regex word boundary excluding the hyphen

Question

i need a regex that matches an expression ending with a word boundary, but which does not consider the hyphen as a boundary. i.e. get all expressions matched by

Answer 1

You can use a lookahead for this, the shortest would be to use a negative lookahead:

type ([a-z])(?![\w-])

(?![\w-]) would mean "fail the match if the next character is in \w or is a -".

Here is an option that uses a normal lookahead:

type ([a-z])(?=[^\w-]|$)

You can read (?=[^\w-]|$) as "only match if the next character is not in the character class [\w-], or this is the end of the string".

See it working: http://www.rubular.com/r/NHYhv72znm

Answer 2

I had a pretty similar problem except I didn't want to consider the '*' as a boundary character. Here's what I did:

\b(?<!\*)([^\s\*]+)\b(?!*)

Basically, if you're at a word boundary, look back one character and don't match if the previous character was an '*'. If you're in the middle, don't match on a space or asterisk. If you're at the end, make sure the end isn't an asterisk. In your case, I think you could use \w instead of \s. For me, this worked in these situations:

*word
wo*rd
word*