[(site_url)] returning .co.uk and not .org.uk

32 Posts

creativecss Reply #1, 17 years, 5 months ago

Hi,

I am defining the path to my CSS files etc like the following

<base href="[(site_url)]"/>

<link charset="utf-8" title="CSS" media="screen" type="text/css" href="[(site_url)]assets/templates/theme/css/style.css" rel="stylesheet"/>

My Domain is .org.uk but I also have .co.uk pointed towards the site.

My problem is that whatever URL you use to visit the site the CSS path that is built from the [(site_url)] is the .co.uk domain. This does not cause me any problems however I am using some javascript developed by a third party that checks the URL that it is called from so the site_url shoud be .org.uk if thats the URL i used or .co.uk if I used that URL.

Hope this makes sense.

Anyone got any ideas or come across this before.

Thanks
Adam

MODX Staff
1,535 Posts

Garry Nutting Reply #2, 17 years, 5 months ago

The site_url is being cached with the page and that’s why it doesn’t update when the TLD changes.

You can circumvent this by creating a snippet called getSiteUrl:

<?php
return MODX_SITE_URL;

And then, calling this snippet uncached in place of the [(site_url)]:

<base href="[!getSiteUrl!]"/>

Garry Nutting
Senior Developer
MODX, LLC

Email: [email protected]
Twitter: @garryn
Web: modx.com

32 Posts

creativecss Reply #3, 17 years, 5 months ago

Thanks for the reply I will try returning the

$_SERVER["REQUEST_NAME"]

This way I should set the base to whatever domain they are using either .co.uk or .org.uk

386 Posts

therebechips Reply #4, 17 years, 5 months ago

$_SERVER["REQUEST_NAME"] will not give you the alias name. You need $_SERVER["HTTP_HOST"], but beware header injection - you could open yourself up to a XSS exploits if you just allow the header to be output to the page.

<?php
// array of safe domains
$domains=array(
	'www.mysite.co.uk',
	'www.mysite.org.uk',
	'mysite.co.uk',
	'mysite.org.uk'
	);
$server = $domains[0];
foreach ($domains as $key => $value) {
	if ($value == $_SERVER["HTTP_HOST"]) {
		$server = $value;
	}
}
return $server;
?>

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