paths for deploying flask/wsgi differ from development - python-3.x

I am developing an app and the development setup was really easy.
I run it locally with:
$ . .venv/bin/activate
(.venv) $
(.venv) $ python -m flask run
* Serving Flask app 'app'
* 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://127.0.0.1:8080
Press CTRL+C to quit
* Restarting with stat
* Debugger is active!
* Debugger PIN: -###-###
and I have configured apache2 on my (ubuntu) laptop with:
ProxyPass / http://127.0.0.1:8080
My code is structured like:
app.py
pages/scc_main/scc.html
...
The code has this:
import re
import jinja2
from flask import Flask
from flask import request
import data
app = Flask(__name__)
env = jinja2.Environment(loader=jinja2.FileSystemLoader("pages"))
#app.route('/')
def hello_world():
return '<h2>Hello, World!</h2>'
#app.route('/contracts/scc')
#app.route('/contracts/scc/')
def contracts_main():
main = env.get_template('scc_main/scc.html')
context = data.build('scc_main')
return main.render(**context)
And everything works great. As in:
$ curl 'http://localhost/'
<h2>Hello, World!</h2>$
But when I deploy. Wow. I set my site's root to point to the app. That is actually working. I can hit https://opencalaccess.org/ and it gets my static content.
I have:
import sys
import logging
logging.basicConfig(
level=logging.DEBUG,
filename='/var/www/<full-path>/logs/contracts_scc.log',
format='^(asctime)s %(message)s')
sys.path.insert(0, '/var/www/<full-path>')
sys.path.insert(0, '/var/www/<full-path>/.venv/lib/python3.8/site-packages')
And https://opencalaccess.org/contracts/scc works. But only after I change the Environment call above to:
env = jinja2.Environment(loader=jinja2.FileSystemLoader("/var/www/full-path>/pages"))
Now, any link which is just a link is fine. But anything that looks at the flask.request.path gives me:
The browser (or proxy) sent a request that this server could not understand.
What the heck? Setting up the dev environment was so easy. What do you have to do to get this working in deployment? Any suggestions?
ADDED:
Well, it seems clear that it is the WSGI part that is having the problem. My script is not receiving the request structure and so it cannot read any parameters. I have all my parameters on the URL, so my data building method reads the request.path to see what to do.
So, where to go from here. We will see.

I am no longer able to reproduce this.

Related

Python Flask app is not running on heroku servers

I am trying to deploy my flask app on heroku but it shows a runtime error on the heroku server, but when i test it on my end, AKA localhost, it seems to work fine
My Code:
import flask
from flask import request
from flask import jsonify
from GoogleNews import GoogleNews
from apscheduler.schedulers.background import BackgroundScheduler
import multiprocessing
googlenews = GoogleNews()
news_latest = []
googlenews.set_lang('en')
googlenews.set_period('1d')
googlenews.set_encode('utf-8')
def get_latest_news():
global news_latest
googlenews.get_news('TECHNOLOGY')
news_latest = googlenews.result()
my_scheduler = BackgroundScheduler()
my_scheduler.add_job(func=get_latest_news, trigger="interval", seconds=5)
my_scheduler.start()
app=flask.Flask(__name__)
#app.route('/', methods=["GET"])
def home():
return jsonify(news_latest[0: 3])
app.run()
requirements.txt:
gunicorn
GoogleNews
APScheduler
PROCFILE:
web: gunicorn app:appweb: gunicorn app:app
runtime.txt:
python-3.9.4
But the server seems to crash when you open it,
And what is remarkable is in the logs, Flask says it is running on a DEVELOPMENT server. It prints the location where it has started the app as 127.0.0.0.5000
I was able to overcome this problem, The right way of doing this is:
import flask
from flask import request
from flask import jsonify
from GoogleNews import GoogleNews
from apscheduler.schedulers.background import BackgroundScheduler
import multiprocessing
googlenews = GoogleNews()
news_latest = []
googlenews.set_lang('en')
googlenews.set_period('1d')
googlenews.set_encode('utf-8')
def get_latest_news():
global news_latest
googlenews.get_news('TECHNOLOGY')
news_latest = googlenews.result()
my_scheduler = BackgroundScheduler()
my_scheduler.add_job(func=get_latest_news, trigger="interval", seconds=5)
my_scheduler.start()
app=flask.Flask(__name__)
#app.route('/', methods=["GET"])
def home():
return jsonify(news_latest[0: 3])
The change was to remove app.run() since heroku servers will already be running your code, if you put the app.run() heroku will consider this as default code and will run that too, hence the interpretation process will stop at app.run() and will never complete the interpretation. Hence the app will crash with no return response.
On localhost is fine to use the default 5000 port but on Heroku you need to use the one provided by the platform in the $PORT env variable
server_port = int(os.environ.get("PORT", 5000))
app.run(debug=False, port=server_port, host='0.0.0.0')

Flask 404 Error when running on my local machine

I have successfully created my virtual environment and installed flask but everytime I run the program I get "Not Found
The requested URL was not found on the server. If you entered the URL manually please check your spelling and try again." when I run it on REPL it works.
from flask import Flask, render_template
app = Flask(__name__)
app.route('/')
def index():
return '<p>Hello Flask</p>'
if __name__ == '__main__':
app.run(debug=True)
* Detected change in 'c:\\Users\\mavhu\\Desktop\\Flask_Blogging_App\\main.py', reloading
* Restarting with stat
* Debugger is active!
* Debugger PIN: 398-868-074
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
127.0.0.1 - - [03/Dec/2020 16:59:06] "?[33mGET / HTTP/1.1?[0m" 404 -
127.0.0.1 - - [03/Dec/2020 16:59:11] "?[33mGET / HTTP/1.1?[0m" 404 -
CODE SCREENSHOT
Check the following steps:
Did you use this command before running the Flask app : set FLASK_APP=main.py
If you are running on Windows, disable any Firewall or AntiVirus solution. Sometimes, they block all such connections on localhost.
Also try using the URL http://localhost:5000

current_app application context in flask. Get working instance of flask

I am trying to setup simple flask application to perceive app_context() feature.
This is my code:
# config.py
UPDATE_PERIOD = 100.0
# init.py
from flask import Flask
import config
app = Flask(__name__)
app.config.from_object(config)
app.config['DEBUG'] = True
#app.route('/')
def hello_world():
return 'Hello, World!'
if __name__ == "__main__":
app.run()
# ex.py
from flask import current_app
from flask import Flask
app = Flask(__name__)
with app.app_context() as app:
print(current_app.config['DEBUG'])
print(current_app.config['UPDATE_PERIOD'])
Now, I am running my init.py:
python3 init.py
* Serving Flask app "init" (lazy loading)
* Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
Use a production WSGI server instead.
* Debug mode: on
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
* Restarting with stat
* Debugger is active!
* Debugger PIN: 666-659-182
which is okay, how it should be.
Here I expect that app.config['DEBUG'] = True and app.config['UPDATE_PERIOD'] = 100.0
And I am trying to extract this vars in ex.py.
The problem is: when I execute: python3 ex.py I expect this:
True
100.0
but eventually I am getting this:
False
Traceback (most recent call last):
File "ex.py", line 10, in <module>
print(current_app.config['UPDATE_PERIOD'])
KeyError: 'UPDATE_PERIOD'
so I cannot get how app_context() feature is working whereas I think I am doing the same things as written in docs.
There is an option to import in ex.py:
from init import app
but I want to get working instance without complex import since my application is quite big.
Can anyone please help me with this..?
Thank you in advance!

Not able to import module server from my testing folder

I am writing a flask app which follows the following folder structure.
backend
server.py
__init__.py
test
__init__.py
servertest.py
"""Server ping test for flask"""
import flask
import pytest
from server import app
#pytest.fixture()
def client():
yield testing.TestClient(app)
def test_ping_resource(client):
doc = "Flask GET service is working"
result = client.simulate_get("/api/ping")
assert result.status_code == 200
assert result.json == doc
And this is my test file. When I am running the file. it gives
from server import app
E ImportError: No module named 'server'
What am I doing wrong, which is making the server module invisible to test module?
When you run a python file, its parent folder is added to the python path, but the parent of its parent (here, your root folder) is not!
What you could do is:
running your tests from the root folder (preferred)
Using relative imports, here: from ..server import app
Modify your python path (not advised)
Python imports modules from a list of path:
import sys
sys.path
You have to provide Python a way to find server.py:
you can run your code from the backend directory, because the current path is automatically happened to sys.path (cd backend; pytest)
you can add the full path to backend to PYTHONPATH
At some point, you will need to read the doc about import.
Also, to use pytest naturally, your test files must start with test.

Flask uWSGI application structure and python3 imports

I have a basic flask-restful application with a structure that looks like this as recommended on the flask website.
/application
/application
/config.py
/__init__.py
/wsgi.ini
Slightly irrelevant, but config.py is generated by our CD server.
wsgi.ini looks likes this
[uwsgi]
module = application:app
master = true
processes =5
socket = /tmp/application.sock
chmod-socket = 660
vacuum = true
die-on-term = true
and __init__.py looks something like this
import config
from flask import Flask, request, g
from flask_restful import Resource, Api, abort
app = Flask(__name__)
api = Api(app)
if __name__ == '__main__':
app.run(host='0.0.0.0', debug=True)
Now when I try to startup the application using uwsgi --ini wsgi.ini I get this error:
File "./application/__init__.py", line 2, in <module>
import config
Originally __init__.py was called main.py and I executed that to debug, which is why the import is now wrong. I'm guess I will need to change the import to be from .config import *
My question is two fold:
Could I have avoided the import problem completely. ie is there a way in python3 to import sibling modules that will work for both approaches
is my wsgi.ini in the correct place, or should it be in the inner application directory?
Python 3.x dropped support for infra-package relative imports. You need to use an absolute import:
from application import config
or the new spelling of relative imports:
from . import config
If you need to also support legacy versions of Python you can enable this behavior with:
from __future__ import absolute_import

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