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Home/ Questions/Q 8138903
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T11:39:05+00:00 2026-06-06T11:39:05+00:00

I’m in the process of rewriting a jQuery plugin to be used in an

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I’m in the process of rewriting a jQuery plugin to be used in an RSS reader I’m building during an internship. This plugin uses Google’s Feed API to pull a JSON-formatted RSS feed and return it to the developer, allowing them fine-tuned control over how that feed is displayed on the webpage. I have been following the official jQuery Plugin Authoring page as a reference.

On the reference page, code examples say that you need to add your plugin to jQuery’s prototype: $.fn. Here’s what I’ve done:

(function($) {
    "use strict";

    $.fn.rssObj = function(newUrl) {
        var RSSFeed = function(newUrl) {
            /*
             * An object to encapsulate a Google Feed API request.
             */

            this.feedUrl = newUrl;
        };

        RSSFeed.prototype.load = function() {
            var feed = new google.feeds.Feed(this.feedUrl);
            feed.load(function(result) {
                console.log(result);
            });
        };

        return new RSSFeed(newUrl);
    };

})(jQuery);

When I attempt to use this plugin by executing $.rssObj("http://rss.test.com"), my browser gives me this error:

$.rssObj() is not a function

What am I doing wrong?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-06T11:39:06+00:00Added an answer on June 6, 2026 at 11:39 am

    You add to $.fn if you want your function to be available on jQuery instances (e.g., the objects you get back from $("your selector here") and such). If you want your function available from the $ object directly, you add it directly to it.

    Here’s an example showing each:

    // Creating the plugin
    (function($) {
      
      // This will be on *instances*
      $.fn.green = function() {
        // `this` is the jQuery instance we were called on
        return this.css("color", "green");
      };
      
      // This will be on the $/jQuery object itself
      $.blue = function(selector) {
        // You don't use `this` here (you could if you want,
        // it will be === $/jQuery, but there's no reason to)
        $(selector).css("color", "blue");
        return this;
      };
      
    })(jQuery);
    
    // Usage
    jQuery(function($) {
      
      // Make all divs green with a border
      $("div").green().css("border", "1px solid green");
      
      // Make all paragraphs blue
      $.blue("p");
      
    });
    <div>I'm a div</div>
    <p>I'm a paragraph</p>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
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