How to make correct TCP request using python

Question

I am trying to make request but google.com returns status 400, but It should be 302. What's wrong with my request? Do i need additional request header? Any ideas?

Current code:

import socket
host = "www.google.com"
port = 80

client = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
client.connect((host,port))
client.send("GET / HTTP1.1\r\nHost: www.google.com\r\n\r\n")

response = client.recv(4096)
print response

Response:

HTTP/1.0 400 Bad Request
Content-Type: text/html; charset=UTF-8
Content-Length: 1504
Date: Mon, 07 Sep 2015 16:25:02 GMT
Server: GFE/2.0

<!DOCTYPE html>
<html lang=en>
  <meta charset=utf-8>
  <meta name=viewport content="initial-scale=1, minimum-scale=1, width=device-width">
  <title>Error 400 (Bad Request)!!1</title>
  <style>
    *{margin:0;padding:0}html,code{font:15px/22px arial,sans-serif}html{background:#fff;color:#222;padding:15px}body{margin:7% auto 0;max-width:390px;min-height:180px;padding:30px 0 15px}* > body{background:url(//www.google.com/images/errors/robot.png) 100% 5px no-repeat;padding-right:205px}p{margin:11px 0 22px;overflow:hidden}ins{color:#777;text-decoration:none}a img{border:0}@media screen and (max-width:772px){body{background:none;margin-top:0;max-width:none;padding-right:0}}#logo{background:url(//www.google.com/images/logos/errorpage/error_logo-150x54.png) no-repeat;margin-left:-5px}@media only screen and (min-resolution:192dpi){#logo{background:url(//www.google.com/images/logos/errorpage/error_logo-150x54-2x.png) no-repeat 0% 0%/100% 100%;-moz-border-image:url(//www.google.com/images/logos/errorpage/error_logo-150x54-2x.png) 0}}@media only screen and (-webkit-min-device-pixel-ratio:2){#logo{background:url(//www.google.com/images/logos/errorpage/error_logo-150x54-2x.png) no-repeat;-webkit-background-size:100% 100%}}#logo{display:inline-block;height:54px;width:150px}
  </style>
  <a href=//w
ww.google.com/><span id=logo aria-label=Google></span></a>
  <p><b>400.</b> <ins>That’s an error.</ins>
  <p>Your client has issued a malformed or illegal request.  <ins>That’s all we know.</ins>

If you want to use http, do yourself a favour and use requests. That will save you a lot of headaches. — Roland Smith
– Roland Smith, Commented Sep 7, 2015 at 20:36
@RolandSmith Thanks, you'e right, I use requests and also pycurl in my backend applications but i want to learn socket programming, do penetration tests and make some TCP/UDP based tasks. HTTP requests library is not enough for me at the moment :) — Cyprian Gepfert
– Cyprian Gepfert, Commented Sep 7, 2015 at 20:50

primero · Accepted Answer · 2015-09-07 20:25:46Z

2

In the string you use in the send function, you have missed a slash, when specifying the HTTP protocol version.

client.send("GET / HTTP1.1\r\nHost: www.google.com\r\n\r\n")

should be:

client.send('GET / HTTP/1.1\r\nHost: www.google.com\r\n\r\n')

answered Sep 7, 2015 at 20:25

primero

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2 Comments

Cyprian Gepfert Over a year ago

Solved, thanks man. I also noticed that to solve these kind of problems I can use wireshark and compare content with packets generated by higher level libraries or software, like Advanced Rest Client etc.

primero Over a year ago

No problem. I wanted to test it with ngrep on linux, but fortunately I saw the typo. It's a good plan to check using packet traces because you can see a lot more info.

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