changing a given url to return only the domain

Question

I'm getting a url from a form, this way:

$input_website = isset($_POST['website']) ? check_plain($_POST['website']) : 'None';

I need to get back a naked domain name(for some API integration), for example: http://www.example.com will return as example.com and www.example.com will return example.com etc.

I have this code now, that returns the correct url for the first case http://www.example.com but returns nothing for www.example.com or even example.com:

function get_domain($url)
{
    $pieces = parse_url($url);
    $domain = isset($pieces['host']) ? $pieces['host'] : '';
    if (preg_match('/(?P<domain>[a-z0-9][a-z0-9\-]{1,63}\.[a-z\.]{2,6})$/i', $domain, $regs)) {
        return $regs['domain'];
    }
    return false;
}

Can you please advice on the matter?

Murad Hasan · Answer 1 · 2016-05-03T11:55:26.970

2

As per discussion with you:

$url = 'www.noamddd.com';
$arrUrl = explode("/", $url);
echo $arrUrl[0];

Old Answer:

Make a function with the following code block and get the domain names.

Try this

more about parse_url

$url = 'http://google.com/dhasjkdas/sadsdds/sdda/sdads.html';
$parse = parse_url($url);
print $parse['host']; //google.com

Also you can do this in another way:

echo $domain = str_ireplace('www.', '', parse_url($url, PHP_URL_HOST));//google.com

edited May 03 '16 at 11:55

answered May 03 '16 at 09:43

Murad Hasan

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thanks. iv'e changed the code to be tis 'code' function get_domain($url) { $parse = parse_url($url); return $parse['host']; } $webSite = get_domain($input_website); 'code' but it's coming back empty again. – Noam Gur May 03 '16 at 10:17
can you please post your `url`?? – Murad Hasan May 03 '16 at 10:21
I'm receiving the url from an html form, this is a url for example that ive tried and came back empty www.noamddd.com – Noam Gur May 03 '16 at 11:32
use the url with `http://` like `http://www.noamddd.com`. otherwise it will not a url, just a string. – Murad Hasan May 03 '16 at 11:35
As i've said on the question, the function is returning the domain name when the user is return a url with http, my problem is when the user brings the url without http – Noam Gur May 03 '16 at 11:51
Unfortunlly, still not working. the new code you wrote brings back www.noamddd.com instead of noamddd.com, and for http://www.noamddd.com it returns http: – Noam Gur May 03 '16 at 14:15

score 0 · Answer 2 · edited May 23 '17 at 12:16

0

If you just have the URL (and not want the current domain name like frayne-konok suggests) and want to extract the server name, you can use a regular expression like this:

$serverName = preg_replace('|.*?://(.*?)/.*|', '$1', $url);

edited May 23 '17 at 12:16

Community

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answered May 03 '16 at 09:46

JSchirrmacher

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score 0 · Accepted Answer · answered May 05 '16 at 12:03

I ended up doing something a bit different - checking if there is http and if not, i'm adding it using this function:

function addHttp($website) {
    if (!preg_match("~^(?:f|ht)tps?://~i", $url)) {
        $url = "http://" . $url;
    }
    return $website;
}

and only then i'm sending it to my other function that return the domain. For sure not the best way, but it works.

changing a given url to return only the domain

3 Answers3