PHP - Drop down not showing all items in mysql - php

Question

I am Trying to make a drop down menu, and list multiple items as well.
like for example:
here is my database:
if you see in that image you can see only one of the rows have a parent if i want multiple rows to have a parent of 'Far Far Away123' it only shows one of the items
<?php
function build_dropdown ($parent, $pages){
$items = "";
foreach($pages as $page){
// $items = "";
if($page['parent'] == $parent){
$items = $page;
} // END if
} // END foreach
if(is_array($items)){ // If a sub
echo '<ul id="sub_menu" class="sub_navagation'. $items['id'] .'">';
echo '<li>'.$items['page_name'].'</li>';
echo '</ul>';
} // END if
}// End Function
$sql = "SELECT * FROM pages ORDER by item_order";
$result = mysqli_query($db, $sql);
confirm_query($result);
while ($row = mysqli_fetch_assoc($result)) {
$pages[] = $row; // Add each row to $pages array to use later
}
foreach($pages as $key => $page){
if($page['parent'] == 'none'){ ?>
<li id = "<?php echo $page['id']; ?>">
<a href="page.php?id=<?php echo $page['id']; ?>" title="<?php echo $page['page_title']; ?>">
<?php echo $page['page_name']; ?>
</a>
<?php ?>
<?php
} // END if
build_dropdown($page['page_name'], $pages); // If there are child items then build them out
echo "</li> ";
} // END foreach
?>
Thanks

Answer 1

It is because you add only one.
You have a foreach searching all pages for subitems and remember only one of them for the latter if (is_array())
function build_dropdown($parent, $pages)
{
// item is an array
$items = array();
foreach($pages as $page) {
if ($page['parent'] == $parent) {
// add an element to the array
$items[] = $page;
} // END if
} // END foreach
if ($items) {
echo '<ul id="sub_menu" class="sub_navagation">';
foreach ($items as $item) {
echo '<li>'.$item['page_name'];
build_dropdown($item['page_name'], $pages);
echo '</li>';
} // END foreach
echo '</ul>';
} // END if
}// End Function
By the way, it would be better to start with the function giving build_dropdown($parent = 'none', $pages)

Answer 2

Try this inside your buildDropdown() function..
$items = array();
foreach($pages as $page)
{
if($page['parent'] == $parent)
{
$items[] = $page;
}
}
if (count($items) > 0)
{
echo '<ul id="sub_menu_'.$parent.'" class="sub_navagation">';
foreach($items as $item)
{
echo '<li>'.$item['page_name'].'</li>';
}
echo '</ul>';
}