The regular expression \s+$ means "one or more whitespace characters at the end of the line" so it will never match whitespaces in between other characters.

Whenever I want to strip excessive whitespace from a string, I usually go about it in a three-step way:

1. Combine multiple whitespaces into a single space using s/\s+/ /g;
2. Remove the leading space (if any) with s/^\s//;
3. Remove the trailing space (if any) with s/\s$//; (Remember that this assumes no CR/LF character, adjust as needed)

Add a few comments and the job is done.

Perhaps this can be combined into a single-shot regex but it will probably be harder to read and waste more CPU cycles to accomplish exactly the same thing. I never bothered to find out because this approach works for me.