Regex to find last occurrence of pattern in a string

Question

My string being of the form:

"as.asd.sd fdsfs. dfsd  d.sdfsd. sdfsdf sd   .COM"

I only want to match against the last segment of whitespace before the last period(.)

So far I am able to capture whitespace but not the very last occurrence using:

\s+(?=\.\w)

How can I make it less greedy?

Answer 1

In a general case, you can match the last occurrence of any pattern using the following scheme:

pattern(?![\s\S]*pattern)
(?s)pattern(?!.*pattern)
pattern(?!(?s:.*)pattern)

where [\s\S]* matches any zero or more chars as many as possible. (?s) and (?s:.) can be used with regex engines that support these constructs so as to use . to match any chars.

In this case, rather than \s+(?![\s\S]*\s), you may use

\s+(?!\S*\s)

See the regex demo. Note the \s and \S are inverse classes, thus, it makes no sense using [\s\S]* here, \S* is enough.

Details:

• \s+ - one or more whitespace chars
• (?!\S*\s) - that are not immediately followed with any 0 or more non-whitespace chars and then a whitespace.

Answer 2

You can try like so:

(\s+)(?=\.[^.]+$)

(?=\.[^.]+$) Positive look ahead for a dot and characters except dot at the end of line.

Demo:

https://regex101.com/r/k9VwC6/3