Docker run error - Failed to find Flask application or factory in module "app". Use "FLASK_APP=app:name to specify one

Question

trying to dockerize this flask app... running the following

docker build --tag flask-website .

works, output successfully built, successfully tagged.

edit: the next command works

$ docker run --publish 5000:5000 flask-website
 * Environment: production
   WARNING: This is a development server. Do not use it in a production deployment.
   Use a production WSGI server instead.
 * Debug mode: off

ok, so then I run curl localhost:5000

which gives this error

curl: (7) Failed to connect to localhost port 5000: Connection refused

ok straight forward enough, so then I try this

docker-compose up

and this results

Creating network "app_default" with the default driver
Creating app_web_1 ... done
Attaching to app_web_1
web_1  |  * Environment: production
web_1  |    WARNING: This is a development server. Do not use it in a production deployment.
web_1  |    Use a production WSGI server instead.
web_1  |  * Debug mode: off

however trying to navigate to localhost:5000 I get

This site can’t be reachedThe webpage at http://localhost:5000/ 
might be temporarily down or it may have moved permanently 
to a new web address.
ERR_SOCKET_NOT_CONNECTED

directory structure looks like this

├──app_folder/
    └── app/
    |    ├── static/
    |    |      └── css/
    |    |          └──app.css
    |    |      └── js/
    |    |          └──app.js
    |    └── templates/
    |    |   └── app.html
    |    |   └── subapp.html
    |    |   └── subapp1.html
    |    |   └── subapp2.html
    |    |   └── subapp3.html
    |    └── app.py
    |    └── util.py
    |    └── pickle_file.pickle
    |    └── requirements.txt
    |    └── Dockerfile
    |    └── Makefile
    |    └── docker-compose.yml

dockerfile looks like this

FROM python:3.8

ENV PYTHONUNBUFFERED=1

WORKDIR /

COPY requirements.txt requirements.txt
COPY . .

RUN pip install -r requirements.txt

# EXPOSE 5000

CMD [ "python", "-m" , "flask", "run", "--host=0.0.0.0"]

I had tried with EXPOSE 5000 uncommented and commented, making no difference

I also updated the directory structure and dockerfile, which got rid of the command line error I was seeing

docker-compose looks like this

version: "3.7"
services:
  web:
    image: flask-website:latest
    ports:
      - 5000:5000

I tried with the dockerfile, docker-compose, makefile, and requirements outside of the app directory and a slightly modified dockerfile on the WORKDIR line, that resulted in this error

Error: Failed to find Flask application or factory in module "app". Use "FLASK_APP=app:name to specify one.

not sure what else to try? I can run it locally with python -m flask run , but I cannot seem to dockerize it, seems like this should not be this difficult?

for completeness sake, app.py looks like this

from flask import Flask, request, jsonify
from flask import render_template, redirect
import json
import util

app = Flask(__name__, template_folder="templates", static_folder="static")


@app.route("/", methods=["GET", "POST"])
def index():
    return render_template("app.html")


@app.route("/predict_home_price", methods=["GET", "POST"])
def make_prediction():
   x = request.form["x"]
   y = float(request.form["y"]
   response = jsonify(
      {
         "prediction": util.prediction(x, y)
      }
   )
   response.headers.add("Access-Control-Allow-Origin", "*")
   
   return response

if __name__ == "__main__":
    from waitress import serve

    serve(app, host="0.0.0.0", port=5000)

util.py looks like this

import pickle
import pandas as pd
from scipy.special import inv_boxcox

# to run locally uncomment out the following
# with open("/path/to/pickle/app/pickle_mod.pickle", "rb") as f:
# to run in docker use the following
with open("app/pickle_mod.pickle", "rb") as f:
    __model = pickle.load(f)

def prediction(x, y):
   lambs = 0.205
   a = [[x, y]]
   cols = ["x", "y"]
   my_data = pd.DataFrame(data=a, columns=cols)
   pred = inv_boxcox(__model.predict(my_data)[0], lambs)
   return f"${round(pred)}"

if __name__ == "__main__":
   print(prediction(5, 4))

I have also tried both ways with the pickle import in util, same result - I thought because I was building it in a docker container that the second import was correct

I have also tried this block for app.run as well with the same result

if __name__ == "__main__":
    app.run()

Answer 1

I have ran a simple flask app in docker and here is the result. In your docker compose, you do not need to add the command: python app/app.py as this line was added in the Dockerfile.

The only thing you need in your Docker Compose is ports and image name

Answer 2

ok the following changes were required to get this image to build and the container to run

dockerfile:

FROM python:3.8

WORKDIR /

COPY requirements.txt requirements.txt

RUN pip install -r requirements.txt

COPY . .

EXPOSE 5000

ENTRYPOINT ["python", "./app.py"]

then the following change to the main application app.py

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

then to util.py, which had an error I did not see until running docker run [TAG]

with open("pickle_mod.pickle", "rb") as f:
    __model = pickle.load(f)

then run docker build -t [TAG] . then run docker-compose up

then navigate to localhost listening on port 5000 and there is the running container