[FIXED] Python Selenium - get href value

Issue

I am trying to copy the href value from a website, and the html code looks like this:

<p class="sc-eYdvao kvdWiq">
 <a href="https://www.iproperty.com.my/property/setia-eco-park/sale- 
 1653165/">Shah Alam Setia Eco Park, Setia Eco Park
 </a>
</p>

I've tried driver.find_elements_by_css_selector(".sc-eYdvao.kvdWiq").get_attribute("href") but it returned 'list' object has no attribute 'get_attribute'. Using driver.find_element_by_css_selector(".sc-eYdvao.kvdWiq").get_attribute("href") returned None. But i cant use xpath because the website has like 20+ href which i need to copy all. Using xpath would only copy one.

If it helps, all the 20+ href are categorised under the same class which is sc-eYdvao kvdWiq.

Ultimately i would want to copy all the 20+ href and export them out to a csv file.

Appreciate any help possible.

Solution

You want driver.find_elements if more than one element. This will return a list. For the css selector you want to ensure you are selecting for those classes that have a child href

elems = driver.find_elements_by_css_selector(".sc-eYdvao.kvdWiq [href]")
links = [elem.get_attribute('href') for elem in elems]

You might also need a wait condition for presence of all elements located by css selector.

elems = WebDriverWait(driver,10).until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, ".sc-eYdvao.kvdWiq [href]")))

Answered By - QHarr