I am working with selenium in python and try to automating google sign-in in chrome but I am in a problem

Question

It is giving an error of could not sign in.

See

Please Help me

Answer 1

Use this link:

https://accounts.google.com/o/oauth2/v2/auth/oauthchooseaccount?redirect_uri=https%3A%2F%2Fdevelopers.google.com%2Foauthplayground&prompt=consent&response_type=code&client_id=407408718192.apps.googleusercontent.com&scope=email&access_type=offline&flowName=GeneralOAuthFlow

I don't know what's the problem but google has it's own security protocols and it is hard to by pass it.

To use it you must turn off two step verification and other account security. In my case when I tried signing in, for the first 5-6 times it asked me to verify by tapping a number in my smartphone(hope you know what's that), then it is working fine for me.

Now, if you want to avoid error you can use Explicit Wait in selenium.

from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as ec

WebDriverWait(driver, 100).until(ec.visibility_of_element_located((By.XPATH, '###')))
#Place the xpath of  you element in place of ###

By using this, you can make selenium wait until an element loads and then proceed.

If problem continues to occur, you can use third party sites to sign in, which include stack overflow.

from selenium import webdriver 
from time import sleep

username=raw_input("username: ") 
password=raw_input("password: ") 
driver=webdriver.Chrome('...')  #path of your chrome driver         
driver.get('https://stackoverflow.com/users/signup') 
sleep(3) 
driver.find_element_by_xpath('//*[@id="openid-buttons"]/button[1]').click() 
driver.find_element_by_id('identifierId').send_keys(username) 
driver.find_element_by_id('identifierNext').click() 
sleep(3) 
driver.find_element_by_name('password').send_keys(password) 
driver.find_element_by_id('passwordNext').click() 
sleep(2) 
driver.get('https://mail.google.com/mail/u/0/#inbox')

Now, you can get any of your URL and continue.