Quickly login to the Django Admin screen

admin-login.py

# Python standard library
import os
import getpass
from urlparse import urlparse

# Third-party libraries
import requests
from bs4 import BeautifulSoup

__author__ = 'Andy Burton'

# The id used in a django suit admin login form.
FORM_ID = 'login-form'

# Holds the data from the program that will go in the POST request.
FILL_OUT_FORM = {}


def env_or_getpass(var_name, msg):
    """ Check to see if the env var exists. If not let the user type it in. """
    if os.getenv(var_name):
        return os.getenv(var_name)
    return getpass.getpass(msg)


def generate_login_post_data(form_obj):
    """ Create the data that with be sent to the login form as a POST. """
    REQUEST_DATA = {}
    for tag in [input for input in form.find_all('input') if input.get('name')]:
        REQUEST_DATA[tag['name']] = FILL_OUT_FORM.get(tag['name'], tag.get('value'))
    return REQUEST_DATA


if __name__ == '__main__':
    # Get the needed information from the environment or user.
    ADMIN_ID = env_or_getpass('ADMIN_ID', 'Enter your admin username:')
    ADMIN_PW = env_or_getpass('ADMIN_PW', 'Enter your admin password:')
    ADMIN_URL = env_or_getpass('ADMIN_URL', 'Enter the admin login URL:')

    # Set the initial state of the data that will go into the request
    FILL_OUT_FORM = {
        'username': ADMIN_ID,
        'password': ADMIN_PW,
    }

    s = requests.Session()

    html = s.get(ADMIN_URL, verify=False)

    print('Completed GET request {}.'.format(ADMIN_URL))

    # If you can't GET to the login form initially then there's no need to proceed.
    if html.status_code != 200:
        raise ValueError('Status should be 200, but it is {}.'.format(html.status_code))

    bs = BeautifulSoup(html.content, 'html.parser')

    # Grab the <form>...</form> section of the page.
    form = bs.find(id=FORM_ID)

    # Replace the initial path with the action from the form tag.
    url_parts = urlparse(ADMIN_URL)
    post_url = '{}://{}{}'.format(url_parts.scheme, url_parts.netloc, form.get('action'))

    print('Starting POST request {}.'.format(post_url))

    REQUEST_DATA = generate_login_post_data(form)

    # Post the login data. Django wants a referer address as a security feature.
    resp = s.post(post_url, headers={'referer': html.url}, data=REQUEST_DATA, timeout=15, verify=False)

    print('Finished with status code {}.'.format(resp.status_code))