Regex to split string at any white-space character but to isolate any line break character

Question

It's always such a brainteaser for me but I simply can't find that out on my own...

I'd like to split a string with any white-space character BUT to isolate any \n occurrence.

For instance:

input:

"Regex expressions make your life...↵easier!"

ouput:

"Regex"
"expressions"
"make"
"your"
"life..."
"↵"
"easier!"

So far I've tried something like /s|[^\n] but it didn't turn out to be very conclusive.

Any clue?

Answer 1

This works:

[^\S\n]+

Not (NOT-Whitespace or Newline) wich equals Whitespace AND not newline (DeMorgan)

And if i get your "isolating" right, try this: ([^\S\n]|\n)

input:

this is a test
regex

preg_split:

Array
(
[0] => this
[1] => is
[2] => a
[3] => test
[4] => 
[5] => regex
)

hmm, but that will be the same as \s (not \n or \n is obsolet, leaves ^\S - which is \s - so what you mean by "isolating"?

Answer 2

Try this: /(\S+|\n)/g

var s = 'Regex expressions make your life...\neasier!';
s.match(/(\S+|\n)/g)