Parsing links within a html file.

[Shriphani]

Hello,
I have a html file over here by the name guide_ind.html and it
contains links to other html files like guides.html#out line . How do I
point BeautifulSoup (I want to use this module) to
guides.html#out line ?
Thanks
Shriphani P.

[Hai Vu]

On Jan 14, 9:59 am, Shriphani <shripha...@gma il.comwrote:

Hello,
I have a html file over here by the name guide_ind.html and it
contains links to other html files like guides.html#out line . How do I
point BeautifulSoup (I want to use this module) to
guides.html#out line ?
Thanks
Shriphani P.

Try Mark Pilgrim's excellent example at:
http://www.diveintopython.org/http_w...ces/index.html

From the above link, you can retrieve openanything.py which I use in
my example:

# list_url.py
# created by Hai Vu on 1/16/2008

from openanything import fetch
from sgmllib import SGMLParser

class RetrieveURLs(SG MLParser):
def reset(self):
SGMLParser.rese t(self)
self.urls = []

def start_a(self, attributes):
url = [v for k, v in attributes if k.lower() == 'href']
self.urls.exten d(url)
print '\t%s' % (url)

#
--------------------------------------------------------------------------------------------------------------
# main
def main():
site = 'http://www.google.com'

result = fetch(site)
if result['status'] == 200:
# Extracts a list of URLs off the top page
parser = RetrieveURLs()
parser.feed(res ult['data'])
parser.close()

# Display the URLs we just retrieved
print '\nURL retrieved from %s' % (site)
print '\t' + '\n\t'.join(par ser.urls)
else:
print 'Error (%d) retrieving %s' % (result['status'], site)

if __name__ == '__main__':
main()