How to dismiss "Sign in to Google" popup using Selenium and Python

Question

When I enter the google website, this small login window appears:

I'm trying to dismiss it using selenium with pyhton

Here is the code i am using:

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

driver = webdriver.Chrome()

driver.get('https://www.google.com.br')
driver.maximize_window()

# wait
wait = WebDriverWait(driver, 10)
not_now_button = wait.until(EC.element_to_be_clickable((By.XPATH, '//*[@id="yDmH0d"]/c-wiz/div/div/c-wiz/div/div/div/div[2]/div[2]/button')))

# click
not_now_button.click()

I get the error below:

Traceback (most recent call last):
  File "C:\Users\sec-info\Desktop\python\googlr.py", line 13, in <module>
    not_now_button = wait.until(EC.element_to_be_clickable((By.XPATH, '//*[@id="yDmH0d"]/c-wiz/div/div/c-wiz/div/div/div/div[2]/div[2]/button')))
  File "C:\Users\sec-info\AppData\Local\Programs\Python\Python311\Lib\site-packages\selenium\webdriver\support\wait.py", line 95, in until
    raise TimeoutException(message, screen, stacktrace)
selenium.common.exceptions.TimeoutException: Message: 
Stacktrace:
Backtrace:
    (No symbol) [0x0037DCE3]
    (No symbol) [0x003139D1]
    (No symbol) [0x00224DA8]
    (No symbol) [0x0025019F]
    (No symbol) [0x002503AB]
    (No symbol) [0x0027EE62]
    (No symbol) [0x0026AF14]
    (No symbol) [0x0027D57C]
    (No symbol) [0x0026ACC6]
    (No symbol) [0x00246F68]
    (No symbol) [0x002480CD]
    GetHandleVerifier [0x005F3832+2506274]
    GetHandleVerifier [0x00629794+2727300]
    GetHandleVerifier [0x0062E36C+2746716]
    GetHandleVerifier [0x00426690+617600]
    (No symbol) [0x0031C712]
    (No symbol) [0x00321FF8]
    (No symbol) [0x003220DB]
    (No symbol) [0x0032C63B]
    BaseThreadInitThunk [0x76CBFA29+25]
    RtlGetAppContainerNamedObjectPath [0x77E57A9E+286]
    RtlGetAppContainerNamedObjectPath [0x77E57A6E+238]

Python: 3.11.2 Selenium: 4.8.2 Chromedriver: 111.0.5563.64

Are you sure the code you have shared here is the code you executed? In the screenshot which you have provided, I see the browser is in incognito mode. However, I do not see you setting the driver to incognito mode in your code. — Shawn, Mar 09 '23 at 15:33

gords · Answer 1 · 2023-03-09T17:19:31.287

0

Right-click > Inspect

Then use the simplest css (that xpath looks overly complicated)

If you could post the html, I could help

In Chrome, press F12 > Elements > Ctrl+F

Type in the css from above

button.M6CB1c.rr4y5c[aria-label='Agora não']

check if Chrome sees the element

edited Mar 09 '23 at 17:19

answered Mar 09 '23 at 15:01

gords

51
7

Agora não is the google website – exu_caveirinha Mar 09 '23 at 16:19
chrome saw element – exu_caveirinha Mar 09 '23 at 17:30
clickable is visible and enabled could you wait for displayed? – gords Mar 09 '23 at 17:52
Thanks a lot for the help! The solution was to change the iframe – exu_caveirinha Mar 09 '23 at 17:58

Harish · Answer 2 · 2023-03-09T17:45:49.900

0

You could use xpath or css selector to locate that particular button.

You need to switch to iframe since the button you're trying to click is inside an iframe.

Use below code to switch to iframe:

driver.switch_to.frame('callout')

Using xpath :

not_now_button = wait.until(EC.element_to_be_clickable((By.XPATH,"//button[@class='M6CB1c rr4y5c' and @aria-label='Agora não']")))

Using css selector:

not_now_button =  wait.until(EC.element_to_be_clickable((By.CSS_SELECTOR, "button.M6CB1c.rr4y5c[aria-label='Agora não']")))

Once you're done clicking the button, do not forget to switch back to default content using below code :

driver.switch_to.default_content()

Your code should be as below:

driver.switch_to.frame('callout')
not_now_button = wait.until(EC.element_to_be_clickable((By.XPATH,"//button[@class='M6CB1c rr4y5c' and @aria-label='Agora não']")))
not_now_button.click()
driver.switch_to.default_content()

If you want to use css selector instead of xpath to locate the button, use the css locator strategy as shown above .

edited Mar 09 '23 at 17:45

answered Mar 09 '23 at 16:47

Harish

306
1
4
14

Thanks for the help, but the error persists – exu_caveirinha Mar 09 '23 at 17:05
@exu_caveirinha Same error which you mentioned in your question or different error you get? – Harish Mar 09 '23 at 17:20
the same error appears – exu_caveirinha Mar 09 '23 at 17:31
Looks like , this particular button is inside iframe. Hence, you need to switch to iframe first before locating the button. Please try the edited answer. – Harish Mar 09 '23 at 17:36
It worked. THANK YOU VERY MUCH. The solution was to change the iframe. Thank you very much for the knowledge – exu_caveirinha Mar 09 '23 at 17:57

score 0 · Answer 3 · answered Mar 09 '23 at 23:37

The element...

NoThanks

...is within an <iframe>

So to click on the element to dismiss the popup you have to:

Induce WebDriverWait for the desired frame to be available and switch to it.
Induce WebDriverWait for the desired element to be clickable.

You can use either of the following locator strategies:

driver.get("https://www.google.com.br/")
WebDriverWait(driver, 20).until(EC.frame_to_be_available_and_switch_to_it((By.CSS_SELECTOR,"iframe[name=callout]")))
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "button[aria-label='No thanks']"))).click()

Note : You have to add the following imports :

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

Reference

You can find a couple of relevant discussions in:

Switch to an iframe through Selenium and python

How to dismiss "Sign in to Google" popup using Selenium and Python

3 Answers3

Reference