Reputation: 15122
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)$/
But how do you specify "space or end of string" and "space or start of string" using a regular expression?
Upvotes: 200
Views: 197733
Reputation: 9332
You can use any of the following:
\b #A word break and will work for both spaces and end of lines.
(^|\s) #the | means or. () is a capturing group.
/\b(stackoverflow)\b/
Also, if you don't want to include the space in your match, you can use lookbehind/aheads.
(?<=\s|^) #to look behind the match
(stackoverflow) #the string you want. () optional
(?=\s|$) #to look ahead.
Upvotes: 265
Reputation: 75272
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.
Upvotes: 29
Reputation: 208715
\b
matches at word boundaries (without actually matching any characters), so the following should do what you want:
\bstackoverflow\b
Upvotes: 12
Reputation: 9806
(^|\s)
would match space or start of string and ($|\s)
for space or end of string. Together it's:
(^|\s)stackoverflow($|\s)
Upvotes: 101