CODEWITHSUNDEEP
pythonurllib2
Jijoy
Jijoy

Reputation: 12714

urllib2 is throwing an error for an url , while it's opening properly in browser

I am trying to open an url through python like this

  import urllib2
  f = urllib2.urlopen('http://www.futurebazaar.com/Search/laptop')

It's throwing following error

File "C:\Python26\lib\urllib2.py", line 1134, in do_open r = h.getresponse() File "C:\Python26\lib\httplib.py", line 986, in getresponse response.begin() File "C:\Python26\lib\httplib.py", line 391, in begin version, status, reason = self._read_status() File "C:\Python26\lib\httplib.py", line 355, in _read_status raise BadStatusLine(line) httplib.BadStatusLine

But this url is opening via browser.

Upvotes: 0

Views: 1818

Answers (1)

Glenn Maynard
Glenn Maynard

Reputation: 57464

The website is broken. If the optional "Accept" header isn't supplied, the site closes the connection without responding; this is invalid behavior.

Workaround:

import urllib2
req = urllib2.Request('http://www.futurebazaar.com/Search/laptop')
req.add_header('Accept', '*/*')
f = urllib2.urlopen(req)

Upvotes: 5

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