Reputation: 16840
How do I read the response headers returned from a PyCurl request?
Upvotes: 25
Views: 18870
Reputation: 679
import pycurl
from StringIO import StringIO
headers = StringIO()
c = pycurl.Curl()
c.setopt(c.URL, url)
c.setopt(c.HEADER, 1)
c.setopt(c.NOBODY, 1) # header only, no body
c.setopt(c.HEADERFUNCTION, headers.write)
c.perform()
print headers.getvalue()
Add any other curl setopts as necessary/desired, such as FOLLOWLOCATION.
Upvotes: 17
Reputation: 391
Anothr alternate, human_curl usage: pip human_curl
In [1]: import human_curl as hurl
In [2]: r = hurl.get("http://stackoverflow.com")
In [3]: r.headers
Out[3]:
{'cache-control': 'public, max-age=45',
'content-length': '198515',
'content-type': 'text/html; charset=utf-8',
'date': 'Thu, 01 Sep 2011 11:53:43 GMT',
'expires': 'Thu, 01 Sep 2011 11:54:28 GMT',
'last-modified': 'Thu, 01 Sep 2011 11:53:28 GMT',
'vary': '*'}
Upvotes: 6
Reputation: 17004
This might or might not be an alternative for you:
import urllib
headers = urllib.urlopen('http://www.pythonchallenge.com').headers.headers
Upvotes: 1
Reputation: 35459
There are several solutions (by default, they are dropped). Here is an example using the option HEADERFUNCTION which lets you indicate a function to handle them.
Other solutions are options WRITEHEADER (not compatible with WRITEFUNCTION) or setting HEADER to True so that they are transmitted with the body.
#!/usr/bin/python
import pycurl
import sys
class Storage:
def __init__(self):
self.contents = ''
self.line = 0
def store(self, buf):
self.line = self.line + 1
self.contents = "%s%i: %s" % (self.contents, self.line, buf)
def __str__(self):
return self.contents
retrieved_body = Storage()
retrieved_headers = Storage()
c = pycurl.Curl()
c.setopt(c.URL, 'http://www.demaziere.fr/eve/')
c.setopt(c.WRITEFUNCTION, retrieved_body.store)
c.setopt(c.HEADERFUNCTION, retrieved_headers.store)
c.perform()
c.close()
print retrieved_headers
print retrieved_body
Upvotes: 30