如何创建HTTP GET请求Scapy？

Z时代
2024-01-10
分类：问答

我需要创建HTTP GET请求并保存数据响应。我试图用这个：

    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()

但是，当我打印时reply，没有看到任何数据响应。另外，当我签入“ Wireshark”时，我收到了SYN，SYN / ACK，但没有收到ACK。

图片：

编辑：

我现在尝试这样做：

# 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!'