6.6 KiB
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ECB
(ECB) Electronic Code Book - symmetric encryption scheme which replaces each block of the clear text by the block of ciphertext. It is the simplest encryption scheme. The main idea is to split the clear text into blocks of N bits (depends on the size of the block of input data, encryption algorithm) and then to encrypt (decrypt) each block of clear text using the only key.
Using ECB has multiple security implications:
- Blocks from encrypted message can be removed
- Blocks from encrypted message can be moved around
Detection of the vulnerability
Imagine you login into an application several times and you always get the same cookie. This is because the cookie of the application is <username>|<password>
.
Then, you generate to new users, both of them with the same long password and almost the same username.
You find out that the blocks of 8B where the info of both users is the same are equals. Then, you imagine that this might be because ECB is being used.
Like in the following example. Observe how these** 2 decoded cookies** has several times the block \x23U\xE45K\xCB\x21\xC8
\x23U\xE45K\xCB\x21\xC8\x23U\xE45K\xCB\x21\xC8\x04\xB6\xE1H\xD1\x1E \xB6\x23U\xE45K\xCB\x21\xC8\x23U\xE45K\xCB\x21\xC8+=\xD4F\xF7\x99\xD9\xA9
\x23U\xE45K\xCB\x21\xC8\x23U\xE45K\xCB\x21\xC8\x04\xB6\xE1H\xD1\x1E \xB6\x23U\xE45K\xCB\x21\xC8\x23U\xE45K\xCB\x21\xC8+=\xD4F\xF7\x99\xD9\xA9
This is because the username and password of those cookies contained several times the letter "a" (for example). The blocks that are different are blocks that contained at least 1 different character (maybe the delimiter "|" or some necessary difference in the username).
Now, the attacker just need to discover if the format is <username><delimiter><password>
or <password><delimiter><username>
. For doing that, he can just generate several usernames with similar and long usernames and passwords until he find the format and the length of the delimiter:
Username length: | Password length: | Username+Password length: | Cookie's length (after decoding): |
---|---|---|---|
2 | 2 | 4 | 8 |
3 | 3 | 6 | 8 |
3 | 4 | 7 | 8 |
4 | 4 | 8 | 16 |
7 | 7 | 14 | 16 |
Exploitation of the vulnerability
Removing entire blocks
Knowing the format of the cookie (<username>|<password>
), in order to impersonate the username admin
create a new user called aaaaaaaaadmin
and get the cookie and decode it:
\x23U\xE45K\xCB\x21\xC8\xE0Vd8oE\x123\aO\x43T\x32\xD5U\xD4
We can see the pattern \x23U\xE45K\xCB\x21\xC8
created previously with the username that contained only a
.
Then, you can remove the first block of 8B and you will et a valid cookie for the username admin
:
\xE0Vd8oE\x123\aO\x43T\x32\xD5U\xD4
Moving blocks
In many databases it is the same to search for WHERE username='admin';
or for WHERE username='admin ';
(Note the extra spaces)
So, another way to impersonate the user admin
would be to:
- Generate a username that:
len(<username>) + len(<delimiter) % len(block)
. With a block size of8B
you can generate username called:username
, with the delimiter|
the chunk<username><delimiter>
will generate 2 blocks of 8Bs. - Then, generate a password that will fill an exact number of blocks containing the username we want to impersonate and spaces, like:
admin
The cookie of this user is going to be composed by 3 blocks: the first 2 is the blocks of the username + delimiter and the third one of the password (which is faking the username): username |admin
Then, just replace the first block with the last time and will be impersonating the user admin
: admin |username
References
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Learn & practice AWS Hacking:HackTricks Training AWS Red Team Expert (ARTE)
Learn & practice GCP Hacking: HackTricks Training GCP Red Team Expert (GRTE)
Support HackTricks
- Check the subscription plans!
- Join the 💬 Discord group or the telegram group or follow us on Twitter 🐦 @hacktricks_live.
- Share hacking tricks by submitting PRs to the HackTricks and HackTricks Cloud github repos.