Regex: Specify "space or start of string" and "space or end of string"

Question

Imagine you are trying to pattern match "stackoverflow".

You want the following:

 this is stackoverflow and it rocks [MATCH]

 stackoverflow is the best [MATCH]

 i love stackoverflow [MATCH]

 typostackoverflow rules [NO MATCH]

 i love stackoverflowtypo [NO MATCH]

I know how to parse out stackoverflow if it has spaces on both sites using:

/\s(stackoverflow)\s/

Same with if its at the start or end of a string:

/^(stackoverflow)\s/

/\s(stackoverflow)$/

But how do you specify "space or end of string" and "space or start of string" using a regular expression?

Answer 1

You can use any of the following:

\b      #A word break and will work for both spaces and end of lines.
(^|\s)  #the | means or. () is a capturing group. 


/\b(stackoverflow)\b/

Also, if you don't want to include the space in your match, you can use lookbehind/aheads.

(?<=\s|^)         #to look behind the match
(stackoverflow)   #the string you want. () optional
(?=\s|$)          #to look ahead.

Answer 2

(^|\s) would match space or start of string and ($|\s) for space or end of string. Together it's:

(^|\s)stackoverflow($|\s)

Answer 3

Here's what I would use:

 (?<!\S)stackoverflow(?!\S)

In other words, match "stackoverflow" if it's not preceded by a non-whitespace character and not followed by a non-whitespace character.

This is neater (IMO) than the "space-or-anchor" approach, and it doesn't assume the string starts and ends with word characters like the \b approach does.

Answer 4

\b matches at word boundaries (without actually matching any characters), so the following should do what you want:

\bstackoverflow\b