Javascript reg ex to match whole word only, bound only by whitespace

Matching all:

\bBlah\b

Debuggex Demo

You can try: \sBlah\s.

Or if you allow beginning and end anchors, (^|\s)Blah(\s|$)

This will match "Blah" by itself, or each Blah in "Blah and Blah"

See also

• regular-expressions.info/Character classes and Anchors
  • \s stands for "whitespace character".
  • The caret ^ matches the position before the first character in the string
  • Similarly, $ matches right after the last character in the string

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Lookahead variant

If you want to match both Blah in "Blah Blah", then since the one space is "shared" between the two occurrences, you must use assertions. Something like:

(^|\s)Blah(?=\s|$)

See also

• regular-expressions.info/Lookarounds

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Capturing only Blah

The above regex would also match the leading whitespace.

If you want only Blah, ideally, lookbehind would've been nice:

(?<=^|\s)Blah(?=\s|$)

But since Javascript doesn't support it, you can instead write:

(?:^|\s)(Blah)(?=\s|$)

Now Blah would be captured in \1, with no leading whitespace.

See also

• regular-expressions.info/Grouping and flavor comparison

extracting all words in a string

words_array = str.match(/\b(\w|')+\b/gim)   //only single qout allowed

Try \sBlah\s — that will match any form of whitespace on either side.