CODEWITHSUNDEEP
pythonimageseleniumweb-scrapingbeautifulsoup
iAM BEAUTIFUL
iAM BEAUTIFUL

Reputation: 13

How can I select the image link from src html part using python selenium

I'm trying to get all the links of the images from the below mentioned html code type:

<a href="#!" class="elevatezoom-gallery" data-update="" data-image="https://img.xyz.com/rcmshopping/PROD_IMAGES/fffff_2018_05_10_10_49_23.jpg" data-zoom-image="https://img.xyz.com/rcmshopping/PROD_IMAGES/fffff_2018_05_10_10_49_23.jpg"> <img src="https://img.xyz.com/rcmshopping/PROD_IMAGES/fffff_2018_05_10_10_49_23.jpg"> </a>

I am not able to fetch any link from this part of html out f 3 any 1 is good for me. Im new to python please guide.

Upvotes: 1

Views: 244

Answers (2)

undetected Selenium
undetected Selenium

Reputation: 193088

To print the value of the src attribute you have to induce WebDriverWait for the visibility_of_element_located() and you can use either of the following Locator Strategies:

  • Using XPATH:

    print(WebDriverWait(browser, 20).until(EC.visibility_of_element_located((By.XPATH, "//a[@class='elevatezoom-gallery']/img"))).get_attribute("src"))

  • Using CSS_SELECTOR:

    print(WebDriverWait(browser, 20).until(EC.visibility_of_element_located((By.CSS_SELECTOR, "a.elevatezoom-gallery>img"))).get_attribute("src"))

  • 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

Upvotes: 1

Aleksander Ikleiw
Aleksander Ikleiw

Reputation: 2675

You can achieve that by using the find_elements_by_css_selector() function.

links = self.webdriver.find_elements_by_css_selector('img[src="https://img.xyz.com/rcmshopping/PROD_IMAGES/fffff_2018_05_10_10_49_23.jpg"])

Upvotes: 0

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