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03.29.2006 at 07:29AM PST, ID: 21793651
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get domain name in php

Tags: php, domain, get, name
Im want to verify the domain name where my scripts are running

what I started with is

$_SERVER['HTTP_HOST']

the problem with this is that the result could be www.domain.com or domain.com

where I would like it to be domain.com always at all times only because I will be comparing the domain name with a field in the database to verify if they are allowed to be using my application anyone knows if there is a predefined variable for this or do i have to cut off the www.  in code ?

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Question Stats
Zone: Web Development
Question Asked By: prowebinteractive
Solution Provided By: tolgaong
Participating Experts: 6
Solution Grade: A
Views: 563
Translate:
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03.29.2006 at 07:32AM PST, ID: 16322604

Rank: Sage

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03.29.2006 at 07:33AM PST, ID: 16322609

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03.29.2006 at 07:47AM PST, ID: 16322769

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03.29.2006 at 07:51AM PST, ID: 16322830

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03.29.2006 at 08:05AM PST, ID: 16322997

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03.29.2006 at 08:14AM PST, ID: 16323083

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03.29.2006 at 08:24AM PST, ID: 16323193

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03.29.2006 at 08:39AM PST, ID: 16323375

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03.29.2006 at 08:45AM PST, ID: 16323444

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03.29.2006 at 08:46AM PST, ID: 16323451

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03.29.2006 at 07:13PM PST, ID: 16328981

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04.19.2006 at 01:11AM PDT, ID: 16485289

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03.29.2006 at 07:32AM PST, ID: 16322604

Rank: Sage

Nope, I think you have to cut it off in the code.
$domain = preg_replace("/^www\./", "", $_SERVER['HTTP_HOST']);
 
03.29.2006 at 07:33AM PST, ID: 16322609
there is no such variable,
I think you should cut off the "www." in the code.
 
03.29.2006 at 07:47AM PST, ID: 16322769
Just simply use
$domain = str_replace("www.","",$_SERVER['HTTP_HOST']);
 
03.29.2006 at 07:51AM PST, ID: 16322830

Rank: Sage

Problem with str_replace is if your domain is something like 'mysiteonwww.com', you'll come away with just 'mysiteoncom'
 
03.29.2006 at 08:05AM PST, ID: 16322997
and what if its a subdomain, or there is no www.
 
03.29.2006 at 08:14AM PST, ID: 16323083

Rank: Sage

if there's no www, no problem.

The below will return 'domain.com' for all of the following:
www.domain.com
domain.com
subdomain.domain.com
www.subdomain.domain.com

$domain = preg_replace("/^(.*\.)?([^.]*\..*)$/", "$2", $_SERVER['HTTP_HOST']);
 
03.29.2006 at 08:24AM PST, ID: 16323193

Rank: Sage

Here are a few other ways to do it:

preg_match("/[^.]*\.[^.]*$/", $_SERVER['HTTP_HOST'], $domain);
echo $domain[0];

$pieces = explode(".", strrev($_SERVER['HTTP_HOST']));
echo strrev($pieces[0] . "." . $pieces[1]);
 
03.29.2006 at 08:39AM PST, ID: 16323375
holy sweet jesus  nice but confusing, it works and thats what counts, I have tried it on every type of domain yet though  such as domain.qc.ca   dpo you know if it will work on this domain name ?
 
03.29.2006 at 08:45AM PST, ID: 16323444

Rank: Master



Just a quick pointer - I achieve the same using the IP Address of the server:

   $_SERVER['SERVER_ADDR']

  ..and I compare this with the IP address to which the domain name should resolve.

Basically, this avoids you having to do *any* additional comparison or string massaging.
 
03.29.2006 at 08:46AM PST, ID: 16323451

Rank: Sage

No, that would just return qc.ca
 
03.29.2006 at 07:13PM PST, ID: 16328981

Rank: Master

So, you only want to cut off "www." if "www." is at the beginning?
That's an easy one:

<?php

$_domains = array(
    "www.domain.qc.ca",
    "www.domain.com",
    "www.subdomain.domain.com",
    "domain.qc.ca",
    "www.domain.org",
    "domain.org",
    "subdomain.company.com",
    "mysiteonwww.com",
    "www.mysiteonwww.com",
    "www.www.com"
);

foreach ($_domains as $_domain) {
    echo preg_replace("/^www\./", "", $_domain);
    echo "<BR>\n";
}

?>

--brian
 
04.19.2006 at 01:11AM PDT, ID: 16485289

Rank: Master

$subject="http://www.experts-exchange.com";
if (preg_match('/\\b(https?|ftp):\/\/www\.(?P<domain>[-A-Z0-9.]+)\\z/i', $subject, $regs)) {
      $result = $regs['domain'];
      }
echo($result);
Accepted Solution
 
 
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