Apache webserver and Flask app

Question

I have apache2 web server running on Ubuntu 14.04. By default when I start apache web server on http://localhost I can see "Apache2 Ubuntu Default Page" which is launched from /var/www/html/index.html file.

I have a flask app at /home/ubuntu/myprojects/ location. The flask app is running on virtualenv and have proper folder structure to render a html file. Below is the folder structure

/home/ubuntu/myprojects/hello/hello.py (flask app rendering a html) /home/ubuntu/myprojects/hello/templates/hello.html

The flask app code is as below:

from flask import Flask
from flask import render_template
from flask import request

app = Flask(__name__)

@app.route('/')
def my_form():
        return render_template("hello.html")

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

When I run http://localhost:5000 the hello.html is rendered. I want to render hello.html from hello.py flask app when http://localhost is called without any port number specified. To do that I added below code:

app.run(host='0.0.0.0', port=80)

But then, when I run the flask app, is exists with error:

 * Running on http://0.0.0.0:80/ (Press CTRL+C to quit)
Traceback (most recent call last):
  File "hello.py", line 21, in <module>
    app.run(host='0.0.0.0', port=80)
  File "/home/ubuntu/myproject/venv/local/lib/python2.7/site-packages/flask/app.py", line 772, in run
    run_simple(host, port, self, **options)
  File "/home/ubuntu/myproject/venv/local/lib/python2.7/site-packages/werkzeug/serving.py", line 618, in run_simple
    test_socket.bind((hostname, port))
  File "/usr/lib/python2.7/socket.py", line 224, in meth
    return getattr(self._sock,name)(*args)
socket.error: [Errno 13] Permission denied

I don't know what I am doing wrong. On http://localhost, the index.html from /var/www/html/ is getting rendering.

In addition to above, when I used mod_wsgi I added below code Added application.wsgi

import os, sys, logging
logging.basicConfig(stream=sys.stderr)

PROJECT_DIR = '/home/ubuntu/myproject/hello'

activate_this = os.path.join(PROJECT_DIR, 'bin', 'activate_this.py')
execfile(activate_this, dict(__file__=activate_this))
sys.path.append(PROJECT_DIR)

from hello import app as application

In /etc/apache2/sites-available/000-default.conf, I added below:

<VirtualHost *:80>
        WSGIDaemonProcess FunRoute user=ubuntu group=root threads=5
        WSGIScriptAlias / /home/ubuntu/myproject/FunRoute/application.wsgi

        <Directory /home/ubuntu/myproject/FunRoute/>
                WSGIProcessGroup FunRoute
                WSGIApplicationGroup %{GLOBAL}
                Order deny,allow
                Allow from all
        </Directory>
</VirtualHost>

Also I changed the hello.py back to :

app.run(host='0.0.0.0')

But then I received 403 Forbidden error when I try to launch http://localhost

Forbidden

You don't have permission to access / on this server.

Apache/2.4.7 (Ubuntu) Server at 52.8.217.39 Port 80

Rookie Rookie · Accepted Answer · 2015-08-07T06:51:41

After little bit for googling I got it resolved.

The Apache 2.4 and above have additional security enabled, where the syntax has changed:

In Apache 2.2 if we want to allow a folder to be accessed, e.g. the site files, then we give below options:

  <Directory /path/to/site/files>
    Order deny,allow
    Allow from all
  </Directory>

But in case of Apache 2.4 we need to give below option:

<Directory /path/to/site/files>
        Options Indexes FollowSymLinks
        AllowOverride None
        Require all granted
</Directory>

This allowed me the access and 403 forbidden error was resolved and also mod_wsgi helped me to create a flask app, which I can access via http://localhost and not http://localhost:5000

Apache webserver and Flask app

2 Answers