Use cURL from your Pwnbox (not the target machine) to obtain the source code of the “ https://www.inlanefreight.com ” website and filter all unique paths of that domain. Submit the number of these paths as the answer.

the right command it’s this
curl https://www.inlanefreight.com > htb.txt && cat htb.txt | tr " " “\n” | cut -d"‘" -f2 | cut -d’"’ -f2 | grep “ www.inlanefreight.com ” | sort -u | wc -l 2>/dev/null

first do a curl and redirect the output to htb.txt and(&&) use a cat to htb.txt and the rest is filter

if you don’t know what it’s a path in a domain:
For example, in the url https://cloudflare.com/learning/,cloudflare.com is the domain name, while https is the protocol and /learning/ is the path to a specific page on the website

function guesser(){
document.getElementById(“answer430”).value = x
document.getElementById(“btnAnswer430”).click()

setInterval(guesser,1000)

This shorter version worked for me:

curl https://www.inlanefreight.com | tr " " “\n” | cut -d"‘" -f2 | cut -d’"’ -f2 | grep www.inlanefreight.com | sort -u | wc -l

Thank you for the hint anyway. I was overwhelmed at first, but now it’s totally comprehensible.

If you are used to using regex, then you don’t need to use so many commands - grep can handle all the filtering for you:

curl https://www.inlanefreight.com/ | grep -Po "https://www.inlanefreight.com/[^'\"]*" | sort -u | wc -l

Could you clarify what you did on that grep filter? I am new to this, I would like to understand what [^'"]* this does.
Thanks!

Sure - but bear in mind that it’s not been covered in the material, I was just saying if you already know regex then you can use that instead.

[^'"]* means any character ([ ] lists characters) that isn’t (^ = isn’t) a ’ or a " . The * at the end means match if there are 0 or more of them.

So basically what that expression is doing is searching for https://www.inlanefreight.com/ including all characters after until it reaches a ’ or a " (which would be the closing of the anchor link)

We need to check for both ’ and ", as some of the links are written:
<a href='https://www.inlanefreight.com/'> and others are written <a href="https://www.inlanefreight.com/">

It should be more like grep -Po "https?://www.inlanefreight.com[^\"'?#]*" shouldn’t it ?
The url path ends at the first ? (query parameters) and # (fragment). Thus the correct answer should be one less.
( https://www.inlanefreight.com/index.php/wp-json/oembed/1.0/embed is present in the result with different query parameters)