I am trying to include a specific css file depending on the URL provided by the browser.
My website has the following format:
Header information
php include for content
Footer information
I am doing this so when I change my header menu links, I don't have to change every page on the website.
The php include code is as follows:
<?php
$page = $_GET['page'];
$pages = array('main', 'contact');
if (!empty($page)) {
if(in_array($page,$pages)) {
$page .= '.php';
include($page);
}
else {
include('404.php');
}
}
else {
include('main.php');
}
?>
The URL is made up as follows
index.php?page= for example index.php?page=contact
As part of the php include code, when the page=???? element isn't within the array it loads page 404.php which is an error page.
As can be seen, the array currently contains two pages, main (the home page content) and contact (the contact form). If a user tries to load a page called test (index.php?page=test) my webpage will display the 404 page (however doesn't load index.php?page=404 - the URL in browser stays as index.php?page=test.
What I'm trying to achieve is to load 404.css file when page=test(or any other page name not included in the array) is loaded.
Does anyone know how i'd go about achieving this? I've tried writing an if statement in the header to load 404.css however I was using strstr that fetches the browser URL (so test, not 404), therefore the correct css file doesn't get loaded.
Any help is much appreciated (or a better way to achieve what i'm trying to do).
Thanks in advance
Iain