CODEWITHSUNDEEP
pythonregex
sirjay
sirjay

Reputation: 1766

How regex until last occurrence?

I am using python, I need regex to get contacts link of web page. So, I made <a (.*?)>(.*?)Contacts(.*?)</a> and result is:

href="/ru/o-nas.html"  id="menu263" title="About">About</a></li><li><a href="/ru/photogallery.html" id="menu645" title="Photo">Photo</a></li><li  class="last"><a href="/ru/kontakt.html" class="last" id="menu583" title="">Contacts

,but I need on last <a ... like

href="/ru/kontakt.html" class="last" id="menu583" title="">Contacts

What regex pattern should I use?

python code:

match = re.findall('<a (.*?)>(.*?)Contacts(.*?)</a>', body)
if match:
    for m in match:
        print ''.join(m)

Upvotes: 1

Views: 177

Answers (2)

Susheel
Susheel

Reputation: 1679

Try BeautifulSoup

from BeautifulSoup import BeautifulSoup
import urllib2
import re
links = []
urls  ['www.u1.com','www.u2.om'....]
for url in urls:
    page = urllib2.urlopen(url)
    soup = BeautifulSoup(page)
    for link in soup.findAll('a'):
        if link.string.lower() == 'contact':
            links.append(link.get('href'))

Upvotes: 0

AKS
AKS

Reputation: 19801

Since you are parsing HTML, I would suggest to use BeautifulSoup

# sample html from question
html = '<li><a href="/ru/o-nas.html"  id="menu263" title="About">About</a></li><li><a href="/ru/photogallery.html" id="menu645" title="Photo">Photo</a></li><li  class="last"><a href="/ru/kontakt.html" class="last" id="menu583" title="">Contacts</a></li>'

from bs4 import BeautifulSoup
doc = BeautifulSoup(html)
aTag = doc.find('a', id='menu583') # id for Contacts link
print(aTag['href'])
# '/ru/kontakt.html'

Upvotes: 3

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