dynamic multi level dropdown menu php sql - php

Question

I have nav menu in html and i want to create a loop everything is fine but the item that have children echo twice! i know where is the problem but i can't figure out how to solve it :(
my sql table named menu
and this is my php:
$db = mysqli_connect('localhost', 'root', 'password', 'aftab');
<?php
$get = mysqli_query($db , "SELECT * from menu where parent_id is NULL");
while ($rowmenu = mysqli_fetch_assoc($get)) {
echo '<li class="menu-item menu-item-type-post_type menu-item-object-page menu-item-656"><a href="'. $rowmenu['link'] . '" >' . $rowmenu['name'] .'</a>' ;
$id = $rowmenu['id'] ;
$check = mysqli_query($db , "SELECT * from menu where parent_id = '$id'");
if ( mysqli_num_rows($check) ) {
echo '<li class="menu-item menu-item-type-post_type menu-item-object-page menu-item-has-children menu-item-656"><a href="'. $rowmenu['link'] . '" >' . $rowmenu['name'] .'</a>' ;
echo '<ul class="sub-menu">' ;
while ( $row2 = mysqli_fetch_assoc($check) ) {
echo '<li class="menu-item-302">' . $row2['name'] . '</li>' ;
}
echo '</ul>' ;
} else {
echo '</li>' ;
}
}
?>
and this the result:enter image description here
i know it happen because the father item that hold the sub menu's called in $get once and another time when it need other css class.i tried if , foreach , while and many things.
i need that item that holds submenus should have "menu-item-has-children" class otherwise its not show the sub menus.

Answer 1

When a menu item, have childrens then echo menu with menu-item-has-children class, otherwise echo a simple menu and move on.
<?php
$get = mysqli_query($db , "SELECT * from menu where parent_id is NULL");
while ($rowmenu = mysqli_fetch_assoc($get)) {
$id = $rowmenu['id'] ;
$check = mysqli_query($db , "SELECT * from menu where parent_id = '$id'");
$haveSubMenu = mysqli_num_rows($check);
if($haveSubMenu)
echo '<li class="menu-item menu-item-type-post_type menu-item-object-page menu-item-656 menu-item-has-children"><a href="'. $rowmenu['link'] . '" >' . $rowmenu['name'] .'</a>' ;
else
echo '<li class="menu-item menu-item-type-post_type menu-item-object-page menu-item-656"><a href="'. $rowmenu['link'] . '" >' . $rowmenu['name'] .'</a>' ;
if ($haveSubMenu)
{
echo '<ul class="sub-menu">' ;
while ( $row2 = mysqli_fetch_assoc($check) ) {
echo '<li class="menu-item-302">' . $row2['name'] . '</li>' ;
}
echo '</ul>' ;
} else {
echo '</li>' ;
}
}
?>

Answer 2

Here is see that Our Products from the database is printed twice.This is because you are fetching and echoing $row2['name'] twice but you can handle the first coming array by changing your SQL query.
Change your first SQL query to
$get = mysqli_query($db , "SELECT * from menu where parent_id is NULL AND name != 'OUR PRODUCTS'");
Querying this to database you will get all the NULL values in the array excluding OUR PRODUCTS and thus it will be echoed only once by the second query.

Answer 3

I tried the above to answer all is working only a few level dropdown menu.
If you want to really create your menu for multi-level dropdown as like tree structure just use the bellow function. for more detailing once go through
https://bootstrapfriendly.com/blog/dynamic-multi-level-dropdown-sticky-menu-in-php-mysql-using-bootstrap/
<?php
include_once("connection.php");
$query = "SELECT id, label, link, parent FROM menus ORDER BY sort ASC, label";
$result = mysqli_query($conn, $query) or die("database error:" . mysqli_error($conn));
// Create an array to conatin a list of items and parents
$menus = array(
'items' => array(),
'parents' => array()
);
// Builds the array lists with data from the SQL result
while ($items = mysqli_fetch_assoc($result)) {
// Create current menus item id into array
$menus['items'][$items['id']] = $items;
// Creates list of all items with children
$menus['parents'][$items['parent']][] = $items['id'];
//echo $items;
}
// function to create dynamic treeview menus
function createMenu($parent, $menu)
{
$html = "";
if (isset($menu['parents'][$parent])) {
// $html .= '<ul class="sina-menu sina-menu-right" data-in="fadeInLeft" data-out="fadeInOut">';
foreach ($menu['parents'][$parent] as $itemId) {
if (!isset($menu['parents'][$itemId])) {
$html .= "<li >
<a href='" . $menu['items'][$itemId]['link'] . "'>" . $menu['items'][$itemId]['label'] . "</a>
</li>";
}
if (isset($menu['parents'][$itemId])) {
$html .= "<li class='dropdown'>
<a class='dropdown-toggle' data-toggle='dropdown' href='" . $menu['items'][$itemId]['link'] . "'>" . $menu['items'][$itemId]['label'] . "</a>";
$html .= '<ul class="dropdown-menu">';
$html .= createMenu($itemId, $menu);
$html .= '</ul>';
$html .= "</li>";
}
}
// $html .= "</ul>";
}
return $html;
}