How to get domain name from URL

I once had to write such a regex for a company I worked for. The solution was this:

• Get a list of every ccTLD and gTLD available. Your first stop should be IANA. The list from Mozilla looks great at first sight, but lacks ac.uk for example so for this it is not really usable.
• Join the list like the example below. A warning: Ordering is important! If org.uk would appear after uk then example.org.uk would match org instead of example.

Example regex:

.*([^\.]+)(com|net|org|info|coop|int|co\.uk|org\.uk|ac\.uk|uk|__and so on__)$

This worked really well and also matched weird, unofficial top-levels like de.com and friends.

The upside:

• Very fast if regex is optimally ordered

The downside of this solution is of course:

• Handwritten regex which has to be updated manually if ccTLDs change or get added. Tedious job!
• Very large regex so not very readable.

A little late to the party, but:

const urls = [
  'www.abc.au.uk',
  'https://github.com',
  'http://github.ca',
  'https://www.google.ru',
  'http://www.google.co.uk',
  'www.yandex.com',
  'yandex.ru',
  'yandex'
]

urls.forEach(url => console.log(url.replace(/.+\/\/|www.|\..+/g, '')))

/^(?:www\.)?(.*?)\.(?:com|au\.uk|co\.in)$/

I don't know of any libraries, but the string manipulation of domain names is easy enough.

The hard part is knowing if the name is at the second or third level. For this you will need a data file you maintain (e.g. for .uk is is not always the third level, some organisations (e.g. bl.uk, jet.uk) exist at the second level).

The source of Firefox from Mozilla has such a data file, check the Mozilla licensing to see if you could reuse that.

It is not possible without using a TLD list to compare with as their exist many cases like http://www.db.de/ or http://bbc.co.uk/ that will be interpreted by a regex as the domains db.de (correct) and co.uk (wrong).

But even with that you won't have success if your list does not contain SLDs, too. URLs like http://big.uk.com/ and http://www.uk.com/ would be both interpreted as uk.com (the first domain is big.uk.com).

Because of that all browsers use Mozilla's Public Suffix List:

https://en.wikipedia.org/wiki/Public_Suffix_List

You can use it in your code by importing it through this URL:

http://mxr.mozilla.org/mozilla-central/source/netwerk/dns/effective_tld_names.dat?raw=1

Feel free to extend my function to extract the domain name, only. It won't use regex and it is fast:

http://www.programmierer-forum.de/domainnamen-ermitteln-t244185.htm#3471878

Basically, what you want is:

google.com        -> google.com    -> google
www.google.com    -> google.com    -> google
google.co.uk      -> google.co.uk  -> google
www.google.co.uk  -> google.co.uk  -> google
www.google.org    -> google.org    -> google
www.google.org.uk -> google.org.uk -> google

Optional:

www.google.com     -> google.com    -> www.google
images.google.com  -> google.com    -> images.google
mail.yahoo.co.uk   -> yahoo.co.uk   -> mail.yahoo
mail.yahoo.com     -> yahoo.com     -> mail.yahoo
www.mail.yahoo.com -> yahoo.com     -> mail.yahoo

You don't need to construct an ever-changing regex as 99% of domains will be matched properly if you simply look at the 2nd last part of the name:

(co|com|gov|net|org)

If it is one of these, then you need to match 3 dots, else 2. Simple. Now, my regex wizardry is no match for that of some other SO'ers, so the best way I've found to achieve this is with some code, assuming you've already stripped off the path:

 my @d=split /\./,$domain;                # split the domain part into an array
 $c=@d;                                   # count how many parts
 $dest=$d[$c-2].'.'.$d[$c-1];             # use the last 2 parts
 if ($d[$c-2]=~m/(co|com|gov|net|org)/) { # is the second-last part one of these?
   $dest=$d[$c-3].'.'.$dest;              # if so, add a third part
 };
 print $dest;                             # show it

To just get the name, as per your question:

 my @d=split /\./,$domain;                # split the domain part into an array
 $c=@d;                                   # count how many parts
 if ($d[$c-2]=~m/(co|com|gov|net|org)/) { # is the second-last part one of these?
   $dest=$d[$c-3];                        # if so, give the third last
   $dest=$d[$c-4].'.'.$dest if ($c>3);    # optional bit
 } else {
   $dest=$d[$c-2];                        # else the second last
   $dest=$d[$c-3].'.'.$dest if ($c>2);    # optional bit 
 };
 print $dest;                             # show it

I like this approach because it's maintenance-free. Unless you want to validate that it's actually a legitimate domain, but that's kind of pointless because you're most likely only using this to process log files and an invalid domain wouldn't find its way in there in the first place.

If you'd like to match "unofficial" subdomains such as bozo.za.net, or bozo.au.uk, bozo.msf.ru just add (za|au|msf) to the regex.

I'd love to see someone do all of this using just a regex, I'm sure it's possible.

Could you just look for the word before .com (or other) (the order of the other list would be the opposite of the frequency see here

and take the first matching group i.e.

window.location.host.match(/(\w|-)+(?=(\.(com|net|org|info|coop|int|co|ac|ie|co|ai|eu|ca|icu|top|xyz|tk|cn|ga|cf|nl|us|eu|de|hk|am|tv|bingo|blackfriday|gov|edu|mil|arpa|au|ru)(\.|\/|$)))/g)[0]

You can test it could by copying this line into the developers' console on any tab

This example works in the following cases:

All answers here are very nice, but all will fails sometime. So i know it is not common to link something else, already answered elsewhere, but you'll find that you have to not waste your time into impossible thing. This because domains like mydomain.co.uk there is no way to know if an extracted domain is correct. If you speak about to extract by URLs, something that ever have http or https or nothing in front (but if it is possible nothing in front, you have to remove

filter_var($url, filter_var($url, FILTER_VALIDATE_URL))

here below, because FILTER_VALIDATE_URL do not recognize as url a string that do not begin with http, so may remove it, and you can also achieve with something stupid like this, that never will fail:

$url = strtolower('hTTps://www.example.com/w3/forum/index.php');

if( filter_var($url, FILTER_VALIDATE_URL) && substr($url, 0, 4) == 'http' )
{
// array order is !important
$domain = str_replace(array("http://www.","https://www.","http://","https://"), array("","","",""), $url);
$spos = strpos($domain,'/');
if($spos !== false)
{
 $domain = substr($domain, 0, $spos);
} } else { $domain = "can't extract a domain"; }

echo $domain;

Check FILTER_VALIDATE_URL default behavior here

But, if you want to check a domain for his validity, and ALWAYS be sure that the extracted value is correct, then you have to check against an array of valid top domains, as explained here: https://stackoverflow.com/a/70566657/6399448 or you'll NEVER be sure that the extracted string is the correct domain. Unfortunately, all the answers here sometime will fails.

P.s the unique answer that make sense here seem to me this (i did not read it before sorry. It provide the same solution, even if do not provide an example as mine above mentioned or linked): https://stackoverflow.com/a/569219/6399448

I know you actually asked for Regex and were not specific to a language. But In Javascript you can do this like this. Maybe other languages can parse URL in a similar way.

Easy Javascript solution

const domain = (new URL(str)).hostname.replace("www.", "");

Leave this solution in js for completeness.

You need a list of what domain prefixes and suffixes can be removed. For example:

Prefixes:

• www.

Suffixes:

• .com
• .co.in
• .au.uk

I know the question is seeking a regex solution but in every attempt it won't work to cover everything

I decided to write this method in Python which only works with urls that have a subdomain (i.e. www.mydomain.co.uk) and not multiple level subdomains like www.mail.yahoo.com

def urlextract(url):
  url_split=url.split(".")
  if len(url_split) <= 2:
      raise Exception("Full url required with subdomain:",url)
  return {'subdomain': url_split[0], 'domain': url_split[1], 'suffix': ".".join(url_split[2:])}

Let's say we have this: http://google.com

and you only want the domain name

let url = http://google.com;
let domainName = url.split("://")[1];
console.log(domainName);