How to create HTTP GET request Scapy?
Yair B.
I need to create HTTP GET request and save the data response. I tried to use this:
syn = IP(dst=URL) / TCP(dport=80, flags='S')
syn_ack = sr1(syn)
getStr = 'GET / HTTP/1.1\r\nHost: www.google.com\r\n\r\n'
request = IP(dst='www.google.com') / TCP(dport=80, sport=syn_ack[TCP].dport,
seq=syn_ack[TCP].ack, ack=syn_ack[TCP].seq + 1, flags='A') / getStr
reply = sr1(request)
print reply.show()
But when I print reply I don't see any data response. In addition, when I checked in 'Wireshark' I got SYN, SYN/ACK but I didn't get an ACK.
Image: 
Edit:
I try to do that now:
# Import scapy
from scapy.all import *
# Print info header
print "[*] ACK-GET example -- Thijs 'Thice' Bosschert, 06-06-2011"
# Prepare GET statement
get='GET / HTTP/1.0\n\n'
# Set up target IP
ip=IP(dst="www.google.com")
# Generate random source port number
port=RandNum(1024,65535)
# Create SYN packet
SYN=ip/TCP(sport=port, dport=80, flags="S", seq=42)
# Send SYN and receive SYN,ACK
print "\n[*] Sending SYN packet"
SYNACK=sr1(SYN)
# Create ACK with GET request
ACK=ip/TCP(sport=SYNACK.dport, dport=80, flags="A", seq=SYNACK.ack, ack=SYNACK.seq + 1) / get
# SEND our ACK-GET request
print "\n[*] Sending ACK-GET packet"
reply,error=sr(ACK)
# print reply from server
print "\n[*] Reply from server:"
print reply.show()
print '\n[*] Done!'
but its print me in reply from server;
0000 IP / TCP 192.168.44.130:23181 > 216.58.208.164:http A / Raw ==> IP / TCP 216.58.208.164:http > 192.168.44.130:23181 A / Padding None
And I need Line-based text data: text/html.
wookie919
You are sending a SYN and correctly receiving a SYN_ACK. At this point, you should generate and send an ACK based on the SYN_ACK that you've received, and THEN finally transmit the HTTP GET request. It seems that you are somewhat confused about the TCP 3-way handshake mechanism. In short, you are not supposed to 'get' an ACK, you are supposed to generate and send this yourself.
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