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Installing a Flask webapp on a Digital Ocean (or Linode, etc.) Ubuntu box
How to install a Flask webapp on a Digital Ocean VPS that has Ubuntu Linux on it: https://github.com/jabbalaci/DigitalOceanFlask.
This writing of mine appeared in Import Python Weekly Newsletter No. 65.
gunicorn from upstart fails to start
Today I was trying to put together these: ubuntu + nginx + gunicorn + start gunicorn from upstart + flask + virtualenv. After a whole afternoon, I managed to do it. I will write a detailed post about it, but now I just want to concentrate on one part, which was the most problematic.
Problem
I wanted to start gunicorn from upstart. So I issued the following command:
$ sudo service gunicorn start gunicorn start/running, process 21415 $ sudo service gunicorn status gunicorn stop/waiting
Each time I tried to launch it, it failed.
Solution
First of all, I didn’t know where the upstart log files were located. I know, it’s lame, but after a few hours I relized that it’d be a good idea to google it… So, the log files are in this folder:
/var/log/upstart
The log of gunicorn is here:
/var/log/upstart/gunicorn.log
Now, I started to monitor the end of this file in another terminal with “tail -f /var/log/upstart/gunicorn.log
“. With this information I could easily find the error. I put here my solution and then I add some explanation:
# /etc/init/gunicorn.conf # description "Gunicorn daemon for a Flask project" start on (local-filesystems and net-device-up IFACE=eth0) stop on runlevel [!12345] # If the process quits unexpectadly trigger a respawn respawn setuid demo setgid demo script . "/home/demo/.virtualenvs/myproj/bin/activate" cd /home/demo/projects/myproj/api exec gunicorn --name="my project" --bind=127.0.0.1:9000 --config /etc/gunicorn.d/gunicorn.py api:app --pid=gunicorn_from_nginx.pid end script
And you know what caused the problem? Instead of “.” I tried to use “source”. This is what I found in every tutorial. But as it turned out, this script is executed with “/bin/sh
“, not with “/bin/bash
“, and sh doesn’t know source. Geez…
The normal user on this machine is called “demo
“. My Flask project is within a virtual environment. You can create a script
block and the steps within are executed with “sh”. So first we “source” the virtual environment, then enter the directory where the Python script to be executed is located (here “api
” is the name of the folder, and in there I have a file called “api.py
“, which contains the “app
” variable). gunicorn is also installed in the virtual environment! You don’t need to install it globally with “sudo
“!
Nginx is listening on port 80 and every connection that arrives there are forwarded to this gunicorn instance on port 9000. But I’ll write about it a bit later.
Appendix
The content of “/etc/gunicorn.d/gunicorn.py
“:
"""gunicorn WSGI server configuration.""" from multiprocessing import cpu_count from os import environ def max_workers(): return cpu_count() * 2 + 1 max_requests = 1000 worker_class = 'gevent' workers = max_workers()
pretty JSON output with Flask-RESTful
Problem
Flask-RESTful is an awesome way to write REST APIs. In debug mode, its output is nicely indented, easy to read. However, in production mode the JSON is compressed and hard to read.
If debug is False, how to have a nicely formatted output?
Solution
from flask import Flask from flask_restful import Api app = Flask(__name__) api = Api(app) if production: print("# running in production mode") HOST = '0.0.0.0' DEBUG = False # START: temporary help for the UI developers, remove later settings = app.config.get('RESTFUL_JSON', {}) settings.setdefault('indent', 2) settings.setdefault('sort_keys', True) app.config['RESTFUL_JSON'] = settings # END else: print("# running in development mode") HOST='127.0.0.1' DEBUG = True # ... if __name__ == '__main__': app.run(debug=DEBUG, host=HOST, port=1234)
Note that here I use the development server shipped with Flask, which is not suitable for real production. So this solution is between development and production.
detect duplicate keys in a JSON file
Problem
I want to edit a JSON file by hand but I’m afraid that somewhere I introduce a duplicate key by accident. If it happens, then the second key silently overwrites the first one. Example:
$ cat input.json { "content": { "a": 1, "a": 2 } }
Naive approach:
import json with open("input.json") as f: d = json.load(f) print(d) # {'content': {'a': 2}}
If there is a duplicate key, it should fail! But it remains silent and you have no idea that you just lost some data.
Solution
I found the solution here.
import json def dict_raise_on_duplicates(ordered_pairs): """Reject duplicate keys.""" d = {} for k, v in ordered_pairs: if k in d: raise ValueError("duplicate key: %r" % (k,)) else: d[k] = v return d def main(): with open("input.json") as f: d = json.load(f, object_pairs_hook=dict_raise_on_duplicates) print(d)
Now you get a nice error message:
Traceback (most recent call last): File "./check_duplicates.py", line 28, in <module> main() File "./check_duplicates.py", line 21, in main d = json.load(f, object_pairs_hook=dict_raise_on_duplicates) File "/usr/lib64/python3.5/json/__init__.py", line 268, in load parse_constant=parse_constant, object_pairs_hook=object_pairs_hook, **kw) File "/usr/lib64/python3.5/json/__init__.py", line 332, in loads return cls(**kw).decode(s) File "/usr/lib64/python3.5/json/decoder.py", line 339, in decode obj, end = self.raw_decode(s, idx=_w(s, 0).end()) File "/usr/lib64/python3.5/json/decoder.py", line 355, in raw_decode obj, end = self.scan_once(s, idx) File "./check_duplicates.py", line 13, in dict_raise_on_duplicates raise ValueError("duplicate key: %r" % (k,)) ValueError: duplicate key: 'a'
If your json file has no duplicates, then the code aboce nicely prints its content.