I just can’t find a way to make it possible to show only current parent items down to current page item in a vertical menu the “wordpress way”.
What I want to achieve is the following dynamic structure, if I visit Page 3.2.2.2:
Page 1 Page 2 Page 3 Page 4
- Page 3.1
- Page 3.2
- Page 3.2.1
- Page 3.2.2
- Page 3.2.2.1
- Page 3.2.2.2
- Page 3.2.2.2.1
- Page 3.2.2.2.2
- Page 3.2.2.3
- Page 3.2.3
- Page 3.3
- Page 3.4
- Page 3.5
So, only submenus for Page 3.2.2.2‘s parents and it’s own (if it has a submenu) will be listed, even if Page 3.4 or Page 3.2.3 also has submenus.
Answers:
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Method 1
Finally, I solved it myself. Here is the solution:
In functions.php:
function show_all_children($parent_id, $post_id, $current_level)
{
$top_parents = array();
$top_parents = get_post_ancestors($post_id);
$top_parents[] = $post_id;
$children = get_posts(
array(
'post_type' => 'page'
, 'posts_per_page' => -1
, 'post_parent' => $parent_id
, 'order_by' => 'title'
, 'order' => 'ASC'
));
if (empty($children)) return;
echo '<ul class="children level-'.$current_level.'-children">';
foreach ($children as $child)
{
echo '<li';
if (in_array($child->ID, $top_parents))
{
echo ' class="current_page_item"';
}
echo '>';
echo '<a href="'.get_permalink($child->ID).'" rel="nofollow noreferrer noopener">';
echo apply_filters('the_title', $child->post_title);
echo '</a>';
// now call the same function for child of this child
if ($child->ID && (in_array($child->ID, $top_parents)))
{
show_all_children($child->ID, $post_id, $current_level+1);
}
echo '</li>';
}
echo '</ul>';
}
In sidebar.php:
<?php $parents_ids = get_post_ancestors($post->ID); $top_parent_id = (count($parents_ids) > 0) ? $parents_ids[count($parents_ids)-1] : $post->ID; show_all_children($top_parent_id, $post->ID, 1); ?>
Method 2
Try using this snippet from WordPress Codex:
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0