Using Selenium in Python to click all buttons in a webpage

Question

I am trying to click on all the buttons on a webpage. I'd like to be able to click them all. webpage i can click on one of them by using css selector

browser.find_element_by_css_selector('li.clickable_area:nth-child(1) > div:nth-child(3)').click()

these are the css selector for the 5 buttons

the 5 buttons follow this pattern:

Button 1: li.clickable_area: nth - child(1) > div:nth - child(3)

Button 2: li.clickable_area: nth - child(2) > div:nth - child(3)

Button 3: li.clickable_area: nth - child(3) > div:nth - child(3)

Button 4: li.clickable_area: nth - child(4) > div:nth - child(3)

Button 5: li.clickable_area: nth - child(5) > div:nth - child(3)

How i can i click them all using css selector without writing a code for each one?

then loop over button selector by different variables z=browser.find(button1) --> z.click() and so on — Zaraki Kenpachi
– Zaraki Kenpachi, Commented Feb 16, 2020 at 16:21

supputuri · Accepted Answer · 2020-02-16 19:05:02Z

1

You can use the loop to iterate through and click on the number of buttons.

number_of_buttons = 5
for x in range(number_of_buttons):
    button = browser.find_element_by_css_selector("li.clickable_area:nth-child(" + str(x+1) + ") > div:nth-child(3)")
    button.click()

If you want to click on all the li(x) > div:nth-child(3) then you can use the below.

number_li_elems=len(WebDriverWait(browser,30).until(EC.visibility_of_all_elements_located((By.CSS_SELECTOR, "li.clickable_area"))))
for x in range(number_li_elems):
    # you have to get the element by index every time, otherwise you will get StaleElement Exception
    button = browser.find_element_by_css_selector("li.clickable_area:nth-child(" + str(x+1) + ") > div:nth-child(3)")
    button.click()

edited Feb 16, 2020 at 19:05

answered Feb 16, 2020 at 18:24

supputuri

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6 Comments

BGFX Over a year ago

i've tried both but im getting AttributeError: 'NoneType' object has no attribute 'click'

supputuri Over a year ago

try with the updated answer, you might got the error in the last line in both the answers. You can check if the button element is pointing to the right div that you want to interact (in the interactive console manually)before clicking.

BGFX Over a year ago

now I'm not getting an error message, but it's not clicking on the buttons and for the second code it's scrolling without clicking

supputuri Over a year ago

try replacing the .click() line with browser.execute_script("arguments[0].click();", button)

BGFX Over a year ago

Thank you it worked flawlessly, do you mind if you show me how to make it wait half a second before each click?

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Muzzamil · Accepted Answer · 2020-02-16 19:17:04Z

0

Make a list of all buttons and iterate it. Please try below code:

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

    buttons=WebDriverWait(browser,30).until(EC.visibility_of_all_elements_located((By.CSS_SELECTOR, "li.clickable_area > div:nth-child(3)")))
        for x in range(0,len(buttons)):
            if buttons[x].is_displayed():
                buttons[x].click()

OR

buttons=WebDriverWait(browser,30).until(EC.visibility_of_all_elements_located((By.XPATH, "//div[contains(text(), 'Button ')]")))
        for x in range(0,len(buttons)):
           button = WebDriverWait(driver, 50).until(EC.presence_of_element_located((By.XPATH, "//div[contains(text(), 'Button ')]")))
                   button.click()

edited Feb 16, 2020 at 19:17

answered Feb 16, 2020 at 16:32

Muzzamil

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6 Comments

BGFX Over a year ago

i'm getting this TypeError: __init__() takes 2 positional arguments but 3 were given

supputuri Over a year ago

EC.visibility_of_all_elements_located(By. you should have another ( before By, that's why you are getting the error. However, this logic will through StaleElement exception, because of button[x].click() will refresh the element references and you can't use the older selenium references. Check my answer to know how to better handle this case.

Muzzamil Over a year ago

@supputuri Thanks for point out. I am just typing from mobile keypad so I have missed it. It will be throw staleElement if page will be loaded after click but from mobile I can see after click on button page is not loading or refreshing, Else we have to find element for each iteration. I am not sure but it seems a dummy webpage for testing. sorry it is an image.

supputuri Over a year ago

Not sure which url used to run the script, it's always safe to get the element with in loop when clicking on elements as OP did not provided the application url.

Muzzamil Over a year ago

You are right. It is an image I though it is app url.

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