Dynamic menu from PHP array - php

Question

Hey I have question about Dynamic menu which is created in php.
Code is from stackoverflow, what I want is to get my parent styled with red color if children of those parent is selected, here is code:
$menu = Array(
Array(
'title' => 'Home',
'link' => 'a'
),
Array(
'title' => 'Parent',
'link' => 'b',
'children' => Array(
Array(
'title' => 'Sub 1',
'link' => 'c'
),
Array(
'title' => 'Sub 2',
'link' => 'd'
),
)
)
);
function buildMenu($menuArray)
{
foreach ($menuArray as $node)
{
$selected = ($node['link']== $_GET['menu']) ? $selected = 'style="color: red;"' : null;
echo "<li ".$selected."><a href='?menu=".$node['link']."'/>" . $node['title'] . "</a>";
if ( ! empty($node['children'])) {
echo "<ul>";
buildMenu($node['children']);
echo "</ul>";
}
echo "</li>";
}
}
buildMenu($menu);
So how it needs to go:
Home
Parent - selected
Sub 1 - selected
Sub 2
or
Home
Parent - selected
Sub 1
Sub 2 - selected
Hope someone understand what i want? If my children under parent is selected also parent needs to be selected.

Answer 1

I have added one function to check element in children array. May be something better solution are there. But at this its quick solution for you :)
$menu = array(
array(
'title' => 'Home',
'link' => 'a'
),
array(
'title' => 'Parent',
'link' => 'b',
'children' => array(
array(
'title' => 'Sub 1',
'link' => 'c'
),
array(
'title' => 'Sub 2',
'link' => 'd'
),
)
)
);
function buildMenu($menuArray) {
foreach ($menuArray as $node) {
$getMenu = isset($_GET['menu']) ? $_GET['menu'] : '';
$checkParent = (isset($node['children']) && !empty($node['children'])) ? checkInChildArray($getMenu, $node['children']) : '';
$parentSelected = ($checkParent) ? $selected = 'style="color: red;"' : null;
echo "<li " . $parentSelected . "><a href='?menu=" . $node['link'] . "'>" . $node['title'] . "</a></li>";
if (isset($node['children']) && !empty($node['children'])) {
echo "<ul>";
foreach ($node['children'] as $subMenu) {
$childSelected = ($subMenu['link'] == $getMenu) ? $selected = 'style="color: red;"' : null;
echo "<li " . $childSelected . "><a href='?menu=" . $subMenu['link'] . "'>" . $subMenu['title'] . "</a></li>";
}
echo "</ul>";
}
echo "</li>";
}
}
// Checking if selected menu inside children array.
function checkInChildArray($needle, $haystack, $strict = false) {
foreach ($haystack as $item) {
if (($strict ? $item['link'] === $needle : $item == $needle) || (is_array($item) && checkInChildArray($needle, $item, $strict))) {
return true;
}
}
return false;
}
echo buildMenu($menu);
Working Demo

Answer 2

Use jQuery to add parent li a background color
$('li.selected').parent().closest('li').css("color","red");

Answer 3

Pass every menu level to the url. You can add the "selected" class for every menu level. So:
$current_menu_level_1 = (isset($_GET['menu_level_1'])) ? $_GET['menu_level_1'] : false;
$current_menu_level_2 = (isset($_GET['menu_level_2'])) ? $_GET['menu_level_2'] : false;
When building your menu, compare the 'to be build item' to the $current_menu_level1/2 variable and add echo the class when they are the same.

Answer 4

please, consider create a css class selected for <li> elements and <ul> elements.
With PHP insert this styles when needed, Like this:
$selected = ($node['link']== $_GET['menu']) ? $selected = 'selected' : '';
echo "<li class='".$selected."'>";
echo "<ul class='".$selected."'>";
buildMenu($node['children']);
echo "</ul>";