Get all child elements

PythonSeleniumSelenium Webdriver

Python Problem Overview


In Selenium with Python is it possible to get all the children of a WebElement as a list?

Python Solutions


Solution 1 - Python

Yes, you can achieve it by find_elements_by_css_selector("*") or find_elements_by_xpath(".//*").

However, this doesn't sound like a valid use case to find all children of an element. It is an expensive operation to get all direct/indirect children. Please further explain what you are trying to do. There should be a better way.

from selenium import webdriver

driver = webdriver.Firefox()
driver.get("http://www.stackoverflow.com")

header = driver.find_element_by_id("header")

# start from your target element, here for example, "header"
all_children_by_css = header.find_elements_by_css_selector("*")
all_children_by_xpath = header.find_elements_by_xpath(".//*")

print 'len(all_children_by_css): ' + str(len(all_children_by_css))
print 'len(all_children_by_xpath): ' + str(len(all_children_by_xpath))

Solution 2 - Python

Yes, you can use find_elements_by_ to retrieve children elements into a list. See the python bindings here: http://selenium-python.readthedocs.io/locating-elements.html

Example HTML:

<ul class="bar">
    <li>one</li>
    <li>two</li>
    <li>three</li>
</ul>

You can use the find_elements_by_ like so:

parentElement = driver.find_element_by_class_name("bar")
elementList = parentElement.find_elements_by_tag_name("li")

If you want help with a specific case, you can edit your post with the HTML you're looking to get parent and children elements from.

Solution 3 - Python

Another veneration of find_elements_by_xpath(".//*") is:

from selenium.webdriver.common.by import By


find_elements(By.XPATH, ".//*")

Solution 4 - Python

You can't use

all_children_by_css = header.find_elements_by_css_selector("*")

You now need to use

all_children_by_css = header.find_elements(By.XPATH, "*')

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questionwalshie4View Question on Stackoverflow
Solution 1 - PythonYi ZengView Answer on Stackoverflow
Solution 2 - PythonRichardView Answer on Stackoverflow
Solution 3 - PythonJortegaView Answer on Stackoverflow
Solution 4 - PythonQuentinJSView Answer on Stackoverflow