how to reference widget variable

• 

  vilignoble · Member · Oct 6, 2013 at 4:58 am

  • Copy link
  • Add topic to favorites

  I want to be able to display different widget options in wordpress options to display based on which dropdown i select. The example shows widget_type dropdown menu, and when I select number one option I want more options to display beneath the widget_type dropdown menu and when two is selected I want some other options to display beneath the widget_type dropdown menu.

  <?php
  
  class test_Widget extends WP_Widget {
  
      function __construct() {
          parent::__construct(
              'test_widget', // Base ID
              __('test Sidebar', 'text_domain'), // Name
              array( 'description' => __( 'test sidebar widget.', 'text_domain' ), ) // Args
          );
      }
      public function widget( $args, $instance ) {
          $widget_type = apply_filters( 'widget_type', $instance['widget_type'] );;
  
          echo $args['before_widget'];
          echo $args['after_widget'];
      }
      public function form( $instance ) {
          if ( isset( $instance[ 'widget_type'] ) ) {
                      $widget_type = $instance[ 'widget_type' ];
                  }
                  else {
                      $widget_type = __( '', 'text_domain' );
                  }
          ?>
                  <p>
                      <label for="<?php echo $this->get_field_id( 'widget_type' ); ?>">Widget Type:</label>
                      <select class="widefat" id="<?php echo $this->get_field_id( 'widget_type' ); ?>" name="<?php echo $this->get_field_name( 'widget_type' ); ?>">
                          <option <?php if ( '1' == $instance['widget_type'] ) echo 'selected="selected"'; ?> value="1" value="">1</option>
                          <option <?php if ( '2' == $instance['widget_type'] ) echo 'selected="selected"'; ?> value="2">2</option>
                      </select>
                  </p>
          <?php
      }
  
      public function update( $new_instance, $old_instance ) {
          $instance = array();
          $instance['widget_type'] = ( $new_instance['widget_type'] );
  
          return $instance;
      }
  }

  how to reference widget variable and add more options based on widget dropdown menu

• 

  1tess · Moderator Emeritus · Oct 6, 2013 at 5:21 am

  • Copy link

  We cannot use the code you posted on WordPress.com, so I’m assuming that you are using WordPress.org.

  This means that you are looking for help in the wrong place: these forums provide assistance exclusively for the wp.com platform.

  Here are details about the different flavors of WordPress:
  http://support.wordpress.com/com-vs-org/
  Here is a link to support for wp.org:
  http://wordpress.org/support/