How to build unlimited level of menu through PHP and mysql

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>

Here is a "developer-friendly" version of the "one query, no recursion" solution for this problem.

SQL:

SELECT id, parent_id, title, link, position FROM menu_item ORDER BY parent_id, position;

PHP:

$html = '';
$parent = 0;
$parent_stack = array();

// $items contains the results of the SQL query
$children = array();
foreach ( $items as $item )
    $children[$item['parent_id']][] = $item;

while ( ( $option = each( $children[$parent] ) ) || ( $parent > 0 ) )
{
    if ( !empty( $option ) )
    {
        // 1) The item contains children:
        // store current parent in the stack, and update current parent
        if ( !empty( $children[$option['value']['id']] ) )
        {
            $html .= '<li>' . $option['value']['title'] . '</li>';
            $html .= '<ul>'; 
            array_push( $parent_stack, $parent );
            $parent = $option['value']['id'];
        }
        // 2) The item does not contain children
        else
            $html .= '<li>' . $option['value']['title'] . '</li>';
    }
    // 3) Current parent has no more children:
    // jump back to the previous menu level
    else
    {
        $html .= '</ul>';
        $parent = array_pop( $parent_stack );
    }
}

// At this point, the HTML is already built
echo $html;

You just need to understand the usage of the $parent_stack variable.

It is a "LIFO" stack (Last In, First Out) - the image in the Wikipedia article worths a thousand words: http://en.wikipedia.org/wiki/LIFO_%28computing%29

When a menu option has sub-options, we store its parent ID in the stack:

array_push( $parent_stack, $parent );

And then, we immediately update $parent, making it be the current menu option ID:

$parent = $option['value']['id'];

After we looped all its sub-options, we can return back to the previous level:

$parent = array_pop( $parent_stack );

This is why we stored the parent ID in the stack!

My suggestion is: contemplate the code snippet above, and understand it.

Questions are welcome!

One of the advantages I see in this approach is that it eliminates the risk of entering into an infinite loop, which can happen when recursion is used.

With a database structure like yours, it is possible to build the whole HTML menu with a single query and without recursion.

Yes - I will repeat:

• ONE QUERY
• NO RECURSION

This is the approach I always use myself.

Pasted the code here - fully functional:

http://pastebin.com/GAFvSew4

Jump to line 67 to see the interesting part ("get_menu_html").

The main loop starts at line 85.

There are five "customizable" HTML snippets:

1. menu wrapper opening (line 83)
2. menu wrapper closing (line 122)
3. menu item with childs opening (line 100)
4. menu item with childs closing (line 92)
5. menu item without childs (line 113)

(The code could be cleaner if I hadn't worried with tabulation.)

SQL to create and populate sample database is available at the end of the script.

You can try and let us know your thoughts.