Website Name extract in Python

Question

I want to extract website names from the url. For e.g. https://plus.google.com/in/test.html should give the output as - "plus google"

Some more testcases are -

WWW.OH.MADISON.STORES.ADVANCEAUTOPARTS.COM/AUTO_PARTS_MADISON_OH_7402.HTML

Output:- OH MADISON STORES ADVANCEAUTOPARTS

WWW.LQ.COM/LQ/PROPERTIES/PROPERTYPROFILE.DO?PROPID=6054

Output:- LQ

WWW.LOCATIONS.DENNYS.COM

Output:- LOCATIONS DENNYS

WV.WESTON.STORES.ADVANCEAUTOPARTS.COM

Output:- WV WESTON STORES ADVANCEAUTOPARTS

WOODYANDERSONFORDFAYETTEVILLE.NET/

Output:- WOODYANDERSONFORFAYETTEVILLE

WILMINGTONMAYFAIRETOWNCENTER.HGI.COM

Output:- WILMINGTONMAYFAIRETOWNCENTER HGI

WHITEHOUSEBLACKMARKET.COM/

Output:- WHITEHOUSEBLACKMARKET

WINGATEHOTELS.COM

Output:- WINGATEHOTELS

string = str(input("Enter the url "))
new_list = list(string)
count=0
flag=0

if 'w' in new_list:
    index1 = new_list.index('w')
    new_list.pop(index1)
    count += 1
if 'w' in new_list:
    index2 = new_list.index('w')
    if index2 != -1 and index2 == index1:
        new_list.pop(index2)
        count += 1
if 'w' in new_list:
    index3= new_list.index('w')
    if index3!= -1 and index3== index2 and new_list[index3+1]=='.':
        new_list.pop(index3)
        count+=1      
        flag = 1
if flag == 0:
    start = string.find('/')
    start += 2
    end = string.rfind('.')

    new_string=string[start:end]
    print(new_string)
elif flag == 1:
    start = string.find('.')
    start = start + 1
    end = string.rfind('.')

    new_string=string[start:end]
    print(new_string)

The above works for some testcases but not all. Please help me with it.

Thanks

hiro protagonist · Accepted Answer · 2018-03-14 08:04:58Z

3

this is something you could build upon; using urllib.parse.urlparse:

from urllib.parse import urlparse

tests = ('https://plus.google.com/in/test.html',
         ('WWW.OH.MADISON.STORES.ADVANCEAUTOPARTS.COM/'
          'AUTO_PARTS_MADISON_OH_7402.HTML'),
         'WWW.LQ.COM/LQ/PROPERTIES/PROPERTYPROFILE.DO?PROPID=6054')

def extract(url):
    # urlparse will not work without a 'scheme'
    if not url.startswith('http'):
        url = 'http://' + url
    parsed = urlparse(url).netloc
    split = parsed.split('.')[:-1] # get rid of TLD
    if split[0].lower() == 'www':
        split = split[1:]
    ret = ' '.join(split)
    return ret

for url in tests:
    print(extract(url))

edited Mar 14, 2018 at 8:04

answered Mar 14, 2018 at 7:24

hiro protagonist

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Comments

JackNorthrup · Accepted Answer · 2018-03-14 08:34:46Z

1

The function strips the url from the double slash to the single slash: the rest is 'clean up'

def stripURL( url, TwoSlashes, OneSlash ):
    try:
        start = url.index(TwoSlashes) + len(TwoSlashes)
        end = url.index( OneSlash, start )
        return url[start:end]
    except ValueError:
        return ""
url= raw_input("URL : ")
if "www." in url:url=url.replace("www.","")
Strip = stripURL( url, "//", "/" )
# Strips anything after the last period found
Stripped = Strip[:Strip.rfind(".")]
# get rid of the any periods used in the name 
Stripped = Stripped.replace("."," ")
print Stripped

answered Mar 14, 2018 at 8:34

JackNorthrup

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3 Comments

Anubhav Sarangi Over a year ago

When I run the above code with input(for e.g.) - "www.google.com", there is no output.

JackNorthrup Over a year ago

The example used the format including http:// . I used the slashes to separate the url. copy a url from the address bar and it works

Anubhav Sarangi Over a year ago

Yeah...now it works fine but I want that the code should support urls without https:// or http://..... Anyways thanks for helping

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