How to get HTTP status message in (py)curl?

Question

spending some time studying pycurl and libcurl documentation, i still can't find a (simple) way, how to get HTTP status message (reason-phrase) in pycurl.

status code is easy:

import pycurl
import cStringIO

curl = pycurl.Curl()
buff = cStringIO.StringIO()
curl.setopt(pycurl.URL, 'http://example.org')
curl.setopt(pycurl.WRITEFUNCTION, buff.write)
curl.perform()

print "status code: %s" % curl.getinfo(pycurl.HTTP_CODE)
# -> 200

# print "status message: %s" % ???
# -> "OK"

score 32 · Accepted Answer · answered Apr 29 '10 at 11:02

i've found a solution myself, which does what i need, but could be more robust (works for HTTP).

it's based on a fact that captured headers obtained by pycurl.HEADERFUNCTION include the status line.

import pycurl
import cStringIO
import re

curl = pycurl.Curl()

buff = cStringIO.StringIO()
hdr = cStringIO.StringIO()

curl.setopt(pycurl.URL, 'http://example.org')
curl.setopt(pycurl.WRITEFUNCTION, buff.write)
curl.setopt(pycurl.HEADERFUNCTION, hdr.write)
curl.perform()

print "status code: %s" % curl.getinfo(pycurl.HTTP_CODE)
# -> 200

status_line = hdr.getvalue().splitlines()[0]
m = re.match(r'HTTP\/\S*\s*\d+\s*(.*?)\s*$', status_line)
if m:
    status_message = m.groups(1)
else:
    status_message = ''

print "status message: %s" % status_message
# -> "OK"

score 8 · Answer 2 · answered Aug 20 '16 at 02:07

8

This is an old thread but I got here looking for similar information. If it is just the status code you're looking for, such as 200, 404, 500 etc. then just do:

your_curl_handle.getinfo(pycurl.RESPONSE_CODE)

which should return a numerical status code :)

answered Aug 20 '16 at 02:07

Kalabaaz

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This is the right way of checking for a status code – ddnomad Jan 22 '20 at 22:07
I am getting `pycurl.error: cannot invoke getinfo() - no curl handle` error. – chanduthedev Nov 24 '20 at 04:16

score 2 · Answer 3 · answered Apr 14 '12 at 08:40

2

I think that you can use human_curl library to create you code simple.

>>> import human_curl as hurl
>>> r = hurl.get('http://example.org')
>>> print r.status_code
200

Full documentation on human_curl you can get on https://github.com/Lispython/human_curl

answered Apr 14 '12 at 08:40

Alexandr

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"reason-phrase" is a term used by the HTTP protocol documention to refer to the human readable string that comes after the status_code. This example just returns the numeric status code. – Mark Lakata Mar 17 '17 at 18:26

score 0 · Answer 4 · answered Apr 04 '22 at 02:49

If you only want the code, you can do, and assuming your pycurl.Curl() instance is called curl (ie. curl = pycurl.Curl()), you can do

curl.getinfo(pycurl.RESPONSE_CODE)

curl.getinfo(pycurl.HTTP_CODE)

But the nice way in my opinion is to parse the header yourself instead of letting libraries spoon-feed you everything.

score -1 · Answer 5 · answered Apr 28 '10 at 09:48

-1

Try BaseHTTPServer.BaseHTTPRequestHandler.responses, it should contain an errorcode dictionnary as explained in this page.

hope this helps.

answered Apr 28 '10 at 09:48

Aif

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that's a possible solution.. i should have pointed out, that i want to have the status message from the server, which may be different, than the standardized one – mykhal Apr 29 '10 at 10:52

How to get HTTP status message in (py)curl?

5 Answers5