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Home/ Questions/Q 6724589
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T09:42:27+00:00 2026-05-26T09:42:27+00:00

Lets say I have the following jQuery plugin (this is just an example to

  • 0

Lets say I have the following jQuery plugin (this is just an example to demonstrate the point):

(function ($) {
    $.fn.colourise = function (options) {
        var settings = $.extend({
            color: "black"
        }, options);

        return this.each(function () {
            $(this).css("color", options.color);
        });
    };
})(jQuery);

which I want to apply to the following markup:

<div data-color="red">
    This text should be red.    
</div>
<div data-color="blue">
    This text should be blue
</div>
<div data-color="green">
    This text should be green
</div>

If the value I want to pass as one of the options of a plugin depends on the element, how do I apply it? At the moment I can only get this working by doing:

$(function () {
    // This feels a bit wrong to have to use a .each() here, but how else do we do it?
    $("div").each(function () {
        $(this).colourise({
            color: $(this).data("color")
        });
    });
});

I.e. by iterating over each one with the .each() method and applying the plugin to each element individually (which kinda makes the this.each() inside the plugin a bit redundant). It feels like I should be able to do something like:

$(function () {
    $("div").colourise({
        color: [get context of this "div" somehow].data("color")
    });
});

But I can’t use $(this) or this here because they refer to the document.

Sorry for the lack of a http://jsfiddle.net/, but the site is really slow for me at the moment, they must be having a few issues.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T09:42:28+00:00Added an answer on May 26, 2026 at 9:42 am

    .each is exactly what you need.

    When you call the method on a multi-element set without a .each callback, you’re creating a single options object that gets used within the plugin for every element in the set.

    You can’t vary it per-element without making a separate object for each element.

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