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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

static_rtti
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kobibo
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  • Can u use an external lib? – Phung Duy Phong May 17 '17 at 10:10
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    Possible duplicate of [how to get domain name from URL](http://stackoverflow.com/questions/569137/how-to-get-domain-name-from-url) – guidot May 17 '17 at 10:15
  • 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. – kobibo May 17 '17 at 10:20
  • Use [**`urllib.parse`**](https://docs.python.org/3/library/urllib.parse.html) – Peter Wood May 17 '17 at 10:42
  • Gather a list of top-level domains, split your url by dots, right-strip your url from TLD, extract name. – Pearley May 17 '17 at 10:10
  • Does this answer your question? [Get protocol + host name from URL](https://stackoverflow.com/questions/9626535/get-protocol-host-name-from-url) – philshem Nov 17 '20 at 20:20

8 Answers8

41

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'
akash karothiya
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    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: https://github.com/john-kurkowski/tldextract#note-about-caching – alphazwest Jul 30 '19 at 15:14
4

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

Mariano Anaya
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    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` – kobibo May 17 '17 at 11:14
  • 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. – Nairum Jul 24 '20 at 18:35
2

Simple solution via regex

import re

def domain_name(url):
    return url.split("www.")[-1].split("//")[-1].split(".")[0]
2

For extracting domain from url

from urllib.parse import urlparse

url = "https://stackoverflow.com/questions/44021846/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
Kapil_Khatik
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0

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.

oddRaven
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0

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]
Denis
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0
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

Jup
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  • 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](/help/how-to-answer). – Community Dec 26 '22 at 02:12
0
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
gies0r
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