[RESOLVED] How to pick domain name from an url string with php regex?

cemlouis

Hello,

I have a problem, I have several url strings like below:

http://www.mysite.com/?action=1234
http://www.mysite.co.uk/doit.php?=abc
http://www.mysite.com/category/myarticle/
http://mysite.com/myresearch/15.html
http://subdomain.mysite.com/thelist/abcd.php
http://mysite.co.uk/book/booking123.php
...
..
.

The list goes on and on, is there any regex func that fits all the endless posibilities to pick up the domain name "mysite.com" or "mysite.co.uk" or "yoursite.co.uk" or any other domain names from the url strings?

Any help greatly appreciated...
Thx

Installer

Why not use parse_url()?

function get_2ld($url)
{
    $parse_arr = parse_url($url);
    $host_arr = explode('.', $parse_arr['host']);
    $ct = count($host_arr);
    $domain = $host_arr[$ct - 2] . '.' . $host_arr[$ct - 1];
    if ($host_arr[$ct - 1] == 'uk') {
        $domain = $host_arr[$ct - 3] . '.' . $domain;
    }
    return $domain;
}

That will take care of the "co.uk" problem, and you can add code to fix any other anomalies you expect to see.

cemlouis

What about the others?

com.cn or com.ca or hk.tg etc.

I need a generic one...

Installer

A generic one? "co.uk", "com.cn", "ht.tg" are not generic. To do it completely you'd need a database.

you can add code

or even alter it:

function get_2ld($url)
{
    $anom_arr = array('uk', 'ca', 'tg');
    $parse_arr = parse_url($url);
    $host_arr = explode('.', $parse_arr['host']);
    $ct = count($host_arr);
    $domain = $host_arr[$ct - 2] . '.' . $host_arr[$ct - 1];
    if (in_array($host_arr[$ct - 1] == 'uk') {
        $domain = $host_arr[$ct - 3] . '.' . $domain;
    }
    return $domain;
}

cemlouis

I don't think so, can be done with below:

$string = 'http://mysite.co.uk/book/booking123.php';
preg_match('/[a-zA-Z0-9\-]+\.[a-zA-Z\.]{2,6}(?=\/)/',$string,$matches);
echo $matches[0];

Thanks anyway...

Installer wrote:

A generic one? "co.uk", "com.cn", "ht.tg" are not generic. To do it completely you'd need a database.

or even alter it:

function get_2ld($url)
{
    $anom_arr = array('uk', 'ca', 'tg');
    $parse_arr = parse_url($url);
    $host_arr = explode('.', $parse_arr['host']);
    $ct = count($host_arr);
    $domain = $host_arr[$ct - 2] . '.' . $host_arr[$ct - 1];
    if (in_array($host_arr[$ct - 1] == 'uk') {
        $domain = $host_arr[$ct - 3] . '.' . $domain;
    }
    return $domain;
}

bradgrafelman

Don't forget to mark this thread resolved.