Having problems with urlparser concatenation

[Gabriel Genellina]

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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