Pass command line parameters to uwsgi script

Question

I'm trying to pass arguments to an example wsgi application, :

config_file = sys.argv[1]

def application(env, start_response):
    start_response('200 OK', [('Content-Type','text/html')])
    return [b"Hello World %s" % config_file]

And run:

uwsgi --http :9090 --wsgi-file test_uwsgi.py  -???? config_file # argument for wsgi script

Any smart way I can achieve it? Couldn't find it in uwsgi docs. Maybe there is another way of providing some parameters to the wsgi application? (env. variables are out of scope)

Answer 1

You could use an .ini file with the pyargv setting that @roberto mentioned. Let's call our config file uwsgi.ini and use the content:

[uwsgi]
wsgi-file=/path/to/test_uwsgi.py
pyargv=human

Then let's create a WGSI app to test it:

import sys
def application(env, start_response):
    start_response('200 OK', [('Content-Type','text/html')])
    return [str.encode("Hello " + str(sys.argv[1]), 'utf-8')]

You can see how to load this file https://uwsgi-docs.readthedocs.io/en/latest/Configuration.html#loading-configuration-files:

 uwsgi --ini /path/to/uwsgi.ini --http :8080

Then when we curl the app, we can see our param echoed back:

$ curl http://localhost:8080
Hello human

If you are trying to pass argparse style arguments to your WSGI app, they work just fine in the .ini too:

pyargv=-y /config.yml

Answer 2

python args:

--pyargv "foo bar"

sys.argv
['uwsgi', 'foo', 'bar']

uwsgi options:

--set foo=bar

uwsgi.opt['foo']
'bar'

Answer 3

I ended up using an env variable but setting it inside a start script:

def start(uwsgi_conf, app_conf, logto):
    env = dict(os.environ)
    env[TG_CONFIG_ENV_NAME] = app_conf
    command = ('-c', uwsgi_conf, '--logto', logto, )
    os.execve(os.path.join(distutils.sysconfig.get_config_var('prefix'),'bin', 'uwsgi'), command, env)