debug Flask server inside Jupyter Notebook

Question

I want to debug small flask server inside jupyter notebook for demo.

I created virtualenv on latest Ubuntu and Python2 (on Mac with Python3 this error occurs as well), pip install flask jupyter.

However, when I create a cell with helloworld script it does not run inside notebook.

from flask import Flask
app = Flask(__name__)

@app.route("/")
def hello():
    return "Hello World!"

if __name__ == "__main__":
    app.run(debug=True,port=1234)

File
“/home/***/test/local/lib/python2.7/site-packages/ipykernel/kernelapp.py”,
line 177, in _bind_socket
s.bind(“tcp://%s:%i” % (self.ip, port)) File “zmq/backend/cython/socket.pyx”, line 495, in
zmq.backend.cython.socket.Socket.bind
(zmq/backend/cython/socket.c:5653) File
“zmq/backend/cython/checkrc.pxd”, line 25, in
zmq.backend.cython.checkrc._check_rc
(zmq/backend/cython/socket.c:10014)
raise ZMQError(errno) ZMQError: Address already in use

NB – I change the port number after each time it fails.

Sure, it runs as a standalone script.

update without (debug=True) it’s ok.

Asked By: chro

||

Source

Answer 1

I installed Jupyter and Flask and your original code works.

The flask.Flask object is a WSGI application, not a server. Flask uses Werkzeug’s development server as a WSGI server when you call python -m flask run in your shell. It creates a new WSGI server and then passes your app as paremeter to werkzeug.serving.run_simple. Maybe you can try doing that manually:

from werkzeug.wrappers import Request, Response
from flask import Flask

app = Flask(__name__)

@app.route("/")
def hello():
    return "Hello World!"

if __name__ == '__main__':
    from werkzeug.serving import run_simple
    run_simple('localhost', 9000, app)

Flask.run() calls run_simple() internally, so there should be no difference here.

Answered By: yorodm

Answer 2

The trick is to run the Flask server in a separate thread. This code allows registering data providers. The key features are

Find a free port for the server. If you run multiple instances of the server in different notebooks they would compete for the same port.
The register_data function returns the URL of the server so you can use it for whatever you need.
The server is started on-demand (when the first data provider is registered)
Note: I added the @cross_origin() decorator from the flask-cors package. Else you cannot call the API form within the notebook.
Note: there is no way to stop the server in this code…
Note: The code uses typing and python 3.
Note: There is no good error handling at the moment

import socket
import threading
import uuid
from typing import Any, Callable, cast, Optional

from flask import Flask, abort, jsonify
from flask_cors import cross_origin
from werkzeug.serving import run_simple

app = Flask('DataServer')


@app.route('/data/<id>')
@cross_origin()
def data(id: str) -> Any:
    func = _data.get(id)
    if not func:
        abort(400)
    return jsonify(func())


_data = {}

_port: int = 0


def register_data(f: Callable[[], Any], id: Optional[str] = None) -> str:
    """Sets a callback for data and returns a URL"""
    _start_sever()
    id = id or str(uuid.uuid4())
    _data[id] = f
    return f'http://localhost:{_port}/data/{id}'


def _init_port() -> int:
    """Creates a random free port."""
    # see https://stackoverflow.com/a/5089963/2297345
    sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    sock.bind(('localhost', 0))

    port = sock.getsockname()[1]
    sock.close()
    return cast(int, port)


def _start_sever() -> None:
    """Starts a flask server in the background."""
    global _port
    if _port:
        return
    _port = _init_port()
    thread = threading.Thread(target=lambda: run_simple('localhost', _port, app))
    thread.start()

Answered By: Michael_Scharf

Answer 3

Although this question was asked long ago, I come up with another suggestion:

The following code is adapted from how PyCharm starts a Flask console.

import sys
from flask.cli import ScriptInfo
app = None
locals().update(ScriptInfo(create_app=None).load_app().make_shell_context())
print("Python %s on %snApp: %s [%s]nInstance: %s" % (sys.version, sys.platform, app.import_name, app.env, app.instance_path))

Now you can access app and use everything described in the Flask docs on working with the shell

Answered By: Peter_B

debug Flask server inside Jupyter Notebook

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