flask restful: how do I pass python function as a parameter to trigger GET request?

Question

in flask-restplus, I want to pass system call object: subprocess.call() to GET request to print out R session information on server endpoint. To do so, I think I need to add system call object to parser, which I hope api endpoint print out R session information. However, when I hit the code, I didn't see any printed session info on server endpoint. How can I correctly pass system.call() to GET request in order to print something out on endpoint? any idea?

my attempt:

here is the system call which can print out R session on python IDE console.

sess = subprocess.call(['C:/Program Files/R/R-3.6.3/bin/Rscript','--vanilla','-e','sessionInfo()'])

my attempt to pass sess to GET request:

from flask import Flask, jsonify
from flask_restplus import Api, Resource, fields, reqparse, inputs
import subprocess

app = Flask(__name__)
api = Api(app)
ns = api.namespace('sess')

sess = subprocess.call(['C:/Program Files/R/R-3.6.3/bin/Rscript','--vanilla','-e','sessionInfo()'])

@ns.route('/')
class AResource(Resource):
    def get(self):
        parser = reqparse.RequestParse()
        parser.add_argument(sess, type= str, required=True)
        #return parser.parse_args()
        try:
            args = parser.parse_args()
            return jsonify(args)
        except:
            return None, 400

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

when I ran the code, I can't see any printed session information on the server endpoint. I think something wrong when I pass sess object to GET request. How can I pass custom python function or object as a parameter to GET request in order to print out something on the server endpoint? any idea?

I am newbie to flask-restplus, so maybe I should not pass sess to GET request like above. I humbly request for possible help. Thanks

note that you're online calling sess once and it is when you where you're initiating your flask app. You want to call subprocess.call for every request you get ? — Bernardo stearns reisen, Apr 22 '20 at 13:21
@Bernardostearnsreisen yes, I want to call subprocess.call for every request you get. how do I make it right? any thought? — jyson, Apr 22 '20 at 13:26
@Bernardostearnsreisen I tried exact same code, but I am getting this in response body: `{ "message": "Internal Server Error" }`. that's strange — jyson, Apr 22 '20 at 13:55
could you edit your post with your updated code ? message": "Internal Server Error" means you got a pythor error on the flask app — Bernardo stearns reisen, Apr 22 '20 at 13:57
@Bernardostearnsreisen I feel like doing `return {'hello123': 'world'}` is not right. am I right? — jyson, Apr 22 '20 at 14:09
this is your response on the browser, what you would like to return as response on the browser ? — Bernardo stearns reisen, Apr 22 '20 at 14:13
@Bernardostearnsreisen correct, so I feel like: `sess = subprocess.call(["echo","I am getting this output!"])`; then `return sess`. is that correct? — jyson, Apr 22 '20 at 14:15
no jyson, sess will return you a string, you need to return a dictionary from your api. we can discuss in the chat — Bernardo stearns reisen, Apr 22 '20 at 15:12
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/212271/discussion-between-bernardo-stearns-reisen-and-jyson). — Bernardo stearns reisen, Apr 22 '20 at 15:12
@Bernardostearnsreisen thanks a lot for your help and time, appreciated. — jyson, Apr 22 '20 at 16:23

Bernardo stearns reisen · Accepted Answer · 2020-04-22T14:02:28.987

Just to be clear, without flask you should be getting at least the output one.

For example running a dummy script:

from flask import Flask
from flask_restplus import Api
import subprocess

app = Flask(__name__)
api = Api(app)

sess = subprocess.call(["echo","I am getting this output!"])

would output

I am getting this output!

How to call subprocess.call in every get request

Simply move the subprocess.call to the get request.
Be careful to not create your API endpoint on '/' as flask_restplus uses it by default as the api reference page, so it would cause confusion. so make sure to use an endpoint like '/hello'

Example code

from flask import Flask
from flask_restplus import Resource, Api
from flask_restplus import Api
import subprocess
import os

app = Flask(__name__)
api = Api(app)

@api.route('/hello')
class HelloWorld(Resource):
    def get(self):
        subprocess.call(["echo","I am getting this output!"])
        subprocess.call(['Rscript','--vanilla','-e','sessionInfo()'])
        return {'hello123': 'world'}

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

When I run the app in on terminal and request http://localhost:5000/hello on my browser I see this output on my terminal

I am getting this output!
R version 3.6.3 (2020-02-29)
Platform: x86_64-apple-darwin18.7.0 (64-bit)
Running under: macOS Mojave 10.14.5

Matrix products: default
BLAS/LAPACK: /usr/local/Cellar/openblas/0.3.9/lib/libopenblasp-r0.3.9.dylib

locale:
[1] en_IE.UTF-8/en_IE.UTF-8/en_IE.UTF-8/C/en_IE.UTF-8/en_IE.UTF-8

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base

loaded via a namespace (and not attached):
[1] compiler_3.6.3
127.0.0.1 - - [22/Apr/2020 15:01:16] "GET /hello HTTP/1.1" 200 -

can you sure you get your output right? why I has internal server error in respond body? — jyson, Apr 22 '20 at 13:56
this is totally sure, you can simply copy and paste and run this code — Bernardo stearns reisen, Apr 22 '20 at 13:57
I assume you don't have R installed, if not do you mind to try can you get R session info on server endpoint? strange thing is, I used your exact same code and run, but didn't get your output message, I got `internal server error` in respond body. — jyson, Apr 22 '20 at 13:59

flask restful: how do I pass python function as a parameter to trigger GET request?

1 Answers1