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Editorial Team
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Asked: 2026-05-12T22:20:48+00:00

>>> import httplib >>> conn = httplib.HTTPConnection(www.google.com) >>> conn.request(HEAD, /index.html) >>> res = conn.getresponse()

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>>> import httplib
>>> conn = httplib.HTTPConnection("www.google.com")
>>> conn.request("HEAD", "/index.html")
>>> res = conn.getresponse()
>>> print res.status, res.reason
200 OK

This code will get the HTTP status code. However, notice that I split up “google.com” and “/index.html” on 2 lines.

And it’s confusing.

What if I want to find the status code of just a general URL???

http://mydomain.com/sunny/boo.avi

http://anotherdomain.com/rss/fee.xml

Can’t I just stick the URL into it, and make it work?

Edit…I cannot use urllib, because I don’t want to downlaod the file

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1 Answer

  1. Editorial Team

    Alternatively, if you expect that actually downloading the data is problematic and you really need the HEAD method, you could parse the URL using urlparse:

    >>> import httplib
>>> import urlparse
>>> url = "http://www.google.com/index.html"
>>> (scheme, netloc, path, params, query, fragment) = urlparse.urlparse(url)
>>> conn = httplib.HTTPConnection(netloc)
>>> conn.request("HEAD", urlparse.urlunparse(('', '', path, params, query, fragment))
>>> res = conn.getresponse()
>>> print res.status, res.reason
302 Found

    And wrap this into a function taking the URL as an argument.

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