Web Crawler To Get Links From New Website

I am trying to get the links from a news website page(from one of its archives). I wrote the following lines of code in Python: main.py contains : import mechanize from bs4 import

Solution 1:

I believe you may want to try accessing the text inside the list item like so:

for tag in soup.findAll('li', attrs={"data-section":"Business"}):
    articletext += tag.string

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Edited: General Comments on getting links from a page

Probably the easiest data type to use to gather a bunch of links and retrieve them later is a dictionary.

To get links from a page using BeautifulSoup, you could do something like the following:

link_dictionary = {}
withurlopen(url_source) as f:
    soup = BeautifulSoup(f)
    for link in soup.findAll('a'):
        link_dictionary[link.string] = link.get('href') 

This will provide you with a dictionary named link_dictionary, where every key in the dictionary is a string that is simply the text contents between the <a> </a> tags and every value is the the value of the href attribute.

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How to combine this what your previous attempt

Now, if we combine this with the problem you were having before, we could try something like the following:

link_dictionary = {}
for tag in soup.findAll('li', attrs={"data-section":"Business"}):
    for link in tag.findAll('a'):
        link_dictionary[link.string] = link.get('href') 

If this doesn't make sense, or you have a lot more questions, you will need to experiment first and try to come up with a solution before asking another new, clearer question.

Solution 2:

You might want to use the powerful XPath query language with the faster lxml module. As simple as that:

import urllib2
from lxml import etree

url = 'http://www.thehindu.com/archive/web/2010/06/19/'
html = etree.HTML(urllib2.urlopen(url).read())

for link in html.xpath("//li[@data-section='Business']/a"):
    print'{} ({})'.format(link.text, link.attrib['href'])

Update for @data-section='Chennai'

#!/usr/bin/pythonimport urllib2
from lxml import etree

url = 'http://www.thehindu.com/template/1-0-1/widget/archive/archiveWebDayRest.jsp?d=2010-06-19'
html = etree.HTML(urllib2.urlopen(url).read())

for link in html.xpath("//li[@data-section='Chennai']/a"):
    print'{} => {}'.format(link.text, link.attrib['href'])