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Html – Selecting Button with Selenium Python

Hi I am practicing and trying to close out of a pop up that occurs when you go to a site. There is an "x" in the top right corner that I want to click once the site opens. I get the site open but can’t figure out how to identify the "x" button ID to click and close it. I am new to HTML and Python. Using MAC/safari.

HTML code when I inspect the button to close the pop up

current code:

from selenium import webdriver

driver = webdriver.Safari()
driver.get('https://littlesleepies.com/collections/vip-early-access?utm_source=facebook&utm_medium=social&utm_campaign=vip_5_16_2024')


import time

def decline_offer():
    decline = WebDriverWait(driver, 10).until(EC.presence_of_element_located((By.XPATH, "//[@id='closeIconSvg']")))
    decline.click()

Go easy on me I am a newbie lol.

Tags:

2

Answers


  1. 0

    Couple of things to try:

    1st, it looks like you are trying to click on the icon element, rather than the button tag. Try //[@id=’closeIconContainer’] instead.

    2nd, I know that you’re using Safari, but this is how I would find the xpath using Chrome.

    Rather than using "//[@id=’closeIconSvg’]" as the xpath, find the element on the page, right click and select inspect. Click on the button element in the html, then right click again, find copy > copy Xpath then drop that in as your xpath.Like So...

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  2. 0

    It’s different for Svg

    Try this: //button[@id='closeIconContainer']/*[name()='svg'][@id='closeIconSvg']

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