Fwd: Hello

[spandana g]

Hello ,

Traceback (most recent call last):
File "C:\Python25\hadi_yahoo.py", line 12, in <module>
file_source.write(urllib2.urlopen(req).read())
File "C:\Python25\lib\urllib2.py", line 124, in urlopen
return _opener.open(url, data)
File "C:\Python25\lib\urllib2.py", line 387, in open
response = meth(req, response)
File "C:\Python25\lib\urllib2.py", line 498, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python25\lib\urllib2.py", line 425, in error
return self._call_chain(*args)
File "C:\Python25\lib\urllib2.py", line 360, in _call_chain
result = func(*args)
File "C:\Python25\lib\urllib2.py", line 506, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
HTTPError: HTTP Error 999: Unable to process request at this time -- error
999

Previously i got the error which I have attached below when I use just
urlopen . But now when I use this http request
user_agent='Mozilla/3.0(compatible;MISE 5.5;Windows NT)'
headers={'User-Agent':user_agent}
req=urllib2.Request(url,None,headers)
file_source.write(urllib2.urlopen(req).read()

its still giving the error mentioned above .. Iam accessing the yahoo search
engine .. link is "http://search.yahoo.com/search?n=20&p=ipod"

I'm attaching the python file i have written just have a look and suggest me
something that works for this query

Thank You,
Spandana.


---------- Forwarded message ----------
From: spandana g <sp***********@gmail.com>
Date: Thu, Jul 3, 2008 at 2:52 PM
Subject: HTTP request error with urlopen
To: py*********@python.org
Hello ,

I have written a code to get the page source of the google search
page .. this is working for other urls. I have this problem with

import re
from urllib2 import urlopen
string='http://www.google.com/search?num=20&hl=en&q=ipod&btnG=Search'
file_source=file("google_source.txt",'w')
file_source.write(urlopen(string).read())
page_content=file_source.readlines()

Traceback (most recent call last) :
File "C:/Python25/google.py", line 5,in <module>
file_source.write(urlopen(string).read())
File "C:\Python25\lib\urllib2.py", line 124 , in urlopen
return__opener.open(url, data)
File "C:\Python25\lib\urllib2.py", line 387 , in open
response =meth(req, response)
File "C:\Python25\lib\urllib2.py", line 498 , in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python25\lib\urllib2.py", line 425, in error
return self._call_chain(*args)
File "C:\Python25\lib\urllib2.py", line 360, in __call_chain
result = func(*args)
File "C:\Python25\lib\urllib2.py", line 506, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
HTTPError: HTTP Error 403: Forbidden

Actually urlopen is working for google labs sets page but not for the
google.com and even I have same problem with wikipedia . Please let me know
... If any one of have any idea about this .

Thank You,
Spandana.