如何使用 selenium 和 python 查找 aria-label

Question

This is my code-->

element = driver.find_elements_by_css_selector("[aria-label='收回讚']")

It doesn't work, it creates an None type. Any help is appreciated! Thanks in advance!

update:

my update code:

try:
     element = WebDriverWait(driver, 5).until(EC.visibility_of_element_located((By.XPATH, "//*[aria-label='收回讚']")))
     #check if unlike path exisit
  

except:
    element = WebDriverWait(driver, 5).until(EC.visibility_of_element_located((By.XPATH, "//*[aria-label='讚']")))
    element.click()
    #if not find like path and click it

Answer 1

To find the element you need to induce WebDriverWait for the visibility_of_element_located() and you can use either of the following Locator Strategies :

• Using CSS_SELECTOR :

   element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.CSS_SELECTOR, "[aria-label='收回讚']")))

• Using XPATH :

   element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.XPATH, "//*[@aria-label='收回讚']")))

• 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

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References

You can find a couple of relevant detailed descroptions in:

• Using aria-label to locate and click an element with Python3 and Selenium