Well, to build my menu my menu I use a db similar structure like this
2 Services 0 3 Photo Gallery 0 4 Home 0 5 Feedback 0 6 FAQs 0 7 News & Events 0 8 Testimonials 0 81 FACN 0 83 Organisation Structure 81 84 Constitution 81 85 Council 81 86 IFAWPCA 81 87 Services 81 88 Publications 81
To assign another submenu for existing submenu I simply assign its parent's id as its value of parent field. parent 0 means top menu
now there is not problem while creating submenu inside another submenu
now this is way I fetch the submenu for the top menu
<ul class="topmenu">
<? $list = $obj -> childmenu($parentid);
//this list contains the array of submenu under $parendid
foreach($list as $menu) {
extract($menu);
echo '<li><a href="#">'.$name.'</a></li>';
}
?>
</ul>
What I want to do is.
I want to check if a new menu has other child menu
and I want to keep on checking until it searches every child menu that is available
and I want to display its child menu inside its particular list item like this
<ul>
<li><a href="#">Home</a>
<ul class="submenu">
........ <!-- Its sub menu -->
</ul>
</li>
</ul>
See Question&Answers more detail:os