I am using the below code to scrape over XFN content from web page http://ajaxian.com but I am gatting the below error:
Traceback (most recent call last): File "C:\Users\Somnath\workspace\us.chakra.social.web.microformat\src\microformats_xfn_scrape.py", line 40, in <module>
page = urllib2.urlopen(URL)
File "C:\Python27\lib\urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "C:\Python27\lib\urllib2.py", line 394, in open
response = self._open(req, data)
File "C:\Python27\lib\urllib2.py", line 417, in _open
'unknown_open', req)
File "C:\Python27\lib\urllib2.py", line 372, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 1232, in unknown_open
raise URLError('unknown url type: %s' % type)
urllib2.URLError: <urlopen error unknown url type: c>
My code is as follows:
'''
Created on Jan 11, 2013
@author: Somnath
'''
# Scraping XFN content from a web page
# -*-coding: utf-8 -*-
import sys
import urllib2
import HTMLParser
from BeautifulSoup import BeautifulSoup
# Try http://ajaxian.com
URL = sys.argv[0]
XFN_TAGS = set([
'colleague',
'sweetheart',
'parent',
'co-resident',
'co-worker',
'muse',
'neighbor',
'sibling',
'kin',
'child',
'date',
'spouse',
'me',
'acquaintance',
'met',
'crush',
'contact',
'friend',
])
#try:
page = urllib2.urlopen(URL)
#except urllib2.URLError:
# print 'Failed to fetch ' + item
#try:
soup = BeautifulSoup(page)
#except HTMLParser.HTMLParseError:
# print 'Failed to parse ' + item
anchorTags = soup.findAll('a')
for a in anchorTags:
if a.has_key('rel'):
if len(set(a['rel'].split()) & XFN_TAGS) > 0:
tags = a['rel'].split()
print a.contents[0], a['href'], tags
I am running PyDev under Eclipse and is using Run As --> Python Run and set the Runtime Configuration with argument "http://ajaxian.com/". Can anybody suggest where I am getting wrong?
One more thing: I have commented the two try blocks in my code because it was giving an error undefined variable : item. If I want to re-include the try-except blocks, should I give a blank definition of variable, item outside the try blocks? How can I get rid of that problem?
As you suggested sys.argv[0]
prints the path of your script, that is because you call your script like
python microformats_xfn_scrape.py <some_argument>
and here index 0 of sys.argv is the name of the script and not the argument.
What you need to do is call your script with the <url>
parameter, like :
python microformats_xfn_scrape.py http://www.ajaxian.com/
and in your script change sys.argv[0]
to sys.argv[1]
as url arguments index is 1.