Problem:

• Splitting of primary and secondary menus.
• Secondary menus are shown when the primary menu item is selected.
• Selecting of all parent items (on all levels) as active class.
• Option to output menu tree (many levels) in submenus (you cannot do this using secondary_links)

Solution:

1. Navigate to: Administer → Site building → Menu
2. Create menu (You can create the necessary menu items right in primary links – menu is already created by default). To make it multilevel, select parent for menu items. There is no need to create two different menus.
3. Open Settings tab and select:

Menu with primary links: your menu (Primary links)
Menu with secondary links: no additional links

4. Insert the following code in page.tpl.php:

   For primary_links

<?php if (isset($primary_links)) { ?><?php print theme(‘links’, $primary_links) ?><?php } ?>

    5. At the lef of page.tpl.php, add this column:

<?php if ($sidebar_left) { ?><?php print $sidebar_left ?><?php } ?>

    6. Create block in admin area and put the following code there:

<?php
$primary_menu_id = variable_get(‘menu_primary_menu’, 0);
if (menu_in_active_trail($primary_menu_id)){
  // get the menu items that lead to the current menu item
  $active_trail =_menu_get_active_trail();
  // get the menu id of the active top-level link
  $mid = $active_trail[1];
  $menu_item = menu_get_item($mid);
  $menu_tree = menu_tree($mid);
print ‘<ul class="YOU CLASS FOR LEFT MENU">’.$menu_tree.'</ul>';
}
?>

   7. To make parent link with class="active", please use TemplatePHP snippet from here. Insert this code to template.php (without closing ?>):

<?php
function phptemplate_links($links, $attributes = array()) {
  if (!count($links)) {
    return ”;
  }
  $level_tmp = explode(‘-‘, key($links));
  $level = $level_tmp[0];
  $output = "<ul class=\"links-$level ".$attributes[‘class’]. "\">\n";
  foreach ($links as $index => $link) {
    $output .= ‘<li';
    if (stristr($index, ‘active’)) {
      $output .= ‘ class="active"';
    }// frontpage AND current-link in menu is <front>
    elseif((drupal_is_front_page()) && ($link[‘href’]=='<front>’)){
      $link[‘attributes’][‘class’] = ‘active';//add class active to <li
      $output .= ‘ class="active"';//add class active to <a
    }
    $output .= ">". l($link[‘title’], $link[‘href’], $link[‘attributes’], $link[‘query’], $link[‘fragment’]) ."</li>\n";
  }
  $output .= ‘</ul>';
  return $output;
}
?>

  8. Then make all parent links look like links with class="active". We will solve this problem with the help of CSS. Such menu generates code from the block:

<ul class="sbtree">
<li class="leaf"><a href="/Company_History">Company History</a></li>
<li class="expanded"><a href="/Business_Units">Business Units</a>
  <ul class="menu">
   <li class="expanded"><a href="/Property_Development">Property Development</a>
    <ul class="menu">
     <li class="leaf"><a href="/board_committees" class="active">Board Committees</a></li >
     <li class="leaf"><a href="/Rules_of_Procedure">Rules of Procedure</a></li>
    </ul>
   </li>
   <li class="leaf"><a href="/Real_Estate_Funds">Real Estate Funds</a></li>
   <li class="leaf"><a href="/Russia">Russia</a></li>
  </ul>
</li>
<li class="leaf"><a href="/Corporate_Governance">Corporate Governance</a></li>
</ul>

You may need these classes to track 4 levels of split menu:

ul.sbtree li a.active – just active menu of the second level
ul.sbtree li.expanded a – parent menu item which contains active item – we wish it to be as active
ul.sbtree li.expanded ul.menu li a – child inactive menu item of the third level – we wish it to show it inactive
ul.sbtree li ul.menu li a.active – child active menu item of the third level
ul.sbtree li.expanded ul.menu li.expanded a – parent item of expanded menu in expanded menu – we wish to show it active
ul.sbtree li.expanded ul.menu li.expanded ul.menu a – child inactive item of expanded menu in expanded menu – we wish to show it inactive
ul.sbtree li.expanded ul.menu li.expanded ul.menu a.active – child active item of 4-level menu