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I created a widget that uses a jquery plugin, and I used is_active_widget to enqueue the script, it's working fine but it's also including the script even on pages that doesn't show this widget. so is there a way to only enqueue the script if this widget is showing ?

thanks in advance.

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4 Answers 4

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You should be able to call wp_enqueue_script() as part of your Widget output.

Edit

Quick-and-dirty, using the bare-bones Widgets API class example:

<?php class wpse48337_Widget extends WP_Widget { public function __construct() { // widget actual processes } public function form( $instance ) { // outputs the options form on admin } public function update( $new_instance, $old_instance ) { // processes widget options to be saved } public function widget( $args, $instance ) { // outputs the content of the widget } } register_widget( 'wpse48337_Widget' ); ?>

Add your wp_enqueue_script() call inline, within your Widget's output - i.e. within the public function widget():

<?php public function widget( $args, $instance ) { // outputs the content of the widget // Enqueue a script needed for // the Widget's output wp_enqueue_script( 'jquery-pluginname', $path, $deps ); // Rest of widget output goes here... } ?>
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  • Actually I'm doing that, and using is_active_widget() but it will enqueue the script on all pages if the widget is active even on one. Commented Apr 9, 2012 at 6:58
  • Using is_active_widget() is not the same as calling wp_enqueue_script() from within your Widget output. See updated answer. Commented Apr 9, 2012 at 11:52
  • thank you for the update, but what do you mean by quick and dirty the example or this way for using wp_enqueue_script() ? and one more off topic question please, I'm seeing a lot of these private and public before functions in widgets so what do they do ? and should we use them widgets specific for themes ? and thank you very much :) Commented Apr 11, 2012 at 6:32
  • By "quick and dirty example", I mean that I'm referencing the standard Widget class construct, as described in the linked Codex article. I'm assuming that you've written your custom Widget using the Widgets API. If not, you should be; but nonetheless, if you need more specific guidance, we'll need to see your Widget code. Commented Apr 11, 2012 at 11:53
  • This doesn't work. Not sure if it used to work since it was marked as an answer. But right now, the widget form is triggered after the wp_enqueue_scripts action Commented Apr 26, 2015 at 12:13
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If the script needs to go in the <head> section:

class Your_Widget extends WP_Widget{ protected static $did_script = false; function __construct(){ // parent::__construct() ... add_action('wp_enqueue_scripts', array($this, 'scripts')); } function scripts(){ if(!self::$did_script && is_active_widget(false, false, $this->id_base, true)){ wp_enqueue_script('your-script'); self::$did_script = true; } } }

Otherwise, Chip Bennett's solution will work for queuing it in the footer.

The $did_script static variable is not required, I used it just to avoid further unnecessary calls to wp_enqueue_script if there are multiple widget instances on the page...

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    Cool idea, but this doesn't actually address the OP's question, since is-active_widget() just lets you know whether a widget has been activated (ie: dragged into one of the sidebar areas in Appearance > Widgets), as opposed to actually appearing on any given page. Commented Aug 19, 2013 at 16:15
  • Condition is wrong: (1) is_active_widget will return a sidebar_id, even if that sidebar is not on displayed the current page. This will add the script on every page. (2) wp_enqueue_script will add a script only once, even if you call it multiple times. no need for self::$did_script (3) because __construct is called on every page request, even if the widget is not even displayed Commented Aug 4, 2016 at 11:05
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Pierre,

The way I handle this is with wp_enqueue_script and wp_dequeue_script, and use a instance variable $is_active in the Your_Widget class

So do the wp_enqueue_script based on is_active_widget which enqueues the script on all pages but with the footer parameter set to true. Note the dequeue is run at a priority to make sure it runs before the scripts are output.

function enqueue_scripts() { if ( is_active_widget( false, $this->id, $this->id_base, true ) ) { wp_enqueue_script( 'your-script-handle', 'your-script-url', array(), '1.0', true ); add_action( 'wp_footer', array($this,'dequeue_redundant_scripts'), 1 ); }

}

Then in the widget function indicate if the widget is active on that page

function widget( $args, $instance ) { // outputs the content of the widget $this->is_active = true; }

Then in the footer dequeue the script if the widget is not active on that page

function dequeue_redundant_scripts() { if (! $this->is_active) { wp_dequeue_script('your-script-handle'); } }

This approach of enqueue then dequeueing if unused also works well for plugins that define shortcodes which require scripts

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I considered 'wp_footer' hook because this hook is executed at footer,and is probably the best way to add scripts only where the widget is used.

class Your_Widget extends WP_Widget{ function widget( $args, $instance ) { add_action('wp_footer',array($this,'front_end_scripts')); } function front_end_scripts(){ ?><script type="text/javascript"> console.log('this works!'); /* Or if you need external scripts then you may use $.getScript([your-script-url] ); */ </script> <?php } }
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    Ugh. wp_enqueue_script() please. No direct printing of JS. Commented Aug 19, 2013 at 15:57
  • I'm afraid wp_enqueue_script() will not work here since the hook besides wp_enqueue_script is fired much before the widget is actually called. Commented Aug 20, 2013 at 5:47
  • You can call wp_enqueue_script() any time before the wp_footer action. The widget is called during the building of the template page. You can either hook onto dynamic_sidebar action, detecting when your widget is actually displayed, or, more simply, just put your enqueue_script() right into the widget() method, as @ChipBennet recommends. Commented Aug 20, 2013 at 13:42

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