Selenium Finding elements by class name in python

Python Selenium Selenium Webdriver Webdriver Classname

Python Problem Overview

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>

Python Solutions

Solution 1 - Python

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.

Solution 2 - Python

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:

Solution 3 - Python

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

Solution 4 - Python

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')

Solution 5 - Python

The most simple way is to use find_element_by_class_name('class_name')

Content Type	Original Author	Original Content on Stackoverflow
Question	Sree	View Question on Stackoverflow
Solution 1 - Python	LittlePanda	View Answer on Stackoverflow
Solution 2 - Python	undetected Selenium	View Answer on Stackoverflow
Solution 3 - Python	ZygD	View Answer on Stackoverflow
Solution 4 - Python	Stan E	View Answer on Stackoverflow
Solution 5 - Python	wanderlust	View Answer on Stackoverflow