Finding elements by class name with Selenium in Python

How can I filter elements that have the same class?

<html>
<body>
<p class="content">Link1.</p>
</body>
</html>
<html>
<body>
<p class="content">Link2.</p>
</body>
</html>
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小开

Use nth-child, for example: http://www.w3schools.com/cssref/sel_nth-child.asp

driver.find_element(By.CSS_SELECTOR, 'p.content:nth-child(1)')

or http://www.w3schools.com/cssref/sel_firstchild.asp

driver.find_element(By.CSS_SELECTOR, 'p.content:first-child')
小开

最简单的方法是使用find_element_by_class_name('class_name')

小开

You can try to get the list of all elements with class = "content" by using find_elements_by_class_name:

a = driver.find_elements_by_class_name("content")

Then you can click on the link that you are looking for.

小开

By.CLASS_NAME was not yet mentioned:

from selenium.webdriver.common.by import By


driver.find_element(By.CLASS_NAME, "content")

This is the list of attributes which can be used as locators in By:

CLASS_NAME
CSS_SELECTOR
ID
LINK_TEXT
NAME
PARTIAL_LINK_TEXT
TAG_NAME
XPATH

小开

As per the HTML:

<html>
<body>
<p class="content">Link1.</p>
</body>
<html>
<html>
<body>
<p class="content">Link2.</p>
</body>
<html>

Two(2) <p> elements are having the same class content.

So to filter the elements having the same class i.e. content and create a list you can use either of the following Locator Strategies:

  • Using class_name:

    elements = driver.find_elements_by_class_name("content")

  • Using css_selector:

     elements = driver.find_elements_by_css_selector(".content")

  • Using xpath:

    elements = driver.find_elements_by_xpath("//*[@class='content']")

Ideally, to click on the element you need to induce WebDriverWait for the visibility_of_all_elements_located() and you can use either of the following Locator Strategies:

  • Using CLASS_NAME:

    elements = WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.CLASS_NAME, "content")))

  • Using CSS_SELECTOR:

    elements = WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.CSS_SELECTOR, ".content")))

  • Using XPATH:

    elements = WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.XPATH, "//*[@class='content']")))

  • 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

References

You can find a couple of relevant discussions in:

小开

The first answer has been deprecated, and the other answers only return one result. This is the correct answer:

driver.find_elements(By.CLASS_NAME, "content")

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