Using Python Flask-restful with mod-wsgi

Question

I am trying to use mod-wsgi with Apache 2.2

I have the following directory structure:

scheduling-algos
-lib
-common
 -config
  -config.json
resources
-Optimization.py
optimization.wsgi
optimization_app.py

My optimization_app.py is the following:

from flask import Flask
from flask_restful import Api
from resources.Optimization import OptimizationAlgo

def optimizeInstances():
    optimization_app = Flask(__name__)
    api = Api(optimization_app)
    api.add_resource(OptimizationAlgo, '/instances')

if __name__ == '__main__':
    optimizeInstances()
    optimization_app.run(host='0.0.0.0', debug=True)

My Optimization.py code looks like the following:

class OptimizationAlgo(Resource):
    def post(self):
       return "success"

When I make a POST request to the url http://<host>:5000/instances, it works just as expected. I want make this work using WSGI. I have mod-wsgi installed with Apache 2.2.

My optimization.wsgi file looks like the following

import sys
sys.path.insert(0, '<path to app>')

from optimization_app import optimizeInstances as application

I get the following error: TypeError: optimizeInstances() takes no arguments (2 given) . Apparently this is not the correct way to use WSGI. What is the correct way to use WSGI? Apparently, this is not the correct way to use WSGI.

Effectively a duplicate of http://stackoverflow.com/questions/36763304/typeerror-takes-no-arguments-2-given-when-using-wsgi-in-python-2-7 — Graham Dumpleton, Apr 21 '16 at 10:09

score 2 · Answer 1 · answered Apr 21 '16 at 10:08

As I told you in your other question, you should perhaps go back and read the Flask documentation again. That way you will learn and understand properly. By ignoring advice and expecting others to tell you, it only annoys people and they will stop helping you. Would suggest you take heed of that rather than leave a trail of separate questions hoping someone will solve your problems for you.

That said, I can't see how the code you give can even work with the Flask development server as you claim. The problem is that optimization_app = Flask(__name__) is setting a local variable within function scope. It isn't setting a global variable. As a result the call of optimization_app.run(host='0.0.0.0', debug=True) should fail with a LookupError as it will not see a variable called optimization_app. Not even sure why you are bothering with the function.

If you go look at the Flask documentation, the pattern it would likely use is:

# optimisation.wsgi

import sys
sys.path.insert(0, '<path to app>')

# We alias 'app' to 'application' here as mod_wsgi expects it to be called 'application'.

from optimization_app import app as application

# optimization_app.py

from flask import Flask
from flask_restful import Api
from resources.Optimization import OptimizationAlgo

app = Flask(__name__)

api = Api(app)
api.add_resource(OptimizationAlgo, '/instances')

if __name__ == '__main__':
    app.run(host='0.0.0.0', debug=True)

Using Python Flask-restful with mod-wsgi

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