Wieder das urllib.Fehler.HTTPError: HTTP Error 400: Bad Request

Hy!
Ich versuchte, Sie zu öffnen web-Seite, die normalerweise das öffnen im browser, aber python nur schwört und nicht arbeiten wollen.

import urllib.request, urllib.error
f = urllib.request.urlopen('http://www.booking.com/reviewlist.html?cc1=tr;pagename=sapphire')

Und einen anderen Weg

import urllib.request, urllib.error
opener=urllib.request.build_opener()
f=opener.open('http://www.booking.com/reviewlist.html?cc1=tr;pagename=sapphi
re')

Beide Möglichkeiten geben, eine Art des Fehlers:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python34\lib\urllib\request.py", line 461, in open
    response = meth(req, response)
  File "C:\Python34\lib\urllib\request.py", line 571, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python34\lib\urllib\request.py", line 493, in error
    result = self._call_chain(*args)
  File "C:\Python34\lib\urllib\request.py", line 433, in _call_chain
    result = func(*args)
  File "C:\Python34\lib\urllib\request.py", line 676, in http_error_302
    return self.parent.open(new, timeout=req.timeout)
  File "C:\Python34\lib\urllib\request.py", line 461, in open
    response = meth(req, response)
  File "C:\Python34\lib\urllib\request.py", line 571, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python34\lib\urllib\request.py", line 499, in error
    return self._call_chain(*args)
  File "C:\Python34\lib\urllib\request.py", line 433, in _call_chain
    result = func(*args)
  File "C:\Python34\lib\urllib\request.py", line 579, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 400: Bad Request

Irgendwelche Ideen?

InformationsquelleAutor Wanu | 2014-11-12