Menu active item with PHP - php

Question

I am trying to make menu with active class for every site.
And I don't know what I am doing wrong. Thnaks for help :D
<li class="nav-item <?php echo $home; ?>">
<a class="nav-link" href="/">Home</a>
</li>
<li class="nav-item <?php echo $hry; ?>">
<a class="nav-link" href="/hry">Hry</a>
</li>
<?php
$host = $_SERVER['SERVER_NAME'] . $_SERVER['REQUEST_URI'];
if($host == 'myurl.com/') {
$home = 'active';
}
elseif ($host == 'myurl.com/hry') {
$hry = 'active';
}
else {
$home = 'non-active';
$hry = 'non-active';
}
?>

Answer 1

Looking at your code, I see problem lies here: You are setting the variables $home $hry after HTML.
You should set them before you output html. Should be like this:
<?php
$host = $_SERVER['SERVER_NAME'] . $_SERVER['REQUEST_URI'];
if($host == 'myurl.com/') {
$home = 'active';
}
elseif ($host == 'myurl.com/hry') {
$hry = 'active';
}
else {
$home = 'non-active';
$hry = 'non-active';
}
?>
<li class="nav-item <?php echo $home; ?>">
<a class="nav-link" href="/">Home</a>
</li>
<li class="nav-item <?php echo $hry; ?>">
<a class="nav-link" href="/hry">Hry</a>
</li>

Answer 2

You can always write a helper function:
function checkIfActive($route){
if(strpos($_SERVER['REQUEST_URI'], $route) !== false){
return "is-active";
}
return null;
}
And insert the helper function into your html for every menu item. It will check to see if its active based on the string you put in
<li class="nav-item <?php checkIfActive('string_for_home_url') ?>">
<a class="nav-link" href="/">Home</a>
</li>
<li class="nav-item <?php checkIfActive('string_for_hry_url'); ?>">
<a class="nav-link" href="/hry">Hry</a>
</li>
This will be a lot easier for you once you application and menu items start growing