Split strings but preserve delimiters in the array of splits

Question

Given a string like the following in JavaScript

var a = 'hello world\n\nbye world\n\nfoo\nbar\n\nfoo\nbaz\n\n';

I want to split it into an array like this

['hello world', '\n\n', 'bye world', '\n\n', 'foo\nbar', '\n\n', 'foo\nbaz', '\n\n'].

If the input is var a = 'hello world\n\nbye world', the result should be ['hello world', '\n\n', 'bye world'].

In other words, I want to split the string around '\n\n' into an array such that the array contains the '\n\n' as well. Is there any neat way to do this in JavaScript?

@AdamZalcman I have tried running a loop over the string and using `indexOf` and `substr` functions to extract what I want but I am wondering if there is a neater way to do solve this. — Lone Learner, Mar 24 '12 at 19:27
@Cameron Either of the two is okay for me: ['a', '\n\n', 'b'] and ['a', '\n\n', '\n\n', 'b']. The output for this is immaterial for my particular case, so we can choose whichever leads to a more elegant code. — Lone Learner, Mar 24 '12 at 19:29
For a similar problem where the separator is to be retained within the split elements, see: http://stackoverflow.com/questions/12001953/javascript-and-regex-split-string-and-keep-the-separator — Jens Jensen, Apr 19 '13 at 19:12

Gumbo · Accepted Answer · 2012-07-08T17:23:56.583

Here’s a one liner:

str.match(/\n\n|(?:[^\n]|\n(?!\n))+/g)

Here’s how it works:

\n\n matches the two consecutive newline characters
(?:[^\n]|\n(?!\n))+ matches any sequence of one or more character of either
- [^\n] not a newline character, or
- \n(?!\n) a newline character but only if not followed by another newline character

This recursive pattern can be applied on any length:

// useful function to quote strings for literal match in regular expressions
RegExp.quote = RegExp.quote || function(str) {
    return (str+"").replace(/(?=[.?*+^$[\]\\(){}|-])/g, "\\");
};
// helper function to build the above pattern recursively
function buildRecursivePattern(chars, i) {
    var c = RegExp.quote(chars[i]);
    if (i < chars.length-1) return "(?:[^" + c + "]|" + c + buildRecursivePattern(chars, i+1) + ")";
    else return "(?!" + c + ")";
}
function buildPattern(str) {
    return RegExp(RegExp.quote(delimiter) + "|" + buildRecursivePattern(delimiter.match(/[^]/g), 0) + "+", "g");
}

var str = 'hello world\n\nbye world\n\nfoo\nbar\n\nfoo\nbaz\n\n',
    delimiter = "\n\n",
    parts;
parts = str.match(buildPattern(delimiter))

Update Here’s a modification for String.prototype.split that should add the feature of containing a matched separator as well:

if ("a".split(/(a)/).length !== 3) {
    (function() {
        var _f = String.prototype.split;
        String.prototype.split = function(separator, limit) {
            if (separator instanceof RegExp) {
                var re = new RegExp(re.source, "g"+(re.ignoreCase?"i":"")+(re.multiline?"m":"")),
                    match, result = [], counter = 0, lastIndex = 0;
                while ((match = re.exec(this)) !== null) {
                    result.push(this.substr(lastIndex, match.index-lastIndex));
                    if (match.length > 1) result.push(match[1]);
                    lastIndex = match.index + match[0].length;
                    if (++counter === limit) break;
                }
                result.push(this.substr(lastIndex));
                return result;
            } else {
                return _f.apply(arguments);
            }
        }
    })();
}

+1000 your regexp mojo is powerful, sensei. How about a breakdown of the regexp for us mere mortals? — Matt Ball, Mar 24 '12 at 19:57
Should the `build()` call in `buildPattern` be `buildRecursivePattern` instead? — Matt Ball, Mar 24 '12 at 22:02

Split strings but preserve delimiters in the array of splits

1 Answers1