Finding an element and clicking on it using Selenium and Python

Question

I am making a script that will navigate through a website and perform some actions for me. But I am stuck trying to make my script click on the following elemnt:

<a href="JavaScript:OnCopy(22291488);" title="Copy Trip"><img src="../Images/copy.gif"></a>

As there is no ID or class name for this element, i tried to find the element by xpath and clicking on it by using the following code:

import selenium
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
from selenium.webdriver.common.keys import Keys
import time

....

copy = driver.find_element_by_xpath('//[@id="divGrid"]/div[2]/div[2]/table/tbody/tr[1]/td[1]/div/span/a[2]')
copy.click()

This isn't working, so I am open to suggestions to how to solve this issue.

Can you supply more of your code and the url with the element you're trying to click on? — Tobias P. G.
– Tobias P. G., Commented Jun 22, 2020 at 18:02

undetected Selenium · Accepted Answer · 2020-06-22 23:02:18Z

1

The desired element is a JavaScript enabled element so to click on the element you have to induce WebDriverWait for the element_to_be_clickable() and you can use either of the following Locator Strategies:

Using CSS_SELECTOR:

WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "a[title='Copy Trip']>img[src$='/Images/copy.gif']"))).click()

Using XPATH:

WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//a[@title='Copy Trip' and contains(@href, 'OnCopy')]/img[contains(@src, '/Images/copy.gif')]"))).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

answered Jun 22, 2020 at 23:02

undetected Selenium

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