Extract domain name from URL in Python

Question

I am tring to extract the domain names out of a list of URLs. Just like in https://stackoverflow.com/questions/18331948/extract-domain-name-from-the-url
My problem is that the URLs can be about everything, few examples:
m.google.com => google
m.docs.google.com => google
www.someisotericdomain.innersite.mall.co.uk => mall
www.ouruniversity.department.mit.ac.us => mit
www.somestrangeurl.shops.relevantdomain.net => relevantdomain
www.example.info => example
And so on..
The diversity of the domains doesn't allow me to use a regex as shown in how to get domain name from URL (because my script will be running on enormous amount of urls from real network traffic, the regex will have to be enormous in order to catch all kinds of domains as mentioned).
Unfortunately my web research the didn't provide any efficient solution.
Does anyone have an idea of how to do this ?
Any help will be appreciated !
Thank you

Controller

Can u use an external lib? – Fusillade

Gather a list of top-level domains, split your url by dots, right-strip your url from TLD, extract name. – Cherish

Possible duplicate of how to get domain name from URL – Seersucker

Yes, I can use external libs. It is not a duplication (I even attached a link to this thread), I couldn't find a satisfying answer there. – Controller

Use urllib.parse – Avulsion

Does this answer your question? Get protocol + host name from URL – Dodds

Answer 1

Use tldextract which is more efficient version of urlparse, tldextract accurately separates the gTLD or ccTLD (generic or country code top-level domain) from the registered domain and subdomains of a URL.

>>> import tldextract
>>> ext = tldextract.extract('http://forums.news.cnn.com/')
ExtractResult(subdomain='forums.news', domain='cnn', suffix='com')
>>> ext.domain
'cnn'

Reservoir

Note: the tldextract library makes an http request upon initial install and creates a cache of the latest tld data. This can raise a permission error for some remote deployments. See here: github.com/john-kurkowski/tldextract#note-about-caching – Dogma

Answer 2

It seems you can use urlparse https://docs.python.org/3/library/urllib.parse.html for that url, and then extract the netloc.

And from the netloc you could easily extract the domain name by using split

Arellano

Thank you for your response, unfortunately, using urlparse on url like m.city.domain.com returned me ParseResult(scheme='', netloc='', path='m.city.domain.com', params='', query='', fragment=''), while the expected output was domain – Controller

Use a valid URL (//m.city.domain.com/), not a something like (m.city.domain.com). Nobody can guess what did you pass when you removed backslashes. – Grainfield

@Controller urlparse follows rfc 1808 syntax which requires // before net_loc docs.python.org/3/library/urllib.parse.html – Only

Answer 3

For extracting domain from url

from urllib.parse import urlparse

url = "https://mcmap.net/q/527702/-extract-domain-name-from-url-in-python"
domain = urlparse(url).netloc
"stackoverflow.com"

For check domain is exist in url

if urlparse(url).netloc in ["domain1", "domain2", "domain3"]:
           do something

Mcshane

pu.netloc may include port. You might want pu.hostname instead (to get the domain without port). – Only

Answer 4

Simple solution via regex

import re

def domain_name(url):
    return url.split("www.")[-1].split("//")[-1].split(".")[0]

Geraldo

Gets the first part of the domain, not the actual domain. Only works for things like www.google.com – Oversell

Unreliable solution, avoid. – Andersonandert

Answer 5

With regex, you could use something like this:

(?<=\.)([^.]+)(?:\.(?:co\.uk|ac\.us|[^.]+(?:$|\n)))

https://regex101.com/r/WQXFy6/5

Notice, you'll have to watch out for special cases such as co.uk.

Intercessor

Answer 6

Check the replace and split methods.

PS: ONLY WORKS FOR SIMPLE LINKS LIKE https://youtube.com (output=youtube) AND (www.user.ru.com) (output=user)

def domain_name(url):

return url.replace("www.","http://").split("//")[1].split(".")[0]

Mammon

Answer 7

import re
def getDomain(url:str) -> str:
    '''
        Return the domain from any url
    '''
    # copy the original url text
    clean_url = url

    # take out protocol
    reg = re.findall(':[0-9]+',url)
    if len(reg) > 0:
        url = url.replace(reg[0],'')
    
    # take out paths routes
    if '/' in url:
        url = url.split('/')

    # select only the domain
    if 'http' in clean_url:
        url = url[2]

    # preparing for next operation
    url = ''.join(url)

    # select only domain
    url = '.'.join(url.split('.')[-2:])

    return url

Internuncio

Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Levitate

Answer 8

from urllib.parse import urlparse
import validators

    hostnames = []
    counter = 0
    errors = 0
    for row_orig in rows:
        try:
            row = row_orig.rstrip().lstrip().split(' ')[1].rstrip()
            if len(row) < 5:
                print(f"Empty row {row_orig}")
                errors += 1
                continue
            if row.startswith('http'):
                domain = urlparse(row).netloc # works for https and http
            else:
                domain = row

            if ':' in domain:
                domain = domain.split(':')[0] # split at port after clearing http/https protocol 

            # Finally validate it
            if validators.domain(domain):
                pass
            elif validators.ipv4(domain):
                pass
            else:
                print(f"Invalid domain/IP {domain}. RAW: {row}")
                errors +=1
                continue

            hostnames.append(domain)
            if counter % 10000 == 1:
                print(f"Added {counter}. Errors {errors}")
            counter+=1
        except:
            print("Error in extraction")
            errors += 1

Unwish

Answer 9

tests = {
  "m.google.com": 'google',
  "m.docs.google.com": 'google',
  "www.someisotericdomain.innersite.mall.co.uk": 'mall',
  "www.ouruniversity.department.mit.ac.us": 'mit',
  "www.somestrangeurl.shops.relevantdomain.net": 'relevantdomain',
  "www.example.info": 'example',
  "github.com": 'github',
}

def get_domain(url, loop=0, data={}):

  dot_count = url.count('.')

  if not dot_count:
    raise Exception("Invalid URL")

  # basic
  if not loop:
    if dot_count < 3:
      data = {
        'main':  url.split('.')[0 if dot_count == 1 else 1]
        }

  # advanced
  if not data and '.' in url:
      if dot_count > 1:
        loop += 1
        start = url.find('.')+1
        end = url.rfind('.') if dot_count != 2 else None
        return get_domain(url[start:end], loop, data)
      else:
        data ={
          'main': url.split('.')[-1]
          }

  return data

for u, v in tests.items():
  print(get_domain(u))

Tupper