python: httplib error: can not send headers

Question

conn = httplib.HTTPConnection('thesite')
conn.request("GET","myurl")
conn.putheader('Connection','Keep-Alive')
#conn.putheader('User-Agent','Mozilla/5.0(Windows; u; windows NT 6.1;en-US) AppleWebKit/533.4 (KHTML, like Gecko) Chrome//5.0.375.126 Safari//5.33.4')
#conn.putheader('Accept-Encoding','gzip,deflate,sdch')
#conn.putheader('Accept-Language','en-US,en;q=0.8')
#conn.putheader('Accept-Charset','ISO-8859-1,utf-8;1=0.7,*;q=0.3')
conn.endheaders()
r1= conn.getresponse()

It raises an error:

  conn.putheader('Connection','Keep-Alive')
  File "D:\Program Files\python\lib\httplib.py", line 891, in putheader
    raise CannotSendHeader()

If I comment out putheader and endheaders, it runs fine. But I need it to be keep alive.

Does anyone know what I did wrong?

sje397 · Accepted Answer · 2010-08-13 03:18:49Z

9

Use putrequest instead of request. Since request also can send headers, it will send a blank line to the server to indicate end of headers, so sending headers afterward will create an error.

Alternatively, you could do as is done here:

import httplib, urllib
params = urllib.urlencode({'spam': 1, 'eggs': 2, 'bacon': 0})
headers = {"Content-type": "application/x-www-form-urlencoded", "Accept": "text/plain"}
conn = httplib.HTTPConnection("musi-cal.mojam.com:80")
conn.request("POST", "/cgi-bin/query", params, headers)
response = conn.getresponse()

edited Aug 13, 2010 at 3:18

answered Aug 13, 2010 at 3:11

sje397

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1 Comment

SingleNegationElimination Over a year ago

what's wrong is you used request instead of putrequest, which completes the entire request. You need to add some headers before you let the request go through, and putrequest does just that.

Pablo Flores Aravena · Accepted Answer · 2016-07-19 16:39:12Z

-1

headers = {"Content-Type":"application/x-www-form-urlencoded", "Connection":"Keep-Alive", "Referer":"http://www.tu.sitio.cl/", "User-Agent":"Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/51.0.2704.103 Safari/537.36"};

code+

conn.request(method="POST", url="/formulario/", body=params, headers=headers)

answered Jul 19, 2016 at 16:39

Pablo Flores Aravena

1

1 Comment

cdomination Over a year ago

Hi, welcome to Stack Overflow! Can you please annotate your answer better?

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