I want to get all the <a>
tags which are children of <li>
:
<div>
<li class="test">
<a>link1</a>
<ul>
<li>
<a>link2</a>
</li>
</ul>
</li>
</div>
I know how to find element with particular class like this:
soup.find("li", { "class" : "test" })
But I don't know how to find all <a>
which are children of <li class=test>
but not any others.
Like I want to select:
<a>link1</a>
This question is related to
python
html
beautifulsoup
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)
"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>
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.
Yet another method - create a filter function that returns True
for all desired tags:
def my_filter(tag):
return (tag.name == 'a' and
tag.parent.name == 'li' and
'test' in tag.parent['class'])
Then just call find_all
with the argument:
for a in soup(my_filter): # or soup.find_all(my_filter)
print a
Source: Stackoverflow.com