retrieve links from web page using python and BeautifulSoup [closed]

Answer 1

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 = 'https://archive.ics.uci.edu/ml/machine-learning-databases/eeg-mld/eeg_full/'
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']
            wget.download(full_path)

Answer 2

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("http://www.gpsbasecamp.com/national-parks")
soup = BeautifulSoup(resp, parser, from_encoding=resp.info().get_param('charset'))

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

or the Python 2 version:

from bs4 import BeautifulSoup
import urllib2

parser = 'html.parser'  # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib2.urlopen("http://www.gpsbasecamp.com/national-parks")
soup = BeautifulSoup(resp, parser, from_encoding=resp.info().getparam('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://www.gpsbasecamp.com/national-parks")
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):
    print(link['href'])

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.

Answer 3

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

try:
    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
                else:
                    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:

getlinks.py http://stackoverflow.com/a/37758066/191246
getlinks.py http://stackoverflow.com/a/37758066/191246 "*users*"
getlinks.py http://fakedomain.mu/somepage.html "*.mp3"

Answer 4

import urllib2
import BeautifulSoup

request = urllib2.Request("http://www.gpsbasecamp.com/national-parks")
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'

Answer 5

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 re.search() 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 = website.read()

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

print links

Answer 6

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("http://www.espncricinfo.com/").read()
soup = BeautifulSoup(url)

for line in soup.find_all('a'):
    print(line.get('href'))