Beautiful Soup to parse url to get another urls data

Question

I need to parse a url to get a list of urls that link to a detail page. Then from that page I need to get all the details from that page. I need to do it this way because the detail page url is not regularly incremented and changes, but the event list page stays the same.

Basically:

example.com/events/
    <a href="http://example.com/events/1">Event 1</a>
    <a href="http://example.com/events/2">Event 2</a>

example.com/events/1
    ...some detail stuff I need

example.com/events/2
    ...some detail stuff I need

Asked By: tim

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Source

Answer 1

Use urllib2 to get the page, then use beautiful soup to get the list of links, also try scraperwiki.com

Edit:

Recent discovery: Using BeautifulSoup through lxml with

from lxml.html.soupparser import fromstring

is miles better than just BeautifulSoup. It lets you do dom.cssselect(‘your selector’) which is a life saver. Just make sure you have a good version of BeautifulSoup installed. 3.2.1 works a treat.

dom = fromstring('<html... ...')
navigation_links = [a.get('href') for a in htm.cssselect('#navigation a')]

Answered By: fmalina

Answer 2

import urllib2
from BeautifulSoup import BeautifulSoup

page = urllib2.urlopen('http://yahoo.com').read()
soup = BeautifulSoup(page)
soup.prettify()
for anchor in soup.findAll('a', href=True):
    print anchor['href']

It will give you the list of urls. Now You can iterate over those urls and parse the data.

inner_div = soup.findAll("div", {"id": "y-shade"})
This is an example. You can go through the BeautifulSoup tutorials.

Answered By: Tauquir

Answer 3

For the next group of people that come across this, BeautifulSoup has been upgraded to v4 as of this post as v3 is no longer being updated..

$ easy_install beautifulsoup4

$ pip install beautifulsoup4

To use in Python…

import bs4 as BeautifulSoup

Answered By: disuse

Answer 4

FULL PYTHON 3 EXAMPLE

Packages

# urllib (comes with standard python distribution)
# pip3 install beautifulsoup4

Example:

import urllib.request
from bs4 import BeautifulSoup

with urllib.request.urlopen('https://www.wikipedia.org/') as f:
    data = f.read().decode('utf-8')

d = BeautifulSoup(data)

d.title.string

The above should print out 'Wikipedia'

Answered By: Robert Johnstone

Beautiful Soup to parse url to get another urls data

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