[FIXED] Run a server from a flask route

Issue

I have a flask server (app1) in file1:

from flask import Flask
app1 = Flask(__name__)

@app1.route('/', methods=['POST']):
    return "Hello!"

if __name__ == '__main__':
    app1.run(port=8080, debug=True)

And another flask server (app2) in file2:

import os
from flask import Flask, request
import subprocess
from time import sleep

app2 = Flask(__name__)

@app2.route('/run-path', methods=['POST'])
def run_path():
    command_str = request.json.get('command')
    command = command_str.split()
    path_to_add = command[1]
    os.environ['PYTHONPATH'] += f":{path_to_add}" # Adding the file that's going to be ran 
    # to PYTHONPATH to prevent import errors 
    
    subprocess.run(
        command,
        env=os.enivron.copy(),
        stdin=subprocess.PIPE,
        stdout=subprocess.PIPE,
        stderr=subprocess.PIPE,
    )
    sleep(600)

if __name__ == '__main__':
    app2.run(port=2222, debug=True)

When someone sends a request to app2 (port 2222), it should run a python command, which is sometimes a flask server. The problem is that anything goes wrong when trying to run this command using subprocess.run.

It seems like it works, but when I'm trying to send a request to app1 (The server which we tried to start with subprocess), there's no response. The server is essentially down and it seems like the command did nothing.

If I set stdout to subprocess.STDOUT, I get this exception:

Traceback (most recent call last):
  File "/Applications/PyCharm CE.app/Contents/helpers/pydev/pydevd.py", line 1758, in <module>
    main()
  File "/Applications/PyCharm CE.app/Contents/helpers/pydev/pydevd.py", line 1752, in main
    globals = debugger.run(setup['file'], None, None, is_module)
  File "/Applications/PyCharm CE.app/Contents/helpers/pydev/pydevd.py", line 1147, in run
    pydev_imports.execfile(file, globals, locals)  # execute the script
  File "/Applications/PyCharm CE.app/Contents/helpers/pydev/_pydev_imps/_pydev_execfile.py", line 18, in execfile
    exec(compile(contents+"\n", file, 'exec'), glob, loc)
  File "/Users/galshahar/Desktop/Projects/PychramProjects/cba/src/api.py", line 48, in <module>
    app.run(port=8080, debug=True)
  File "/Users/galshahar/anaconda3/lib/python3.7/site-packages/flask/app.py", line 943, in run
    run_simple(host, port, self, **options)
  File "/Users/galshahar/anaconda3/lib/python3.7/site-packages/werkzeug/serving.py", line 814, in run_simple
    inner()
  File "/Users/galshahar/anaconda3/lib/python3.7/site-packages/werkzeug/serving.py", line 774, in inner
    fd=fd)
  File "/Users/galshahar/anaconda3/lib/python3.7/site-packages/werkzeug/serving.py", line 660, in make_server
    passthrough_errors, ssl_context, fd=fd)
  File "/Users/galshahar/anaconda3/lib/python3.7/site-packages/werkzeug/serving.py", line 574, in __init__
    socket.SOCK_STREAM)
  File "/Users/galshahar/anaconda3/lib/python3.7/socket.py", line 463, in fromfd
    nfd = dup(fd)
OSError: [Errno 9] Bad file descriptor

Is what I'm trying to do is possible? What am I missing? I'm kinda lost :(

Thanks.

Solution

OK, I can tell you how to work around it. If you look at the traceback, flask calls a package called werkzeug to do the work. werkzeug puts a note in the environment when it has to create a socket, presumably so it can just reuse that socket in forks. In your case, the second process inherits that environment, so it thinks an fd already exists and tries to reuse it, but the fd does not exist in this process.

The ugly solution that should work is to add this in your second script before starting port 8080 (or 2226, based on the original traceback ;):

    del os.environ["WERKZEUG_SERVER_FD"]

Answered By - Tim Roberts