Although, if you did go with the \S route you’d instead need a (?=\S) to perform the same role.

In summary, I was suggesting:

def response = (str =~ /\w{${charLength}}\B/).replaceAll(“\$0\n”)

OR:

def response = (str =~ /\S{${charLength}}(?=\S)/).replaceAll(“\$0\n”)

Also, you might want to consider putting a \b at the end, otherwise if you find something exactly charLength long, you’ll be adding a newline when it probably doesn’t need adding.

Finally, to actually match “substrings that do not contain spaces”, you want \S instead which matches all non-space characters – the \w only matches alphanumeric and underscore, notably excluding hyphenated words.

http://www.silverwareconsulting.com/index.cfm/2011/2/22/Introducing-CFSelenium–A-Native-ColdFusion-Client-Library-for-SeleniumRC