jQuery - Plugins and namespace issues with method chaining

Question

I'm trying to create a simple plugin that I can call by both: $.myPlugin() and $('something').myPlugin()

Heres the code:

(function($) {
  $.fn.myPlugin = function(item) {
    return this;
  };

  $.myPlugin = function(item) {
    return $.fn.myPlugin(item);
  };
}(jQuery));

This works when called either way mentioned above.

However, calling $.myPlugin().hide() does not work. Any chained methods are failing.

Heres a simple JSBin I made showing the issue

Why?

fn is a shortcut for prototype, you're creating different instances and expecting them to somehow be the same ? — adeneo
– adeneo, Commented Nov 16, 2014 at 5:06

JDillon522 · Accepted Answer · 2014-11-16 21:17:36Z

1

You should return $(this), not jquery this:

(function($) {
    $.fn.myMethod = function() {
      this.append('<p>MY METHOD</p>');

      return $(this);
    };

    $.myMethod = function() {
      return $.fn.myMethod();
    };
}(jQuery));

$(function () {
    // Moment of truth
    $('.output').myMethod().hide('slow');

    $('h1').click(function(){
      $('.output').myMethod().hide();
    });
});

edited Nov 16, 2014 at 21:17

JDillon522

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answered Nov 16, 2014 at 7:39

Martynas Barzda

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2 Comments

JDillon522 Over a year ago

I'm not sure what you mean by "return element, not jquery function". I know it works fine by calling $('.output').myMethod().hide() but I want to be able to call it by $.myMethod(), similar to how you can do $.each(item, function(){...})

Martynas Barzda Over a year ago

Then instead return this; use return $('.output'); at 3rd line of you snippet

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