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Home/ Questions/Q 8896999
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T00:13:13+00:00 2026-06-15T00:13:13+00:00

I am trying to reduce my javascript overheads on my site and I am

  • 0

I am trying to reduce my javascript overheads on my site and I am using this plugin:
https://github.com/carhartl/jquery-cookie

This is the code:

/*!
 * jQuery Cookie Plugin v1.3
 * https://github.com/carhartl/jquery-cookie
 *
 * Copyright 2011, Klaus Hartl
 * Dual licensed under the MIT or GPL Version 2 licenses.
 * http://www.opensource.org/licenses/mit-license.php
 * http://www.opensource.org/licenses/GPL-2.0
 */
(function ($, document, undefined) {

  var pluses = /\+/g;

    function raw(s) {
        return s;
    }

    function decoded(s) {
        return decodeURIComponent(s.replace(pluses, ' '));
    }

    var config = $.cookie = function (key, value, options) {

        // write
        if (value !== undefined) {
            options = $.extend({}, config.defaults, options);

            if (value === null) {
                options.expires = -1;
            }

            if (typeof options.expires === 'number') {
                var days = options.expires, t = options.expires = new Date();
                t.setDate(t.getDate() + days);
            }

            value = config.json ? JSON.stringify(value) : String(value);

            return (document.cookie = [
                encodeURIComponent(key), '=', config.raw ? value : encodeURIComponent(value),
                options.expires ? '; expires=' + options.expires.toUTCString() : '', // use expires attribute, max-age is not supported by IE
                options.path    ? '; path=' + options.path : '',
                options.domain  ? '; domain=' + options.domain : '',
                options.secure  ? '; secure' : ''
            ].join(''));
        }

        // read
        var decode = config.raw ? raw : decoded;
        var cookies = document.cookie.split('; ');
        for (var i = 0, l = cookies.length; i < l; i++) {
            var parts = cookies[i].split('=');
            if (decode(parts.shift()) === key) {
                var cookie = decode(parts.join('='));
                return config.json ? JSON.parse(cookie) : cookie;
            }
        }

        return null;
    };

    config.defaults = {};

    $.removeCookie = function (key, options) {
        if ($.cookie(key) !== null) {
            $.cookie(key, null, options);
            return true;
        }
        return false;
    };

})(jQuery, document);

This is my attempt at stripping it down by removing the “write” code

(function ($, document, undefined) {

  var pluses = /\+/g;

function raw(s) {
    return s;
}

function decoded(s) {
    return decodeURIComponent(s.replace(pluses, ' '));
}

var config = $.cookie = function (key, value, options) {

    // read
    var decode = config.raw ? raw : decoded;
    var cookies = document.cookie.split('; ');
    for (var i = 0, l = cookies.length; i < l; i++) {
        var parts = cookies[i].split('=');
        if (decode(parts.shift()) === key) {
            var cookie = decode(parts.join('='));
            return config.json ? JSON.parse(cookie) : cookie;
        }
    }

    return null;
};

config.defaults = {};

$.removeCookie = function (key, options) {
    if ($.cookie(key) !== null) {
        $.cookie(key, null, options);
        return true;
    }
    return false;
};

})(jQuery, document);

If I want to use it just for reading cookies, rather than for deleting, writing and reading, is there any more dead weight I can strip out.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-15T00:13:14+00:00Added an answer on June 15, 2026 at 12:13 am

    Well, using a jQuery plugin just to read a cookie is kind of silly, though I’ve done the same thing with the same plugin!

    What is the shortest function for reading a cookie by name in JavaScript?

    has some shorter code that you could use.

    The core of all these are document.cookie – it kinda sucks that they are just a string, rather than a nice array already – but there you go.

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