retrieve links from web page using python and BeautifulSoup


How can I retrieve the links of a webpage and copy the url address of the links using Python?

Asked By: NepUS



import urllib2
import BeautifulSoup

request = urllib2.Request("")
response = urllib2.urlopen(request)
soup = BeautifulSoup.BeautifulSoup(response)
for a in soup.findAll('a'):
  if 'national-park' in a['href']:
    print 'found a url with national-park in the link'
Answered By: Andrew Johnson

Here’s a short snippet using the SoupStrainer class in BeautifulSoup:

import httplib2
from bs4 import BeautifulSoup, SoupStrainer

http = httplib2.Http()
status, response = http.request('')

for link in BeautifulSoup(response, parse_only=SoupStrainer('a')):
    if link.has_attr('href'):

The BeautifulSoup documentation is actually quite good, and covers a number of typical scenarios:

Edit: Note that I used the SoupStrainer class because it’s a bit more efficient (memory and speed wise), if you know what you’re parsing in advance.

Answered By: ars

just for getting the links, without B.soup and regex:

import urllib2
tag="<a href=""
for item in data:
    if "<a href" in item:
            ind = item.index(tag)
        except: pass
            print item[:end]

for more complex operations, of course BSoup is still preferred.

Answered By: ghostdog74

Others have recommended BeautifulSoup, but it’s much better to use lxml. Despite its name, it is also for parsing and scraping HTML. It’s much, much faster than BeautifulSoup, and it even handles “broken” HTML better than BeautifulSoup (their claim to fame). It has a compatibility API for BeautifulSoup too if you don’t want to learn the lxml API.

Ian Blicking agrees.

There’s no reason to use BeautifulSoup anymore, unless you’re on Google App Engine or something where anything not purely Python isn’t allowed.

lxml.html also supports CSS3 selectors so this sort of thing is trivial.

An example with lxml and xpath would look like this:

import urllib
import lxml.html
connection = urllib.urlopen('')

dom =  lxml.html.fromstring(

for link in dom.xpath('//a/@href'): # select the url in href for all a tags(links)
    print link
Answered By: aehlke

Why not use regular expressions:

import urllib2
import re
url = ""
page = urllib2.urlopen(url)
page =
links = re.findall(r"<a.*?s*href="(.*?)".*?>(.*?)</a>", page)
for link in links:
    print('href: %s, HTML text: %s' % (link[0], link[1]))
Answered By: ahmadh

Under the hood BeautifulSoup now uses lxml. Requests, lxml & list comprehensions makes a killer combo.

import requests
import lxml.html

dom = lxml.html.fromstring(requests.get('').content)

[x for x in dom.xpath('//a/@href') if '//' in x and '' not in x]

In the list comp, the “if ‘//’ and ‘’ not in x” is a simple method to scrub the url list of the sites ‘internal’ navigation urls, etc.

Answered By: cheekybastard

The following code is to retrieve all the links available in a webpage using urllib2 and BeautifulSoup4:

import urllib2
from bs4 import BeautifulSoup

url = urllib2.urlopen("").read()
soup = BeautifulSoup(url)

for line in soup.find_all('a'):
Answered By: Sentient07

For completeness sake, the BeautifulSoup 4 version, making use of the encoding supplied by the server as well:

from bs4 import BeautifulSoup
import urllib.request

parser = 'html.parser'  # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib.request.urlopen("")
soup = BeautifulSoup(resp, parser,'charset'))

for link in soup.find_all('a', href=True):

or the Python 2 version:

from bs4 import BeautifulSoup
import urllib2

parser = 'html.parser'  # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib2.urlopen("")
soup = BeautifulSoup(resp, parser,'charset'))

for link in soup.find_all('a', href=True):
    print link['href']

and a version using the requests library, which as written will work in both Python 2 and 3:

from bs4 import BeautifulSoup
from bs4.dammit import EncodingDetector
import requests

parser = 'html.parser'  # or 'lxml' (preferred) or 'html5lib', if installed
resp = requests.get("")
http_encoding = resp.encoding if 'charset' in resp.headers.get('content-type', '').lower() else None
html_encoding = EncodingDetector.find_declared_encoding(resp.content, is_html=True)
encoding = html_encoding or http_encoding
soup = BeautifulSoup(resp.content, parser, from_encoding=encoding)

for link in soup.find_all('a', href=True):

The soup.find_all('a', href=True) call finds all <a> elements that have an href attribute; elements without the attribute are skipped.

BeautifulSoup 3 stopped development in March 2012; new projects really should use BeautifulSoup 4, always.

Note that you should leave decoding the HTML from bytes to BeautifulSoup. You can inform BeautifulSoup of the characterset found in the HTTP response headers to assist in decoding, but this can be wrong and conflicting with a <meta> header info found in the HTML itself, which is why the above uses the BeautifulSoup internal class method EncodingDetector.find_declared_encoding() to make sure that such embedded encoding hints win over a misconfigured server.

With requests, the response.encoding attribute defaults to Latin-1 if the response has a text/* mimetype, even if no characterset was returned. This is consistent with the HTTP RFCs but painful when used with HTML parsing, so you should ignore that attribute when no charset is set in the Content-Type header.

Answered By: Martijn Pieters
import urllib2
from bs4 import BeautifulSoup
#To get href part alone
print links[0].attrs['href']
Answered By: Tilak Patidar

This script does what your looking for, But also resolves the relative links to absolute links.

import urllib
import lxml.html
import urlparse

def get_dom(url):
    connection = urllib.urlopen(url)
    return lxml.html.fromstring(

def get_links(url):
    return resolve_links((link for link in get_dom(url).xpath('//a/@href')))

def guess_root(links):
    for link in links:
        if link.startswith('http'):
            parsed_link = urlparse.urlparse(link)
            scheme = parsed_link.scheme + '://'
            netloc = parsed_link.netloc
            return scheme + netloc

def resolve_links(links):
    root = guess_root(links)
    for link in links:
        if not link.startswith('http'):
            link = urlparse.urljoin(root, link)
        yield link  

for link in get_links(''):
    print link
Answered By: Ricky Wilson

To find all the links, we will in this example use the urllib2 module together
with the re.module

*One of the most powerful function in the re module is “re.findall()”.
While is used to find the first match for a pattern, re.findall() finds all
the matches and returns them as a list of strings, with each string representing one match*

import urllib2

import re
#connect to a URL
website = urllib2.urlopen(url)

#read html code
html =

#use re.findall to get all the links
links = re.findall('"((http|ftp)s?://.*?)"', html)

print links
Answered By: Mayur Ingle

BeatifulSoup’s own parser can be slow. It might be more feasible to use lxml which is capable of parsing directly from a URL (with some limitations mentioned below).

import lxml.html

doc = lxml.html.parse(url)

links = doc.xpath('//a[@href]')

for link in links:
    print link.attrib['href']

The code above will return the links as is, and in most cases they would be relative links or absolute from the site root. Since my use case was to only extract a certain type of links, below is a version that converts the links to full URLs and which optionally accepts a glob pattern like *.mp3. It won’t handle single and double dots in the relative paths though, but so far I didn’t have the need for it. If you need to parse URL fragments containing ../ or ./ then urlparse.urljoin might come in handy.

NOTE: Direct lxml url parsing doesn’t handle loading from https and doesn’t do redirects, so for this reason the version below is using urllib2 + lxml.

#!/usr/bin/env python
import sys
import urllib2
import urlparse
import lxml.html
import fnmatch

    import urltools as urltools
except ImportError:
    sys.stderr.write('To normalize URLs run: `pip install urltools --user`')
    urltools = None

def get_host(url):
    p = urlparse.urlparse(url)
    return "{}://{}".format(p.scheme, p.netloc)

if __name__ == '__main__':
    url = sys.argv[1]
    host = get_host(url)
    glob_patt = len(sys.argv) > 2 and sys.argv[2] or '*'

    doc = lxml.html.parse(urllib2.urlopen(url))
    links = doc.xpath('//a[@href]')

    for link in links:
        href = link.attrib['href']

        if fnmatch.fnmatch(href, glob_patt):

            if not href.startswith(('http://', 'https://' 'ftp://')):

                if href.startswith('/'):
                    href = host + href
                    parent_url = url.rsplit('/', 1)[0]
                    href = urlparse.urljoin(parent_url, href)

                    if urltools:
                        href = urltools.normalize(href)

            print href

The usage is as follows: "*users*" "*.mp3"
Answered By: ccpizza

Here’s an example using @ars accepted answer and the BeautifulSoup4, requests, and wget modules to handle the downloads.

import requests
import wget
import os

from bs4 import BeautifulSoup, SoupStrainer

url = ''
file_type = '.tar.gz'

response = requests.get(url)

for link in BeautifulSoup(response.content, 'html.parser', parse_only=SoupStrainer('a')):
    if link.has_attr('href'):
        if file_type in link['href']:
            full_path = url + link['href']
Answered By: Blairg23

I found the answer by @Blairg23 working , after the following correction (covering the scenario where it failed to work correctly):

for link in BeautifulSoup(response.content, 'html.parser', parse_only=SoupStrainer('a')):
    if link.has_attr('href'):
        if file_type in link['href']:
            full_path =urlparse.urljoin(url , link['href']) #module urlparse need to be imported

For Python 3:

urllib.parse.urljoin has to be used in order to obtain the full URL instead.

Answered By: AkanKsha Bhardwaj

Links can be within a variety of attributes so you could pass a list of those attributes to select.

For example, with src and href attributes (here I am using the starts with ^ operator to specify that either of these attributes values starts with http):

from bs4 import BeautifulSoup as bs
import requests
r = requests.get('')
soup = bs(r.content, 'lxml')
links = [item['href'] if item.get('href') is not None else item['src'] for item in'[href^="http"], [src^="http"]') ]

Attribute = value selectors


Represents elements with an attribute name of attr whose value is prefixed (preceded) by value.

There are also the commonly used $ (ends with) and * (contains) operators. For a full syntax list see the link above.

Answered By: QHarr

There can be many duplicate links together with both external and internal links. To differentiate between the two and just get unique links using sets:

# Python 3.
import urllib    
from bs4 import BeautifulSoup

url = ""
resp = urllib.request.urlopen(url)
# Get server encoding per recommendation of Martijn Pieters.
soup = BeautifulSoup(resp,'charset'))  
external_links = set()
internal_links = set()
for line in soup.find_all('a'):
    link = line.get('href')
    if not link:
    if link.startswith('http'):

# Depending on usage, full internal links may be preferred.
full_internal_links = {
    urllib.parse.urljoin(url, internal_link) 
    for internal_link in internal_links

# Print all unique external and full internal links.
for link in external_links.union(full_internal_links):
Answered By: Alexander