Java RegExp Split String with saving delimiters

Question

So, I have a simple string that looks like this:

word1 word2! word3? word4; word5, word6
word7 //new line
!word8! word9 word10 word11 word12

And my desire is to split this string with saving whitespace and new line delimiters. Right now I'm using a s.split() method with [\\s\\r\\n] expression as its argument and the output is:

[word1, word2!, word3?, word4;, word5,, word6, , word7, , !word8!, word9, word10, word11, word12]

And I'm okay with a whitespaces not being saved. But what can I do with a \n being saved just as a whitespace?

UPD: I pass this string through RabbitMQ query. In Java it will look like this:

"word1 word2! word3? word4; word5, word6\nword7\n!word8! word9 word10 word11 word12"

Answer 1

You can extract the whitespace and non-whitespace strings (and basically, tokenize the text into whitespace and non-whitespace text chunks) using the \S+|\s+ regex.

See the Java demo:

import java.util.*;
import java.util.regex.*;
 
class Ideone
{
    public static void main (String[] args) throws java.lang.Exception
    {
        String line = "word1 word2! word3? word4; word5, word6\nword7\n!word8! word9 word10 word11 word12";
        Pattern p = Pattern.compile("\\S+|\\s+");
        Matcher m = p.matcher(line);
        List<String> res = new ArrayList<>();
        while(m.find()) {
            res.add(m.group());
        }
        System.out.println(res);
    }
}

Output:

[word1,  , word2!,  , word3?,  , word4;,  , word5,,  , word6, 
, word7, 
, !word8!,  , word9,  , word10,  , word11,  , word12]

where the line breaks are literal line break chars.