How can I get href links from HTML using Python?

You can use the HTMLParser module.

The code would probably look something like this:

from HTMLParser import HTMLParser

class MyHTMLParser(HTMLParser):

    def handle_starttag(self, tag, attrs):
        # Only parse the 'anchor' tag.
        if tag == "a":
           # Check the list of defined attributes.
           for name, value in attrs:
               # If href is defined, print it.
               if name == "href":
                   print name, "=", value


parser = MyHTMLParser()
parser.feed(your_html_string)

Note: The HTMLParser module has been renamed to html.parser in Python 3.0. The 2to3 tool will automatically adapt imports when converting your sources to 3.0.

Look at using the beautiful soup html parsing library.

http://www.crummy.com/software/BeautifulSoup/

You will do something like this:

import BeautifulSoup
soup = BeautifulSoup.BeautifulSoup(html)
for link in soup.findAll("a"):
    print link.get("href")

My answer probably sucks compared to the real gurus out there, but using some simple math, string slicing, find and urllib, this little script will create a list containing link elements. I test google and my output seems right. Hope it helps!

import urllib
test = urllib.urlopen("http://www.google.com").read()
sane = 0
needlestack = []
while sane == 0:
  curpos = test.find("href")
  if curpos >= 0:
    testlen = len(test)
    test = test[curpos:testlen]
    curpos = test.find('"')
    testlen = len(test)
    test = test[curpos+1:testlen]
    curpos = test.find('"')
    needle = test[0:curpos]
    if needle.startswith("http" or "www"):
        needlestack.append(needle)
  else:
    sane = 1
for item in needlestack:
  print item

Here’s a lazy version of @stephen’s answer

import html.parser
import itertools
import urllib.request

class LinkParser(html.parser.HTMLParser):
    def reset(self):
        super().reset()
        self.links = iter([])

    def handle_starttag(self, tag, attrs):
        if tag == 'a':
            for (name, value) in attrs:
                if name == 'href':
                    self.links = itertools.chain(self.links, [value])


def gen_links(stream, parser):
    encoding = stream.headers.get_content_charset() or 'UTF-8'
    for line in stream:
        parser.feed(line.decode(encoding))
        yield from parser.links

Use it like so:

>>> parser = LinkParser()
>>> stream = urllib.request.urlopen('http://stackoverflow.com/questions/3075550')
>>> links = gen_links(stream, parser)
>>> next(links)
'//stackoverflow.com'

Using BS4 for this specific task seems overkill.

Try instead:

website = urllib2.urlopen('http://10.123.123.5/foo_images/Repo/')
html = website.read()
files = re.findall('href="(.*tgz|.*tar.gz)"', html)
print sorted(x for x in (files))

I found this nifty piece of code on http://www.pythonforbeginners.com/code/regular-expression-re-findall and works for me quite well.

I tested it only on my scenario of extracting a list of files from a web folder that exposes the filesfolder in it, e.g.:

and I got a sorted list of the filesfolders under the URL

Using requests with BeautifulSoup and Python 3:

import requests 
from bs4 import BeautifulSoup


page = requests.get('http://www.website.com')
bs = BeautifulSoup(page.content, features='lxml')
for link in bs.findAll('a'):
    print(link.get('href'))

This is way late to answer but it will work for latest python users:

from bs4 import BeautifulSoup
import requests 


html_page = requests.get('http://www.example.com').text

soup = BeautifulSoup(html_page, "lxml")
for link in soup.findAll('a'):
    print(link.get('href'))

Don’t forget to install “requests” and “BeautifulSoup” package and also “lxml“. Use .text along with get otherwise it will throw an exception.

“lxml” is used to remove that warning of which parser to be used. You can also use “html.parser” whichever fits your case.

This answer is similar to others with requests and BeautifulSoup, but using list comprehension.

Because find_all() is the most popular method in the Beautiful Soup search API, you can use soup("a") as a shortcut of soup.findAll("a") and using list comprehension:

import requests
from bs4 import BeautifulSoup

URL = "http://www.yourwebsite.com"
page = requests.get(URL)
soup = BeautifulSoup(page.content, features='lxml')
# Find links
all_links = [link.get("href") for link in soup("a")]
# Only external links
ext_links = [link.get("href") for link in soup("a") if "http" in link.get("href")]

https://www.crummy.com/software/BeautifulSoup/bs4/doc/#calling-a-tag-is-like-calling-find-all

Simplest way for me:

from urlextract import URLExtract
from requests import get

url = "sample.com/samplepage/"
req = requests.get(url)
text = req.text
# or if you already have the html source:
# text = "This is html for ex <a href='http://google.com/'>Google</a> <a href='http://yahoo.com/'>Yahoo</a>"
text = text.replace(' ', '').replace('=','')
extractor = URLExtract()
print(extractor.find_urls(text))

output:

['http://google.com/', 'http://yahoo.com/']