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TryHackMe/Hygine.md
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TryHackMe/Hygine.md
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# TryHackMe-Hygiene
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## NMAP
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```bash
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PORT STATE SERVICE REASON VERSION
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22/tcp open ssh syn-ack ttl 63 OpenSSH 7.6p1 Ubuntu 4ubuntu0.5 (Ubuntu Linux; protocol 2.0)
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| ssh-hostkey:
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37652/tcp open ftp syn-ack ttl 63 ProFTPD 1.3.5e
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| ftp-anon: Anonymous FTP login allowed (FTP code 230)
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|_-rw-r--r-- 1 1000 1000 118 Oct 29 02:21 memo.txt
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Service Info: OSs: Linux, Unix; CPE: cpe:/o:lisnux:linux_kernel
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8080/tcp open http-proxy
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| fingerprint-strings:
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| LDAPBindReq:
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| HTTP/1.1 400
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| Content-Type: text/html;charset=utf-8
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| Content-Language: en
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| Content-Length: 2295
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| Date: Thu, 04 Nov 2021 13:02:11 GMT
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| Connection: close
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| <!doctype html><html lang="en"><head><title>HTTP Status 400
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| Request</title><style type="text/css">h1 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:22px;} h2 {font-f
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amily:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:16px;} h3 {font-family:Tahoma,Arial,sans-serif;color:white;background-c
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olor:#525D76;font-size:14px;} body {font-family:Tahoma,Arial,sans-serif;color:black;background-color:white;} b {font-family:Tahoma,Arial,sans-serif;
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color:white;background-color:#525D76;} p {font-family:Tahoma,Arial,sans-serif;background:white;color:black;font-size:12px;} a {color:black;} a.name
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{color:black;} .line {height:1px;background-color:#525D76;border:none;}</style></head><bod
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| LDAPSearchReq:
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| HTTP/1.1 400
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| Content-Type: text/html;charset=utf-8
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```
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## PORT 37652 (FTP)
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From the nmap scan we can see that anonymous ftp is enabled which means that we can login without specifying the password
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<img src="https://i.imgur.com/NfqZ0l5.png"/>
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from the `memo.txt` file we see that a user named `joe` has sent email with the password hash and on cracking the hash we get the password `nightmare`
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<img src="https://i.imgur.com/Pq9b3UX.png"/>
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## PORT 8080 (HTTP)
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On the webserver there's apache tomcat running
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<img src="https://i.imgur.com/g8vZnOe.png"/>
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if we run `stegcracker` on the png image we can find a easter egg
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<img src="https://i.imgur.com/2ZYN2fO.png"/>
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<img src="https://i.imgur.com/7Rj2hkd.png"/>
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<img src="https://i.imgur.com/5iTa1HU.png"/>
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## Foothold
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### Un-inteded
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We were told to find a username on the page but there wasn't any . All we know is that the username is of 5 characters so let's maybe try to brute force the username with 5 characters
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<img src="https://i.imgur.com/I4eZpa3.png"/>
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<img src="https://i.imgur.com/ivawOEm.png"/>
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We can now then get a shell as `sally`
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<img src="https://i.imgur.com/ircsnLS.png"/>
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### Intended
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Running `gobuster` we can find some directories
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<img src="https://i.imgur.com/U16RHo3.png"/>
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I tried using default creds on `/manager` , `/host-manager` but wasn't succesful so I did a recusive fuzz on `admin`
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<img src="https://i.imgur.com/KCXq3zb.png"/>
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This returned us `staging` so again running gobuster on this
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<img src="https://i.imgur.com/IdCmFhl.png"/>
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<img src="https://i.imgur.com/dnoyOYi.png"/>
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We don't see much here but if we look at the source we can find the username `sally`
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<img src="https://i.imgur.com/Z6O1smj.png"/>
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We can now then get a shell through ssh
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<img src="https://i.imgur.com/ircsnLS.png"/>
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On doing `sudo -l` we can't do run any thing as root as other user since this user isn't in sudoers file
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<img src="https://i.imgur.com/3WIYbmF.png"/>
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## Privilege Escalation (Joe)
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We can the find the user flag in `Desktop` folder of sally and can find another flag in `/home/sally/.local/share/Trash/files`
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<img src="https://i.imgur.com/lFTq9GZ.png"/>
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The hash can be cracked with either `hashcat` or `john` but I'll just use cracksation as I did earlier
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<img src="https://i.imgur.com/7gi9kpi.png"/>
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## Privilege Escalation (root)
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Running `sudo -l` we can see that this user can run all commands
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<img src="https://i.imgur.com/BQuXUDP.png"/>
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## References
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- https://askubuntu.com/questions/911204/how-to-extract-only-7-characters-using-grep
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