything at the nginx level.
Update: You might want to have an exception clause that covers uncaught exceptions, e.g.
if __name__ == '__main__':
try:
app.run(debug=True)
except Exception:
app.logger.exception('Failed')
which should write the traceback of any exception which occurred in app.run() to the log.

I know that this is a VERY old post, but I ran into the issue now, and it took me a bit to find the solution. Flask sends errors to the server. I was running Gunicorn with an upstart script on Ubuntu 14.04 LTS, and the place where I found the error logs was as follows:
/var/log/upstart/myapp.log
http://docs.gunicorn.org/en/stable/deploy.html#upstart
Just in case some other poor soul ends up in this situation.