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Home/ Questions/Q 7710915
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T01:14:35+00:00 2026-06-01T01:14:35+00:00

Given a string like the following in JavaScript var a = ‘hello world\n\nbye world\n\nfoo\nbar\n\nfoo\nbaz\n\n’;

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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?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-01T01:14:37+00:00Added an answer on June 1, 2026 at 1:14 am

    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);
                }
            }
        })();
    }
    
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