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https://github.com/carlospolop/hacktricks
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106 lines
6.6 KiB
Markdown
106 lines
6.6 KiB
Markdown
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{% hint style="success" %}
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Learn & practice AWS Hacking:<img src="/.gitbook/assets/arte.png" alt="" data-size="line">[**HackTricks Training AWS Red Team Expert (ARTE)**](https://training.hacktricks.xyz/courses/arte)<img src="/.gitbook/assets/arte.png" alt="" data-size="line">\
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Learn & practice GCP Hacking: <img src="/.gitbook/assets/grte.png" alt="" data-size="line">[**HackTricks Training GCP Red Team Expert (GRTE)**<img src="/.gitbook/assets/grte.png" alt="" data-size="line">](https://training.hacktricks.xyz/courses/grte)
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<details>
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<summary>Support HackTricks</summary>
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* Check the [**subscription plans**](https://github.com/sponsors/carlospolop)!
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* **Join the** 💬 [**Discord group**](https://discord.gg/hRep4RUj7f) or the [**telegram group**](https://t.me/peass) or **follow** us on **Twitter** 🐦 [**@hacktricks\_live**](https://twitter.com/hacktricks\_live)**.**
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* **Share hacking tricks by submitting PRs to the** [**HackTricks**](https://github.com/carlospolop/hacktricks) and [**HackTricks Cloud**](https://github.com/carlospolop/hacktricks-cloud) github repos.
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</details>
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{% endhint %}
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# ECB
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(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.
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Using ECB has multiple security implications:
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* **Blocks from encrypted message can be removed**
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* **Blocks from encrypted message can be moved around**
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# Detection of the vulnerability
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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>`**.\
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Then, you generate to new users, both of them with the **same long password** and **almost** the **same** **username**.\
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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**.
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Like in the following example. Observe how these** 2 decoded cookies** has several times the block **`\x23U\xE45K\xCB\x21\xC8`**
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```
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\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
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\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
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```
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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).
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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 s**imilar and long usernames and passwords until he find the format and the length of the delimiter:**
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| Username length: | Password length: | Username+Password length: | Cookie's length (after decoding): |
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| ---------------- | ---------------- | ------------------------- | --------------------------------- |
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| 2 | 2 | 4 | 8 |
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| 3 | 3 | 6 | 8 |
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| 3 | 4 | 7 | 8 |
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| 4 | 4 | 8 | 16 |
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| 7 | 7 | 14 | 16 |
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# Exploitation of the vulnerability
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## Removing entire blocks
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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:
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```
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\x23U\xE45K\xCB\x21\xC8\xE0Vd8oE\x123\aO\x43T\x32\xD5U\xD4
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```
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We can see the pattern `\x23U\xE45K\xCB\x21\xC8` created previously with the username that contained only `a`.\
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Then, you can remove the first block of 8B and you will et a valid cookie for the username `admin`:
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```
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\xE0Vd8oE\x123\aO\x43T\x32\xD5U\xD4
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```
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## Moving blocks
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In many databases it is the same to search for `WHERE username='admin';` or for `WHERE username='admin ';` _(Note the extra spaces)_
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So, another way to impersonate the user `admin` would be to:
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* Generate a username that: `len(<username>) + len(<delimiter) % len(block)`. With a block size of `8B` you can generate username called: `username `, with the delimiter `|` the chunk `<username><delimiter>` will generate 2 blocks of 8Bs.
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* Then, generate a password that will fill an exact number of blocks containing the username we want to impersonate and spaces, like: `admin `
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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 `
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**Then, just replace the first block with the last time and will be impersonating the user `admin`: `admin |username`**
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## References
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* [http://cryptowiki.net/index.php?title=Electronic_Code_Book\_(ECB)](http://cryptowiki.net/index.php?title=Electronic_Code_Book_\(ECB\))
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{% hint style="success" %}
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Learn & practice AWS Hacking:<img src="/.gitbook/assets/arte.png" alt="" data-size="line">[**HackTricks Training AWS Red Team Expert (ARTE)**](https://training.hacktricks.xyz/courses/arte)<img src="/.gitbook/assets/arte.png" alt="" data-size="line">\
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Learn & practice GCP Hacking: <img src="/.gitbook/assets/grte.png" alt="" data-size="line">[**HackTricks Training GCP Red Team Expert (GRTE)**<img src="/.gitbook/assets/grte.png" alt="" data-size="line">](https://training.hacktricks.xyz/courses/grte)
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<details>
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<summary>Support HackTricks</summary>
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* Check the [**subscription plans**](https://github.com/sponsors/carlospolop)!
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* **Join the** 💬 [**Discord group**](https://discord.gg/hRep4RUj7f) or the [**telegram group**](https://t.me/peass) or **follow** us on **Twitter** 🐦 [**@hacktricks\_live**](https://twitter.com/hacktricks\_live)**.**
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* **Share hacking tricks by submitting PRs to the** [**HackTricks**](https://github.com/carlospolop/hacktricks) and [**HackTricks Cloud**](https://github.com/carlospolop/hacktricks-cloud) github repos.
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</details>
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{% endhint %}
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