How do I retrieve the link of an image through Selenium


I’m trying to make my program fetch the link of an image and then store it as a string in a variable.
This is the xpath of the image. I need to do it through xpaths because the xpaths on the website are very similar bar the "/article[x]". This allow me to increase the number with a variable so that I can go through all the xpaths on the page.


Picture of the website that I’m trying to retrieve the links of the image
Picture of the website that I'm trying to retrieve the links of the image

My code:

from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from import By
from import WebDriverWait
from import expected_conditions as EC
import tkinter
import time

Anime = input("Enter Anime:")
driver = webdriver.Chrome(executable_path=r"C:\Users\amete\Documents\chromedriver.exe")


search = driver.find_element_by_xpath('//input[@name="q"]')
wait = WebDriverWait(driver, 20)
wait.until(EC.element_to_be_clickable((By.XPATH, '//input[@name="q"]')))
#  Clears the field
search.send_keys(Keys.CONTROL, 'a')

#  The field is now cleared and the program can type whatever it wants

#  Accept the cookies
wait.until(EC.element_to_be_clickable((By.XPATH, '//*[@id="qc-cmp2-ui"]/div[2]/div/button[3]'))).click()

#  Added this wait
wait.until(EC.element_to_be_clickable((By.XPATH,'//h2[@id="anime"]//ancestor::div[@class="content-left"]//article[1]/div[contains(@class, "list")][1]/div[contains(@class, "information")]/a[1]')))
link = driver.find_element_by_xpath('//h2[@id="anime"]//ancestor::div[@class="content-left"]//article[1]/div[contains(@class, "list")][1]/div[contains(@class, "information")]/a[1]').text

piclink = driver.('/html/body/div[2]/div[2]/div[3]/div[2]/div[2]/div[1]/div/article[1]/div[2]/div[1]/a/img')
print (piclink)


you can get it like this (specify the attribute)

piclink = driver.find_element_by_xpath('/html/body/div[2]/div[2]/div[3]/div[2]/div[2]/div[1]/div/article[1]/div[2]/div[1]/a/img').get_attribute('src')

Answered By – CatChMeIfUCan

This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0

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