At Thursday 9/11/2006 20:23, i80and wrote:
I'm working on a basic web spider, and I'm having problems with the
urlparser.
[...]
SpliceStart = Website.find('<a href="', (i+1))
SpliceEnd = (Website.find('">', SpliceStart))
ParsedURL =
urlparse((Website[SpliceStart+9:(SpliceEnd+1)]))
robotparser.set_url(ParsedURL.hostname + '/' +
'robots.txt')
-----
Traceback (most recent call last):
File "C:/Documents and Settings/Andrew/Desktop/ScoutCode-0.09.py",
line 120, in <module>
FindLinks(Website)
File "C:/Documents and Settings/Andrew/Desktop/ScoutCode-0.09.py",
line 84, in FindLinks
robotparser.read()
File "C:\Program Files\Python25\lib\robotparser.py", line 61, in read
f = opener.open(self.url)
File "C:\Program Files\Python25\lib\urllib.py", line 190, in open
return getattr(self, name)(url)
File "C:\Program Files\Python25\lib\urllib.py", line 451, in
open_file
return self.open_local_file(url)
File "C:\Program Files\Python25\lib\urllib.py", line 465, in
open_local_file
raise IOError(e.errno, e.strerror, e.filename)
IOError: [Errno 2] The system cannot find the path specified:
'en.wikipedia.org\\robots.txt'
Note the last line 'en.wikipedia.org\\robots.txt'. I want
'en.wikipedia.org/robots.txt'! What am I doing wrong?
No, you don't want 'en.wikipedia.org/robots.txt'; you want
'http://en.wikipedia.org/robots.txt'
urllib treats the former as a file: request, here the \\ in the
normalized path.
You are parsing the link and then building a new URI using ONLY the
hostname part; that's wrong. Use urljoin(ParsedURL, '/robots.txt') instead.
You may try Beautiful Soup for a better HTML parsing.
--
Gabriel Genellina
Softlab SRL
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