How to find children of nodes using BeautifulSoup

Question

Try this

li = soup.find('li', {'class': 'text'})
children = li.findChildren("a" , recursive=False)
for child in children:
    print(child)

There's a super small section in the DOCs that shows how to find/find_all direct children.

https://www.crummy.com/software/BeautifulSoup/bs4/doc/#the-recursive-argument

In your case as you want link1 which is first direct child:

# for only first direct child
soup.find("li", { "class" : "test" }).find("a", recursive=False)

If you want all direct children:

# for all direct children
soup.find("li", { "class" : "test" }).findAll("a", recursive=False)

Perhaps you want to do

soup.find("li", { "class" : "test" }).find('a')

try this:

li = soup.find("li", { "class" : "test" })
children = li.find_all("a") # returns a list of all <a> children of li

other reminders:

The find method only gets the first occurring child element. The find_all method gets all descendant elements and are stored in a list.

"How to find all a which are children of <li class=test> but not any others?"

Given the HTML below (I added another <a> to show te difference between select and select_one):

<div>
  <li class="test">
    <a>link1</a>
    <ul>
      <li>
        <a>link2</a>
      </li>
    </ul>
    <a>link3</a>
  </li>
</div>

The solution is to use child combinator (>) that is placed between two CSS selectors:

>>> soup.select('li.test > a')
[<a>link1</a>, <a>link3</a>]

In case you want to find only the first child:

>>> soup.select_one('li.test > a')
<a>link1</a>