Logging in to google using python?

Question

I am fairly new to web programing but for the sake of it, I am trying to login to google account not using standard code but as a python application, but it is impossible to do so has anyone tried to this before? can anyone help?

Not an application, the google account page over the web. https://www.google.com/accounts/ — Mir Khashkhash, Jul 19 '11 at 22:14

score 28 · Answer 1 · edited Jul 13 '17 at 18:41

I made a python class that handle google login and the is able to get any google service page that requires the user to be logged in:

class SessionGoogle:
    def __init__(self, url_login, url_auth, login, pwd):
        self.ses = requests.session()
        login_html = self.ses.get(url_login)
        soup_login = BeautifulSoup(login_html.content).find('form').find_all('input')
        my_dict = {}
        for u in soup_login:
            if u.has_attr('value'):
                my_dict[u['name']] = u['value']
        # override the inputs without login and pwd:
        my_dict['Email'] = login
        my_dict['Passwd'] = pwd
        self.ses.post(url_auth, data=my_dict)

    def get(self, URL):
        return self.ses.get(URL).text

The idea is to go to the login page GALX hidden input value and send it back to google + login and password. It requires modules requests and beautifulSoup

Example of use:

url_login = "https://accounts.google.com/ServiceLogin"
url_auth = "https://accounts.google.com/ServiceLoginAuth"
session = SessionGoogle(url_login, url_auth, "myGoogleLogin", "myPassword")
print session.get("http://plus.google.com")

Hope this helps

Thanks! That's a really smart way to deal with all of the hidden elements, definetely gonna keep this 'trick' in my toolbox — MadRabbit, Nov 14 '15 at 14:07
Super slick and makes my life way easier than playing with regex to get the GALX. — dreyco676, Aug 28 '16 at 05:43
You could make a dictionary comprehension to make it prettier/simpler: my_dict = {i.get('name'): i.get('value') for i in lst if i.get('value')} — Sebastian Nielsen, Jun 29 '17 at 19:25
Does this still work, and if so, what are the url_login and url_auth to use? — Noah Covey, Aug 09 '18 at 02:49

score 1 · Answer 2 · edited Apr 14 '21 at 15:52

2020 update for python 3:

import urllib.request

def unread_messages(user, passwd):
    auth_handler = urllib.request.HTTPBasicAuthHandler()
    auth_handler.add_password(
        realm='New mail feed',
        uri='https://mail.google.com',
        user='%s@gmail.com' % user,
        passwd=passwd
    )
    opener = urllib.request.build_opener(auth_handler)
    urllib.request.install_opener(opener)
    feed = urllib.request.urlopen('https://mail.google.com/mail/feed/atom')
    return feed.read()


print(unread_messages('username', 'password'))

score 1 · Answer 3 · edited May 23 '17 at 11:46

Although probably not exactly what you were looking for here I found some code from a similar post that did run from me.


import urllib2

def get_unread_msgs(user, passwd):
    auth_handler = urllib2.HTTPBasicAuthHandler()
    auth_handler.add_password(
        realm='New mail feed',
        uri='https://mail.google.com',
        user='%s@gmail.com' % user,
        passwd=passwd
    )
    opener = urllib2.build_opener(auth_handler)
    urllib2.install_opener(opener)
    feed = urllib2.urlopen('https://mail.google.com/mail/feed/atom')
    return feed.read()

print get_unread_msgs("put-username-here","put-password-here")

reference:
How to auto log into gmail atom feed with Python?

score 0 · Answer 4 · answered Nov 29 '16 at 03:33

You can use urllib, urllib2 and cookielib libraries of python to login.

import urllib, urllib2, cookielib

def test_login():
    username = '' # Gmail Address
    password = '' # Gmail Password
    cookie_jar = cookielib.CookieJar() 
    opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cookie_jar)) 
    login_dict = urllib.urlencode({'username' : username, 'password' :password}) 
    opener.open('https://accounts.google.com/ServiceLogin', login_dict) 
    response = opener.open('https://plus.google.com/explore')
    print response.read()

if __name__ == '__main__':
    test_login()

Is there a version of this that works in Python 3 (which does not use urllib2)? — toryan, Oct 01 '19 at 00:03

Logging in to google using python?

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