"George" <[EMAIL PROTECTED]> wrote:
> I'm very new to python and I have tried to read the tutorials but I am
> unable to understand exactly how I must do this problem.
>
> Specifically, the showIPnums function takes a URL as input, calls the
> read_page(url) function to obtain the entire page for that URL, and
> then lists, in sorted order, the IP addresses implied in the "<A
> HREF=· · ·>" tags within that page.
>
>
> """
> Module to print IP addresses of tags in web file containing HTML
>
> >>> showIPnums('http://22c118.cs.uiowa.edu/uploads/easy.html')
> ['0.0.0.0', '128.255.44.134', '128.255.45.54']
>
> >>> showIPnums('http://22c118.cs.uiowa.edu/uploads/pytorg.html')
> ['0.0.0.0', '128.255.135.49', '128.255.244.57', '128.255.30.11',
> '128.255.34.132', '128.255.44.51', '128.255.45.53',
> '128.255.45.54', '129.255.241.42', '64.202.167.129']
>
> """
>
> def read_page(url):
> import formatter
> import htmllib
> import urllib
>
> htmlp = htmllib.HTMLParser(formatter.NullFormatter())
> htmlp.feed(urllib.urlopen(url).read())
> htmlp.close()
>
> def showIPnums(URL):
> page=read_page(URL)
>
> if __name__ == '__main__':
> import doctest, sys
> doctest.testmod(sys.modules[__name__])
You forgot to mention that you don't want duplicates in the result. Here's a
function that passes
the doctest:
from urllib import urlopen
from urlparse import urlsplit
from socket import gethostbyname
from BeautifulSoup import BeautifulSoup
def showIPnums(url):
"""Return the unique IPs found in the anchors of the webpage at the given
url.
>>> showIPnums('http://22c118.cs.uiowa.edu/uploads/easy.html')
['0.0.0.0', '128.255.44.134', '128.255.45.54']
>>> showIPnums('http://22c118.cs.uiowa.edu/uploads/pytorg.html')
['0.0.0.0', '128.255.135.49', '128.255.244.57', '128.255.30.11',
'128.255.34.132',
'128.255.44.51', '128.255.45.53', '128.255.45.54', '129.255.241.42',
'64.202.167.129']
"""
hrefs = set()
for link in BeautifulSoup(urlopen(url)).fetch('a'):
try: hrefs.add(gethostbyname(urlsplit(link["href"])[1]))
except: pass
return sorted(hrefs)
HTH,
George
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