Running flask app by executing the subprocess.Popen command

Question

Currently I am starting my flask server by executing the FLASK_APP=name_of_my_app.py && flask run (I am on Ubuntu). I need to achieve the outcome of initiating the server as above through different means - namely, by running something of the kind: python start_same_server.py. After some refactoring, this does not work for me because it breaks my socket.io (flask-socketio). So I had an idea of just creating a supplementary start_server.py module that looks like this:

Please ignore the Windows shell command (it may be wrong, I have not tested it yet). But I am getting an error in my shell when I do python start_server.py:

/bin/sh: 1: flask: not found

It is because of my lack of understanding on how Popen runs shell commands. How do I make Popen see my flask command? (I had an idea of adding flask to sys.path? Also can python -m flask help here somehow?)

Sorry for the noob question. And thanks for any help.

Can you run this exact command from the terminal, without Python? — ForceBru
– ForceBru, Commented Mar 5, 2018 at 13:23

davidism · Accepted Answer · 2018-03-05 14:33:39Z

1

You should use the path to the flask command in the virtualenv, like /home/user/venv/bin/site-packages/flask. flask is not installed to the system, it's just module in the Python virtualenv.

edited Mar 5, 2018 at 14:33

davidism

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answered Mar 5, 2018 at 13:26

Oleg Korobov

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2 Comments

naz Over a year ago

hmm. let me try that. That would be equivalent of adding that path to the sys.path I think

naz Over a year ago

there is no flask module there. There is a _main_.py file which is alias for flask run. But when I do python /../../__main__.py it tells me that: ModuleNotFoundError: No module named 'main.cli'; 'main' is not a package. i.e. it does not see the whole package

Harry1992 · Accepted Answer · 2018-03-05 13:42:06Z

0

Try

Python -m pip install flask
#or ...-m easy_install flask

Why not you try this for the batter result

from flask import Flask as f,  Merkup as m
app = f(__name__)
@app.route('/')
def main():
 html = open('index.html').read()
 html = m(html)
 return html

answered Mar 5, 2018 at 13:42

Harry1992

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