How can I close a "pop-up" window with Selenium/Python?

Question

I hope you're doing well!

I want to scrap this website https://proadvisor.intuit.com/app/accountant/search?region=ca&distance=1000, if you click on it the site load normally but if I run my script appears something like a "pop-up" window (I attached the evidence).

pop-up window

I'm using Selenium and Python to do it. I want to do click on the 'x' button to close that and then I can start to scrap the website so I tried with the next two ways with the XPATH:

close_button = driver.find_element(By.XPATH, '/html/body/div[4]/button') close_button.click()

driver.find_element(By.XPATH, '//div[contains(@class,"coach-marks")]//button[@aria-label="close"]').click()

But the result was the same, the next error is launched and the "pop-up" window doesn't close.

selenium.common.exceptions.NoSuchElementException: Message: no such element: Unable to locate element: {"method":"xpath","selector":"/html/body/div[4]/button"}

The HTML code is:

<button class="a--close_button" aria-label="close" tabindex="0">
    <svg class="svg--close_button" width="19" height="19" viewBox="0 0 19 19">
        <path
        fill-rule="evenodd"
        d="M19 1.83L17.17 0 9.5 7.67 1.83 0 0 1.83 7.67 9.5 0 17.17 1.83 19l7.67-7.67L17.17 19 19 17.17 11.33 9.5z"
        ></path>
    </svg>
</button>

All my Python code is:

from selenium import webdriver
from selenium.webdriver.chrome.service import Service
from selenium.webdriver.chrome.options import Options
from selenium.webdriver.common.by import By
from selenium.webdriver.common.keys import Keys
from webdriver_manager.chrome import ChromeDriverManager

chrome_options = Options()
chrome_options.add_experimental_option("detach", True)
driver = webdriver.Chrome(service=Service(ChromeDriverManager().install()), options=chrome_options)

driver.get("https://proadvisor.intuit.com/app/accountant/search?region=ca&distance=1000")

# click to close button
close_button = driver.find_element(By.XPATH, '/html/body/div[4]/button')
close_button.click()

If you can help me I will very grateful with you.

Regards.

Answer 1

Wait for the element to become visible before trying to interact with it. You can do this by using an explicit wait with the expected_conditions module from "selenium.webdriver.support.ui".

from selenium import webdriver
from selenium.webdriver.chrome.service import Service
from selenium.webdriver.chrome.options import Options
from selenium.webdriver.common.by import By
from selenium.webdriver.common.keys import Keys
from webdriver_manager.chrome import ChromeDriverManager
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC

chrome_options = Options()
chrome_options.add_experimental_option("detach", True)
driver = webdriver.Chrome(service=Service(ChromeDriverManager().install()), options=chrome_options)

driver.get("https://proadvisor.intuit.com/app/accountant/search?region=ca&distance=1000")

# wait for the close button to be visible and click it
wait = WebDriverWait(driver, 10)
close_button = wait.until(EC.visibility_of_element_located((By.XPATH, '//button[@aria-label="close"]')))
close_button.click()