How to deal with spaces in regex in Web scraper

How to?
#1

Hi.

I'm using regex for years. And I'm unable to deal with spaces when using from Web Scraper.

I have a Text:

Route MAD / JFK details kms etc.

I only want to extract MAD / JFK

I tried:

\s([\w]{3}\s/\s[\w]{3})

That works nice in de regex simulators across the web and in Pursehub for example

I also tried:

\s[A-Z]{3}\s/\s[A-Z]{3} and not extract nothing

If I use:

\s[A-Z]{3}\s/\s I extract ' MAD / '

But is impossible to add more characters. If I add \w or [A-Z] or . to capture the next character, the result is null

Ideas???

#2

why not just
[A-Z]{3} / [A-Z]{3}

You don't have to escape the spaces.