数据库中的菜单包含无限的子菜单/项目？

Question

I'm trying to build this function that display a menu with sub menus from a database. Something that can show like sub-sub-sub-menus and so on... If that's ever something I need.

In the database i have 2 tables, menu and menuItem.

menu

menuItem

php file

function get_menu($id) {

require(dirname(__FILE__) . '/config.php');

// main base
$menu = "   SELECT id, menu_name, menu_data, menu_order
            FROM mos_menuItem 
            WHERE menu_parent_id='$id' 
            AND menu_sub='0' 
            ORDER BY menu_order";
$menuResult = mysqli_query($conn, $menu);

echo '<ul>';

while ($menuRow = mysqli_fetch_row($menuResult)) {
    // main menu
    echo '<li><a href="'. $menuRow[2] .'" class="menu-item parent-menu">'. $menuRow[1] .'</a></li>';

        // sub base
        $sub = "SELECT id, menu_name, menu_data, menu_order, menu_sub 
                    FROM mos_menuItem 
                    WHERE menu_parent_id='$menuRow[0]' 
                    AND menu_sub>='1' 
                    ORDER BY menu_order";
        $subResult = mysqli_query($conn, $sub);

        echo '<ul>';

        while ($subRow = mysqli_fetch_row($subResult)) {
            echo '<li><a href="'. $subRow[2] .'" class="menu-item sub-menu">'. $subRow[1] .'</a></li>';
        }   

        echo '</ul>';
}
echo '</ul>';

mysqli_close($conn);
}

so I need to loop something inside itself? I'm not sure where to go from here.

Answer 1

So after some trial and error i figure it out. It really helped to look at Sheikh Azad code. The only difference is in my function i have AND menu_sub='$sub' to make it not going loop the main menu item if the id is the same.. If that make sense.

The function

function mos_get_menu_editor($menuId, $sub) {

    require(dirname(__FILE__) . '/config.php');

    $item = "   SELECT id, menu_name, menu_data, menu_order, menu_sub
                FROM mos_menuItem 
                WHERE menu_parent_id='$menuId' 
                AND menu_sub='$sub' 
                ORDER BY menu_order";
    $itemResult = mysqli_query($conn, $item);

        echo '<ul>';

        while ($itemRow = mysqli_fetch_row($itemResult)) {

            echo '<li><a href="'. $itemRow[2] .'" class="menu-item ' .($sub ? 'sub-menu' : '').'">'. $itemRow[1] .'</a></li>';

                mos_get_menu_editor($itemRow[0], 1);

        }
        echo '</ul>';

    mysqli_close($conn);
}