如何让 selenium 单击网站弹出窗口上的按钮？

Question

I'm trying to get selenium to click a button that's part of a popup. I've tried just clicking the button like you normally would but it doesn't work. Here's the code:

button = addButton = browser.find_element_by_id("button id")
button.click()

That wasn't working so I tried just clicking a coordinate but that doesn't seem to be working either:

el = browser.find_elements_by_xpath("/html/body/div[5]")

action = webdriver.common.action_chains.ActionChains(browser)
action.move_to_element_with_offset(el, 1280, 735)
action.click()
action.perform()

Do I need my program to configure for the popup or something? Im lost.

Answer 1

You need to wait for it to become clickable using Selenium's explicit wait:

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

wait = WebDriverWait(driver, 30)
wait.until(EC.element_to_be_clickable(
    (By.ID, "button id")))

Then just click it:

addButton = browser.find_element_by_id("button id")
addButton.click()

Also, your question lacks clarity about the nature of this popup. If the above method won't work, there may be other options to use.

Your code:

el = browser.find_elements_by_xpath("/html/body/div[5]")

action = webdriver.common.action_chains.ActionChains(browser)
action.move_to_element_with_offset(el, 1280, 735)

will not work for a few reasons: 1 You use find_elements_by_xpath for a single element. find_element_by_xpath have to be used. 2 The path like this is very unlikely to be stable: /html/body/div[5] 3 Do not use coordinates unless it is strictly necessary. If you use such a method, it should use move_to_element

from selenium.webdriver import ActionChains

locator = driver.find_element_by_xpath("some xpath")

actions = ActionChains(browser)
actions.move_to_element(locator)
actions.click().perform()

4 ActionChains is unlikely to be used in your case. It is usually used for dropdowns and on hover elements.