WarGame/webhaking.kr
webhacking.kr - old35
m0nd2y
2021. 4. 6. 13:16
처음에 접속하면 위와 같이 입력창이 나옵니다.
sqli문제이기에 소스코드를 확인해보면
<?php
include "../../config.php";
if($_GET['view_source']) view_source();
?><html>
<head>
<title>Challenge 35</title>
<head>
<body>
<form method=get action=index.php>
phone : <input name=phone size=11 style=width:200px>
<input name=id type=hidden value=guest>
<input type=submit value='add'>
</form>
<?php
$db = dbconnect();
if($_GET['phone'] && $_GET['id']){
if(preg_match("/\*|\/|=|select|-|#|;/i",$_GET['phone'])) exit("no hack");
if(strlen($_GET['id']) > 5) exit("no hack");
if(preg_match("/admin/i",$_GET['id'])) exit("you are not admin");
mysqli_query($db,"insert into chall35(id,ip,phone) values('{$_GET['id']}','{$_SERVER['REMOTE_ADDR']}',{$_GET['phone']})") or die("query error");
echo "Done<br>";
}
$isAdmin = mysqli_fetch_array(mysqli_query($db,"select ip from chall35 where id='admin' and ip='{$_SERVER['REMOTE_ADDR']}'"));
if($isAdmin['ip'] == $_SERVER['REMOTE_ADDR']){
solve(35);
mysqli_query($db,"delete from chall35");
}
$phone_list = mysqli_query($db,"select * from chall35 where ip='{$_SERVER['REMOTE_ADDR']}'");
echo "<!--\n";
while($r = mysqli_fetch_array($phone_list)){
echo htmlentities($r['id'])." - ".$r['phone']."\n";
}
echo "-->\n";
?>
<br><a href=?view_source=1>view-source</a>
</body>
</html>
위에 쿼리에서
> mysqli_query($db,"insert into chall35(id,ip,phone) values('{$_GET['id']}','{$_SERVER['REMOTE_ADDR']}',{$_GET['phone']})") or die("query error");
위 부분에서 admin을 id에 넣어주고, ip주소를 맞춰주면 되니까, 쿼리를 우회해서
> 12), ('admin','현재 내 아이피 주소','1234'
로 넣으면 문제를 풀 수 있습니다.
