Python Selenium - get href value
Solution 1
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]")))
Solution 2
As per the given HTML:
<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>
As the href
attribute is within the <a>
tag ideally you need to move deeper till the <a>
node. So to extract the value of the href
attribute you can use either of the following Locator Strategies:
-
Using
css_selector
:print(driver.find_element_by_css_selector("p.sc-eYdvao.kvdWiq > a").get_attribute('href'))
-
Using
xpath
:print(driver.find_element_by_xpath("//p[@class='sc-eYdvao kvdWiq']/a").get_attribute('href'))
If you want to extract all the values of the href
attribute you need to use find_elements*
instead:
-
Using
css_selector
:print([my_elem.get_attribute("href") for my_elem in driver.find_elements_by_css_selector("p.sc-eYdvao.kvdWiq > a")])
-
Using
xpath
:print([my_elem.get_attribute("href") for my_elem in driver.find_elements_by_xpath("//p[@class='sc-eYdvao kvdWiq']/a")])
Dynamic elements
However, if you observe the values of class attributes i.e. sc-eYdvao
and kvdWiq
ideally those are dynamic values. So to extract the href
attribute you have to induce WebDriverWait for the visibility_of_element_located()
and you can use either of the following Locator Strategies:
-
Using
CSS_SELECTOR
:print(WebDriverWait(driver, 10).until(EC.visibility_of_element_located((By.CSS_SELECTOR, "p.sc-eYdvao.kvdWiq > a"))).get_attribute('href'))
-
Using
XPATH
:print(WebDriverWait(driver, 10).until(EC.visibility_of_element_located((By.XPATH, "//p[@class='sc-eYdvao kvdWiq']/a"))).get_attribute('href'))
If you want to extract all the values of the href
attribute you can use visibility_of_all_elements_located()
instead:
-
Using
CSS_SELECTOR
:print([my_elem.get_attribute("innerHTML") for my_elem in WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.CSS_SELECTOR, "p.sc-eYdvao.kvdWiq > a")))])
-
Using
XPATH
:print([my_elem.get_attribute("innerHTML") for my_elem in WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.XPATH, "//p[@class='sc-eYdvao kvdWiq']/a")))])
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
Eric Choi
Updated on July 09, 2022Comments
-
Eric Choi almost 2 years
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'
. Usingdriver.find_element_by_css_selector(".sc-eYdvao.kvdWiq").get_attribute("href")
returnedNone
. 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.
-
ATJ over 3 yearsThere's a typo in the wait condition: it should be WebDriverWait(driver,10).until ... [with a closing bracket between '10' and '.until').
-
Robert Alexander over 2 yearsThank you worked beautifully also for my case although I am using find_elements_by_class_name