I have a custom post type 'Commercials' and custom post status 'Featured'. Now I want only one featured post. What I mean is right now I can make all posts as Featured, but I want only one to be featured. If there is already a featured post and I select a new one as featured then the old one will go back to publish.

I've been trying for hours to get a solution. Here is the code I'm using.

// Register Custom Post Status
function register_custom_post_status_featured(){
    register_post_status( 'Featured', array(
        'label'                     => _x( 'Featured', 'commercials' ),
        'public'                    => true,
        'internal'                  => true,
        'protected'                 => true,
        'private'                   => false,
        'publicly_queryable'        => true,
        'exclude_from_search'       => false,
        'show_in_admin_all_list'    => true,
        'show_in_admin_status_list' => true,
        'label_count'               => _n_noop( 'Featured <span class="count">(%s)</span>', 'Featured <span class="count">(%s)</span>' ),
    ) );
}
add_action( 'init', 'register_custom_post_status_featured' );

// Display Custom Post Status Option in Post Edit
function display_custom_post_status_featured_option(){
  global $post;
  $complete = '';
  $label = '';
  if($post->post_type == 'commercials'){
    if($post->post_status == 'featured'){
        $selected = 'selected';
    }
    echo '<script>
      $(document).ready(function(){
        $("select#post_status").append("<option value=\"featured\" '.$selected.'>Featured</option>");
        $(".misc-pub-section label").append("<span id=\"post-status-display\"> Featured</span>");
        var currentPostStatus = $("select#post_status").find(":selected").text();
        $("#post-status-display").html(currentPostStatus);
        $( "select[name=\"_status\"]" ).append( "<option value=\"featured\">Featured</option>" );
      });
    </script>
    ';
  }
}
add_action('admin_footer', 'display_custom_post_status_featured_option');

// display label
function rudr_display_featured_status_label( $status ) {
  global $post;
  $complete = '';
  $label = '';
  if($post->post_type == 'commercials'){
    if($post->post_status == 'featured'){
            return array('Featured');
        }
    }
    return $status;
}

add_filter( 'display_post_states', 'rudr_display_featured_status_label' );

Any help would be very much appreciated.

I have a custom post type 'Commercials' and custom post status 'Featured'. Now I want only one featured post. What I mean is right now I can make all posts as Featured, but I want only one to be featured. If there is already a featured post and I select a new one as featured then the old one will go back to publish.

I've been trying for hours to get a solution. Here is the code I'm using.

// Register Custom Post Status
function register_custom_post_status_featured(){
    register_post_status( 'Featured', array(
        'label'                     => _x( 'Featured', 'commercials' ),
        'public'                    => true,
        'internal'                  => true,
        'protected'                 => true,
        'private'                   => false,
        'publicly_queryable'        => true,
        'exclude_from_search'       => false,
        'show_in_admin_all_list'    => true,
        'show_in_admin_status_list' => true,
        'label_count'               => _n_noop( 'Featured <span class="count">(%s)</span>', 'Featured <span class="count">(%s)</span>' ),
    ) );
}
add_action( 'init', 'register_custom_post_status_featured' );

// Display Custom Post Status Option in Post Edit
function display_custom_post_status_featured_option(){
  global $post;
  $complete = '';
  $label = '';
  if($post->post_type == 'commercials'){
    if($post->post_status == 'featured'){
        $selected = 'selected';
    }
    echo '<script>
      $(document).ready(function(){
        $("select#post_status").append("<option value=\"featured\" '.$selected.'>Featured</option>");
        $(".misc-pub-section label").append("<span id=\"post-status-display\"> Featured</span>");
        var currentPostStatus = $("select#post_status").find(":selected").text();
        $("#post-status-display").html(currentPostStatus);
        $( "select[name=\"_status\"]" ).append( "<option value=\"featured\">Featured</option>" );
      });
    </script>
    ';
  }
}
add_action('admin_footer', 'display_custom_post_status_featured_option');

// display label
function rudr_display_featured_status_label( $status ) {
  global $post;
  $complete = '';
  $label = '';
  if($post->post_type == 'commercials'){
    if($post->post_status == 'featured'){
            return array('Featured');
        }
    }
    return $status;
}

add_filter( 'display_post_states', 'rudr_display_featured_status_label' );

Any help would be very much appreciated.

• custom-post-types
• post-status
• limit

• May be think about meta data which will mark your post as "specially marked" instead of "featured" as any post of this post type. – Max Yudin Commented Jun 4, 2019 at 17:29
• @Max Wouldn't that create the same issue, I mean after marking a new post as special the older special post has to be unmarked. – Muhammad Russell Commented Jun 4, 2019 at 17:48

1 Answer 1

You can use one of the status transition filter hooks for this:

• {$new_status}_{$post_type}
• transition_post_status

In the function assigned to the hook, you change the status to publish the previously featured post ( using e.g. $wpdb ). Both actions are performed after saving the post, so you have to change the status in posts other than edited.

add_action( 'featured_commercials', 'se339582_single_featured', 10, 2 );
function se339582_single_featured( $post_id, $post )
{
    global $wpdb;

    $sql = $wpdb->prepare( "UPDATE {$wpdb->posts} SET post_status='publish' "
            ." WHERE post_status='featured' AND id <> %d", $post_id );
    $wpdb->query( $sql );
}

Update:

SQL for one featured post per category:

$taxonomy_slug = 'commercials';
$sql = $wpdb->prepare( "UPDATE {$wpdb->posts} p " .
    " INNER JOIN {$wpdb->term_relationships} tr ON tr.object_id = p.id " .
    " INNER JOIN {$wpdb->term_taxonomy} tt ON tt.term_taxonomy_id = tr.term_taxonomy_id " .
    " SET p.post_status='publish' " .
    " WHERE p.id <> %d AND p.post_status='featured' AND tt.taxonomy=%s AND tt.term_id = %d", 

    $post_id, $taxonomy_slug, $term_id
);