Skip to content
Advertisement

is there a way to parse python-flask oauth2

I have code something like below –

app = Flask(__name__)


# access token
access_token = None


@app.route('/getcode')
def get_authorization_url():
    oauth = OAuth2Session(client_id, redirect_uri=redirect_uri, scope=scope)
    authorization_url, _state = oauth.authorization_url(authorization_base_url, access_type="authorization_code")
    print('authorization_url')
    print(authorization_url)
    return redirect(authorization_url)


@app.route('/')
def callback():
    global access_token
    oauth = OAuth2Session(client_id, redirect_uri=redirect_uri, scope=scope)
    token = oauth.fetch_token(token_url, authorization_response=request.url, client_secret=client_secret)
    access_token = token['access_token']
    print('access token is:', access_token)

    ## we will be shutting down the server after getting access_token
    ## the thread created here is copied in if __name__ == '__main__' block
    ## and will run after closing the server

    # th = threading.Thread(target=data_from_resource_server, args=(access_token,))
    # th.start()

    func = request.environ.get('werkzeug.server.shutdown')
    if func:
        print('stoping server')
        func()


    return 'see terminal for logs'


if __name__ == '__main__':
    app.secret_key = 'example'
    app.env = 'development'
    print()
    print('Open this url in browser:', 'http://127.0.0.1/getcode', end='nn')

    app.run(host='127.0.0.1', port='80')

    print('server stopped')

    ## got access_token, closed the server, now running ray integration code
    if access_token:
        th = threading.Thread(target=data_from_resource_server, args=(access_token,))
        th.start()

Here when app.run(host=’127.0.0.1′, port=’80’) runs gives me URL – http://127.0.0.1/getcode. I need to mannually open enter username and password and again then one more window comes to enter YOB, which then give me something like –

127.0.0.1 - - [04/May/2021 21:20:23] "GET /**getcode?code=G7h_QL0Cpo3kEqyyNBZ68DTX3JhQ_6E6sl_Sk1x5iBc.oG4JFQiKyZGupTuJ-bV6qE9lA**&scope=orders&state=M6hdb7EJxgNKkuBqbihg1SKaUGAJ7W HTTP/1.1" 302  

Here My question is there a way to avoid doing this manually opening the browser and enter credentials and get the code. can we parse the entire thing in python?

Advertisement

Answer

Sounds like a job for Selenium! It can open a web browser and parse the required details for you.

Run the following code after starting the server

from selenium import webdriver
from selenium.webdriver.common.keys import Keys

url = 'http://127.0.0.1/getcode'
driver = webdriver.Firefox()  # (Or Chrome())
driver.get(url)

username = driver.find_element_by_id("username")
password = driver.find_element_by_id("password")

# uncomment this code if your text boxes have pre-populated text
#username.clear()
#password.clear()

username.send_keys("YourUsername") # change this to your username
password.send_keys("PassworD")     # change this to your password
driver.find_element_by_name("submit").click()

# we can implicitly wait before the page loads
driver.implicitly_wait(2)

Now, this handles the first part of your question i.e. automate the login process. Now I’m not sure what’s your next goal but I assume you want the code variable in the URL, which I assume is returned by the OAuth2 funtion.

We can achieve this by simply getting the URL and parsing for the code variable

To get the URL

current_url = driver.current_url;

Now, you can simply parse the URL using urlparse.

import urllib.parse as urlparse
from urllib.parse import parse_qs

parsed = urlparse.urlparse(current_url)
OAuth_code = parse_qs(parsed.query)['code']

Some sources you can refer:

  1. https://medium.com/swlh/automate-data-collection-with-selenium-in-python-246a051206e2
  2. Fill username and password using selenium in python
  3. Find URL after clicking a link
  4. https://stackoverflow.com/a/5075477/11029298
  5. https://selenium-python.readthedocs.io/getting-started.html
User contributions licensed under: CC BY-SA
2 People found this is helpful
Advertisement