how to get only domain name from given URL in php

Question

Suppose i have given URL, i want to get only domain name. How do i achieve this in Php.

    +----------------------+------------+
    | input                | output     |
    +----------------------+------------+
    | www.google.com       | google     |
    | www.mail.yahoo.com   | mail.yahoo |
    | www.mail.yahoo.co.in | mail.yahoo |
    | www.abc.au.uk        | abc        |
     www.subdoamin.domain.co.in     // output subdomain

I applied the follwing trick but fails when i have TLD like "co.uk"

     if(isset($project_detail_all[0]->d_name)) {
        $domain_name = $project_detail_all[0]->d_name ;
        $domain_name = explode('.', $domain_name);
        $count = count($domain_name);
        if (top_level_domains($domain_name[$count-1]) && 
            stristr($rss_url, $domain_name[$count-2])) {
            return  isValidXML($rss_url);
        } else {
            return  ['status'=>false , 'invalid_Domain'=>true];
        }
    } else {
        return  ['status'=>false , 'invalid_Domain'=>true];
    }

Kindly help me

Look into this library function, this may help you very much! [parse_url](http://php.net/manual/en/function.parse-url.php) — Shaunak Shukla, May 19 '17 at 11:41
look this answer on [SO](http://stackoverflow.com/a/37987345/2679536), which can extract all parts of the url, from TLDExtract library. — Shaunak Shukla, May 19 '17 at 12:05

Kurt Van den Branden · Accepted Answer · 2017-05-19T12:05:36.740

2

What you are looking for is parse_url()

<?php

    $url = parse_url("http://abc.news18.co.in");
    //echo $url['host'];
    echo preg_replace("/^([a-zA-Z0-9].*\.)?([a-zA-Z0-9][a-zA-Z0-9-]{1,61}[a-zA-Z0-9]\.[a-zA-Z.]{2,})$/", '$2', $url['host']); 

?>

edited May 19 '17 at 12:05

answered May 19 '17 at 11:45

Kurt Van den Branden

11,995
10
76
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$domain_name = 'abc.news18.co.in/abc.php'; – wasim beg May 19 '17 at 11:51
I added a regex which strips the subdomain. – Kurt Van den Branden May 19 '17 at 12:05
its giving me news18.co.in , but i want only new18 and tld ('co.in') respectively – wasim beg May 19 '17 at 12:24
and again if url does not containing " http:// " parse_url will not work – wasim beg May 19 '17 at 12:26
Then split the result: `$pieces = explode(" ", $url['host']); echo $pieces[1];` – Kurt Van den Branden May 19 '17 at 12:32
And if it does not contain "http://", just add it: `$url = "abc.news18.co.in/page-to-article?whateverparameter=something"; if (strpos($url, 'http://') == false) $url = "http://" . $url;` – Kurt Van den Branden May 19 '17 at 12:41
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/144658/discussion-between-kurt-van-den-branden-and-wasim-beg). – Kurt Van den Branden May 19 '17 at 12:46
Thanks for solution , now i can tailor it accordingly – wasim beg May 19 '17 at 12:53

Bhaumik Pandhi · Answer 2 · 2017-05-19T12:06:26.560

1

please use regex to get only domain.

/.*?\.([A-z\d\.]+)(.co|.au)/s

You can see my solution result here. green colored text is result of regex.

Use below code.

$urls = ['www.google.com','www.mail.yahoo.com','www.mail.yahoo.co.in','www.abc.au.uk','www.subdoamin.domain.co.in','abc.news18.co.in/abc.php'];
$result = [];
foreach($urls as $url){
   preg_match('/.*?\.([A-z\d\.]+)(.co|.au)/s',$url,$match);
   $result[] = $match[1];
}
echo '<pre>';print_r($result);

And you'll get result whatever you want.

edited May 19 '17 at 12:06

answered May 19 '17 at 11:49

Bhaumik Pandhi

2,655
2
21
38

1

It was not working before, now I've changed my answer and regex so it will work now, please open new updated link and you can get idea, – Bhaumik Pandhi May 19 '17 at 11:53

how to get only domain name from given URL in php

2 Answers2