Split string to equal length substrings in Java

Question

How to split the string \"Thequickbrownfoxjumps\" to substrings of equal size in Java. Eg. \"Thequickbrownfoxjumps\" of 4 equal size should give th

Answer 1

I asked @Alan Moore in a comment to the accepted solution how strings with newlines could be handled. He suggested using DOTALL.

Using his suggestion I created a small sample of how that works:

public void regexDotAllExample() throws UnsupportedEncodingException {
    final String input = "The\nquick\nbrown\r\nfox\rjumps";
    final String regex = "(?<=\\G.{4})";

    Pattern splitByLengthPattern;
    String[] split;

    splitByLengthPattern = Pattern.compile(regex);
    split = splitByLengthPattern.split(input);
    System.out.println("---- Without DOTALL ----");
    for (int i = 0; i < split.length; i++) {
        byte[] s = split[i].getBytes("utf-8");
        System.out.println("[Idx: "+i+", length: "+s.length+"] - " + s);
    }
    /* Output is a single entry longer than the desired split size:
    ---- Without DOTALL ----
    [Idx: 0, length: 26] - [B@17cdc4a5
     */


    //DOTALL suggested in Alan Moores comment on SO: https://stackoverflow.com/a/3761521/1237974
    splitByLengthPattern = Pattern.compile(regex, Pattern.DOTALL);
    split = splitByLengthPattern.split(input);
    System.out.println("---- With DOTALL ----");
    for (int i = 0; i < split.length; i++) {
        byte[] s = split[i].getBytes("utf-8");
        System.out.println("[Idx: "+i+", length: "+s.length+"] - " + s);
    }
    /* Output is as desired 7 entries with each entry having a max length of 4:
    ---- With DOTALL ----
    [Idx: 0, length: 4] - [B@77b22abc
    [Idx: 1, length: 4] - [B@5213da08
    [Idx: 2, length: 4] - [B@154f6d51
    [Idx: 3, length: 4] - [B@1191ebc5
    [Idx: 4, length: 4] - [B@30ddb86
    [Idx: 5, length: 4] - [B@2c73bfb
    [Idx: 6, length: 2] - [B@6632dd29
     */

}

But I like @Jon Skeets solution in https://stackoverflow.com/a/3760193/1237974 also. For maintainability in larger projects where not everyone are equally experienced in Regular expressions I would probably use Jons solution.