Regex to get a specific part of an URL

Question

Say I've got an URL:

http://www.example.com/foo/bar/yes_no_no.html

I'm looking for a regex that can extract all the characters of the filename up to the first underscore. So in the above URL I want to extract "yes".

So far I managed to get the whole "yes_no_no" bit using:

/([^\/]+)(?=\.\w+$)/, but I can't seem to get it to only match "yes".

Michał Perłakowski · Accepted Answer · 2016-08-20 20:37:34Z

3

Try this:

/\/([^_\/]+)_?[^\/]*\.[a-z]+$/

Example:

console.log('http://www.example.com/foo/bar/yes_no_no.html'.match(/\/([^_\/]+)_?[^\/]*\.[a-z]+$/)[1])
console.log('http://www.example.com/foo/bar/yes.html'.match(/\/([^_\/]+)_?[^\/]*\.[a-z]+$/)[1])

edited Aug 20, 2016 at 20:37

answered Aug 20, 2016 at 11:57

Michał Perłakowski

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