CODEWITHSUNDEEP
python-2.7web-scrapingweb-crawlerurllib2python-requests
arun jadav
arun jadav

Reputation: 5

How can I remove this error?

What does this error mean? How can I get rid of this problem?

Traceback (most recent call last):
  File "bestcodeever.py", line 16, in <module>
    html=urllib2.urlopen('http://' + fulllink).read()
  File "/usr/lib/python2.7/urllib2.py", line 127, in urlopen
    return _opener.open(url, data, timeout)
  File "/usr/lib/python2.7/urllib2.py", line 410, in open
    response = meth(req, response)
  File "/usr/lib/python2.7/urllib2.py", line 523, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib/python2.7/urllib2.py", line 448, in error
    return self._call_chain(*args)
  File "/usr/lib/python2.7/urllib2.py", line 382, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 531, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 503: Service Unavailable

Upvotes: 0

Views: 607

Answers (1)

xnx
xnx

Reputation: 25548

The HTTP 503 status code means that the web server you've tried to contact is unavailable (perhaps for maintenance reasons). It's still able to respond minimally with this code, so it isn't offline and it is listening to requests, it's just not returning the service you want for now: just try again later.

It is also possible that your IP address has been blacklisted by the server if you've been a bit too demanding of it with your Python scripts lately...

Upvotes: 1

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