How to use css selector path for selenium to get an attribute of an span class - python

Question

I would like to get the title attribute from the span class = g47SY lOXF2 but I can't find the correct css path.

Here is what I tried:

Number = self.browser.find_element_by_css_selector('ul li a span').get_attribute('title')

but it does not work.

Here is the HTML:

Answer 1

To print the value of the title attribute i.e 251 you need 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, 20).until(EC.visibility_of_element_located((By.CSS_SELECTOR, "a[href$='followers/']>span[title]"))).get_attribute("title"))
  

• Using XPATH:

  print(WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.XPATH, "//a[contains(@href, 'followers') and contains(., 'followers')]/span"))).get_attribute("title"))
  

• 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