Find out subdomain using Regular Expression in PHP

Question

Sorry if this is too little of a challenge to be suited as a stack overflow question, but I'm kind of new to Regular Expressions.

My question is, what is the regular expression that returns the string "token" for all the examples bellow?

token.domain.com
token.domain.com/
token.domain.com/index.php
token.domain.com/folder/index.php
token.domain.com/folder/subfolder
token.domain.com/folder/subfolder/index.php
(added after edit)
domain.com
domain.com/

Im trying to use inside an preg_replace function in order to find out the subdomain of the current page from the $_SERVER['SERVER_NAME'] variable.

Thank you in advance, titel

Edit note: Sorry chaos and sebnow , I edited my question, what I initially meant, but forgot to write, was that this would work without any subdomain at all - case in which it would return an empty sting or NULL

2 revs user21926 · Accepted Answer · 2009-02-11 04:20:51Z

4

preg_replace('/^(?:([^\.]+)\.)?domain\.com$/', '\1', $_SERVER['SERVER_NAME'])

Edit: What does the following output for you:

<?php
    echo preg_replace('/^(?:([^\.]+)\.)?domain\.com$/', '\1', "sean.domain.com") . "<br />";
    echo preg_replace('/^(?:([^\.]+)\.)?domain\.com$/', '\1', "titel.domain.com") . "<br />";
    echo preg_replace('/^(?:([^\.]+)\.)?domain\.com$/', '\1', "domain.com") . "<br />";
?>

edited Feb 11, 2009 at 4:20

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2 revs
user21926

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2 Comments

titel Over a year ago

:( unfortunately the solution you described does not seem to work in my case

titel Over a year ago

It does work now, I guess I misspelled something while I was passing the script to use my actual domain. Anyways, thanks a lot. Constantin TOVISI

sebnow · Accepted Answer · 2009-02-11 04:05:19Z

3

The following regex would capture any string before a full stop (the subdomain) and the rest of the domain:

^([^.]+)\..*$

I don't see the need for regex to do this though. It would be much easier to split by the full stop and get the first element:

list($subdomain, $rest) = explode('.', $_SERVER['SERVER_NAME'], 2);

answered Feb 11, 2009 at 4:05

sebnow

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chaos · Accepted Answer · 2009-02-11 03:51:46Z

0

If you want to use preg_replace(), then:

preg_replace('/\..*/', '', $_SERVER['SERVER_NAME'])

answered Feb 11, 2009 at 3:51

chaos

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2 revs, 2 users 67% · Accepted Answer · 2012-08-18 11:25:03Z

0

preg_match('@^([a-zA-Z0-9]*\.)?([a-zA-Z0-9]+)\.([a-zA-Z]{2,4})$@',$_SERVER["HTTP_HOST"], $matches);

edited Aug 18, 2012 at 11:25

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2 revs, 2 users 67%
robo

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