BeautifulSoup getting href [duplicate]

This question already has answers here: retrieve links from web page using python and BeautifulSoup [closed] (16 answers) Closed 9 years ago.

I have the fo开发者_运维百科llowing soup:

<a href="some_url">next</a>
<span class="class">...</span>

From this I want to extract the href, "some_url"

I can do it if I only have one tag, but here there are two tags. I can also get the text 'next' but that's not what I want.

Also, is there a good description of the API somewhere with examples. I'm using the standard documentation, but I'm looking for something a little more organized.

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You can use find_all in the following way to find every a element that has an href attribute, and print each one:

# Python2
from BeautifulSoup import BeautifulSoup
    
html = '''<a href="some_url">next</a>
<span class="class"><a href="another_url">later</a></span>'''
    
soup = BeautifulSoup(html)
    
for a in soup.find_all('a', href=True):
    print "Found the URL:", a['href']

# The output would be:
# Found the URL: some_url
# Found the URL: another_url

# Python3
from bs4 import BeautifulSoup

html = '''<a href="https://some_url.com">next</a>
<span class="class">
<a href="https://some_other_url.com">another_url</a></span>'''

soup = BeautifulSoup(html)

for a in soup.find_all('a', href=True):
    print("Found the URL:", a['href'])

# The output would be:
# Found the URL: https://some_url.com
# Found the URL: https://some_other_url.com

Note that if you're using an older version of BeautifulSoup (before version 4) the name of this method is findAll. In version 4, BeautifulSoup's method names were changed to be PEP 8 compliant, so you should use find_all instead.

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If you want all tags with an href, you can omit the name parameter:

href_tags = soup.find_all(href=True)