Flask run with argument

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I have a simple single page flask web page. This app requires a cmd line argument during run as follow: python3 example.py home/dir1/dir2. Now I want to run this app using flask run command. I set FLASK_APP=example.py then executing flask run command will start the flask server but since I am not able to provide the argument, my app will not work. How can I pass this home/dir1/dir2 argument with flask run? I've used argv for the argument instead of argparse.

flask python
1 Answers

From the flask --help I saw you can pass arguments to the app from the command line like this:

flask --app 'example:create_app("argument to the app!", 5)' run

To do that you have to have a file called example.py which has a method create_app which inits the app, something like this:

from flask import Flask

def create_app(arg1, arg2):
    """Create and configure an instance of the Flask application."""
    app = Flask(__name__)

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

    app.add_url_rule("/", endpoint="index")

    print(arg1)  # Prints "argument to the app!"
    print(arg2)  # Prints "5"

    return app

Other option you have is using environment variables, for example in the shell do

export CMD_ARG=56

and you can access that in the app with

import os
cmd_arg = os.environ["CMD_ARG"]  # "56"

In the flask --help they also have an option -e to pass in environment variables from a file.

user1F7m0
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