Flask Error: typeerror run() got an unexpected keyword argument 'host'

Question

I built a website on virtualbox with Flask. The website can be opened on localhost, but I cannot open it through Port forwarding, so I changed the code from manage.run() to manage.run(host='0.0.0.0').

The problem is that I am receiving this error:

typeerror run() got an unexpected keyword argument 'host'. 

The similar error occurs when change manage.run() to manage.run(debug=True). I just followed the Flask documentation.http://flask.pocoo.org/docs/quickstart/#a-minimal-application Could anyone let me know why I'm receiving this error?

#!/usr/bin/env python
#-*- coding:utf-8 -*-

"""Manage Script."""

from sys import stderr, exit

from flask.ext.script import Manager, prompt_bool

from szupa import create_app
from szupa.extensions import db
from szupa.account.models import User
from szupa.context import create_category_db


app = create_app()
manager = Manager(app)


@manager.command
def initdb():
    """Initialize database."""
    db.create_all()
    create_category_db()


@manager.command
def migrate(created, action="up"):
    module_name = "migrates.migrate%s" % created
    try:
        module = __import__(module_name, fromlist=["migrates"])
    except ImportError:
        print >> stderr, "The migrate script '%s' is not found." % module_name
        exit(-1)
    if prompt_bool("Confirm to execute migrate script '%s'" % module_name):
        try:
            action = getattr(module, action)
        except AttributeError:
            print >> stderr, "The given action '%s' is invalid." % action
            exit(-1)
        action(db)
        print >> stderr, "Finished."


@manager.command
def dropdb():
    """Drop database."""
    if prompt_bool("Confirm to drop all table from database"):
        db.drop_all()


@manager.command
def setadmin(email):
    """Promote a user to administrator."""
    user = User.query.filter_by(email=email).first()
    if not user:
        print >> stderr, "The user with email '%s' could not be found." % email
        exit(-1)
    else:
        user.is_admin = True
        db.session.commit()


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

Answer 1

As @fangwz0577 said in a comment, they solved the problem using manager.add_command. The archived version of their link is here.

Next, create the manage.py file. Use this file to load additional Flask-scripts in the future. Flask-scripts provides a development server and shell:

from flask.ext.script import Manager, Server
from tumblelog import app

manager = Manager(app)

 # Turn on debugger by default and reloader 
manager.add_command("runserver", Server(
    use_debugger = True,
    use_reloader = True,
    host = '0.0.0.0') )

Answer 2

The answer of @SuperBiasedMan is the one who worked for me, except that the import from flask.ext.script import Manager, Server doesn't work because it was deprecated (as an issue on flask's github shows: https://github.com/pallets/flask/issues/1135#issuecomment-61860862)

So if you are still struggling, try to use from flask_script import Manager, Server instead.

Also, on Linux the creation of a "bare bones" application is not needed, just create the __init__.py file and make the manage.py file.