url1='www.google.com'
url2='http://www.google.com'
url3='http://google.com'
url4='www.google'
url5='http://www.google.com/images'
url6='https://www.youtube.com/watch?v=6RB89BOxaYY
How to strip http(s) and www from url in Python?
4 Answers
You can use the string method replace:
url = 'http://www.google.com/images'
url = url.replace("http://www.","")
or you can use regular expressions:
import re
url = re.compile(r"https?://(www\.)?")
url = url.sub('', 'http://www.google.com/images').strip().strip('/')
2 Comments
guri
how if i also want to remove https?
stephentgrammer
NB: both of these replacements will remove occurrences of the substring anywhere they are found in the url, which might not be intended. Safer to explicitly specify that these substrings have to be found at the beginning of the url.
A more elegant solution would be using urlparse:
from urllib.parse import urlparse
def get_hostname(url, uri_type='both'):
"""Get the host name from the url"""
parsed_uri = urlparse(url)
if uri_type == 'both':
return '{uri.scheme}://{uri.netloc}/'.format(uri=parsed_uri)
elif uri_type == 'netloc_only':
return '{uri.netloc}'.format(uri=parsed_uri)
The first option includes https or http, depending on the link, and the second part netloc includes what you were looking for.
Could use regex, depending on how strict your data is. Are http and www always going to be there? Have you thought about https or w3 sites?
import re
new_url = re.sub('.*w\.', '', url, 1)
1 to not harm websites ending with a w.
edit after clarification
I'd do two steps:
if url.startswith('http'):
url = re.sub(r'https?:\\', '', url)
if url.startswith('www.'):
url = re.sub(r'www.', '', url)
3 Comments
guri
http will always be there. www could be or not.Sometime it could also be https?In that case how should i modify above code to remove https aslo
guri
you mean like this if url.startswith('http'): new_url = re.sub('.*w\.', '', url, 1)
All Іѕ Vаиітy
it should be
'https?:\\'This will replace when http/https exist and finally if www. exist:
url=url.replace('http://','')
url=url.replace('https://','')
url=url.replace('www.','')
1 Comment
stephentgrammer
This does triple the work necessary (compared to a regex) and would also replace occurrences of those substrings elsewhere in the url, which would not be intented.
"http://www.google.com/images"[11:]?are arguments (also calledquery) - uparse can keep it in separeted variables.