获取所有子元素

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))

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.

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

from selenium.webdriver.common.by import By




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

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, "*')

You can use get_attribute and BeautifulSoup

html_str = el.get_attribute('innerHTML')
bs = BeautifulSoup(html_str, 'lxml')

In 2022, with selenium==4.2.0, @Richard's answer will need to be rewritten as:

from selenium.webdriver.common.by import By


parentElement = driver.find_element(By.CLASS_NAME,"bar")
elementList = parentElement.find_elements(By.TAG_NAME,"li")