regex word boundary excluding the hyphen

regex

Solution

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

Problem

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 ``` type ([a-z])\b ``` but do not match e.g. ``` type a-1 ``` to rephrase: i want an equivalent of the word boundary operator \b which instead of using the word character class `[A-Za-z0-9_]`, uses the extended class: `[A-Za-z0-9_-]`

Original source