Extracting a URL in Python

Question:

In regards to: Find Hyperlinks in Text using Python (twitter related)

How can I extract just the url so I can put it into a list/array?

Edit

Let me clarify, I don’t want to parse the URL into pieces. I want to extract the URL from the text of the string to put it into an array. Thanks!

Asked By: Kyle Hayes

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Source

Answers:

Misunderstood question:

>>> from urllib.parse import urlparse
>>> urlparse('http://www.ggogle.com/test?t')
ParseResult(scheme='http', netloc='www.ggogle.com', path='/test',
        params='', query='t', fragment='')

or py2.* version:

>>> from urlparse import urlparse
>>> urlparse('http://www.cwi.nl:80/%7Eguido/Python.html')
ParseResult(scheme='http', netloc='www.cwi.nl:80', path='/%7Eguido/Python.html',
        params='', query='', fragment='')

ETA: regex are indeed are the best option here:

>>> s = 'This is my tweet check it out http://tinyurl.com/blah and http://blabla.com'
>>> re.findall(r'(https?://S+)', s)
['http://tinyurl.com/blah', 'http://blabla.com']
Answered By: SilentGhost

In response to the OP’s edit I hijacked Find Hyperlinks in Text using Python (twitter related) and came up with this:

import re

myString = "This is my tweet check it out http://example.com/blah"

print(re.search("(?P<url>https?://[^s]+)", myString).group("url"))
Answered By: Andrew Hare

Regarding this:

import re
myString = "This is my tweet check it out http:// tinyurl.com/blah"
print re.search("(?P<url>https?://[^s]+)", myString).group("url")

It won’t work well if you have multiple urls in the string.
If the string looks like:

myString = "This is my tweet check it out http:// tinyurl.com/blah and http:// blabla.com"

You may do something like this:

myString_list = [item for item in myString.split(" ")]
for item in myString_list:
    try:
        print re.search("(?P<url>https?://[^s]+)", item).group("url")
    except:
        pass
Answered By: bogdan

Don’t forget to check for whether the search returns a value of None—I found the posts above helpful but wasted time dealing with a None result.

See Python Regex "object has no attribute".

i.e.

import re
myString = "This is my tweet check it out http://tinyurl.com/blah"
match = re.search("(?P<url>https?://[^s]+)", myString)
if match is not None: 
    print match.group("url")
Answered By: Adam

[note: Assuming you are using this on Twitter data (as indicated in question), the simplest way of doing this is to use their API, which returns the urls extracted from tweets as a field]

Answered By: kyrenia

Here’s a file with a huge regex:

#!/usr/bin/python
# -*- coding: utf-8 -*-
"""
the web url matching regex used by markdown
http://daringfireball.net/2010/07/improved_regex_for_matching_urls
https://gist.github.com/gruber/8891611
"""
URL_REGEX = r"""(?i)b((?:https?:(?:/{1,3}|[a-z0-9%])|[a-z0-9.-]+[.](?:com|net|org|edu|gov|mil|aero|asia|biz|cat|coop|info|int|jobs|mobi|museum|name|post|pro|tel|travel|xxx|ac|ad|ae|af|ag|ai|al|am|an|ao|aq|ar|as|at|au|aw|ax|az|ba|bb|bd|be|bf|bg|bh|bi|bj|bm|bn|bo|br|bs|bt|bv|bw|by|bz|ca|cc|cd|cf|cg|ch|ci|ck|cl|cm|cn|co|cr|cs|cu|cv|cx|cy|cz|dd|de|dj|dk|dm|do|dz|ec|ee|eg|eh|er|es|et|eu|fi|fj|fk|fm|fo|fr|ga|gb|gd|ge|gf|gg|gh|gi|gl|gm|gn|gp|gq|gr|gs|gt|gu|gw|gy|hk|hm|hn|hr|ht|hu|id|ie|il|im|in|io|iq|ir|is|it|je|jm|jo|jp|ke|kg|kh|ki|km|kn|kp|kr|kw|ky|kz|la|lb|lc|li|lk|lr|ls|lt|lu|lv|ly|ma|mc|md|me|mg|mh|mk|ml|mm|mn|mo|mp|mq|mr|ms|mt|mu|mv|mw|mx|my|mz|na|nc|ne|nf|ng|ni|nl|no|np|nr|nu|nz|om|pa|pe|pf|pg|ph|pk|pl|pm|pn|pr|ps|pt|pw|py|qa|re|ro|rs|ru|rw|sa|sb|sc|sd|se|sg|sh|si|sj|Ja|sk|sl|sm|sn|so|sr|ss|st|su|sv|sx|sy|sz|tc|td|tf|tg|th|tj|tk|tl|tm|tn|to|tp|tr|tt|tv|tw|tz|ua|ug|uk|us|uy|uz|va|vc|ve|vg|vi|vn|vu|wf|ws|ye|yt|yu|za|zm|zw)/)(?:[^s()<>{}[]]+|([^s()]*?([^s()]+)[^s()]*?)|([^s]+?))+(?:([^s()]*?([^s()]+)[^s()]*?)|([^s]+?)|[^s`!()[]{};:'".,<>?«»“”‘’])|(?:(?<!@)[a-z0-9]+(?:[.-][a-z0-9]+)*[.](?:com|net|org|edu|gov|mil|aero|asia|biz|cat|coop|info|int|jobs|mobi|museum|name|post|pro|tel|travel|xxx|ac|ad|ae|af|ag|ai|al|am|an|ao|aq|ar|as|at|au|aw|ax|az|ba|bb|bd|be|bf|bg|bh|bi|bj|bm|bn|bo|br|bs|bt|bv|bw|by|bz|ca|cc|cd|cf|cg|ch|ci|ck|cl|cm|cn|co|cr|cs|cu|cv|cx|cy|cz|dd|de|dj|dk|dm|do|dz|ec|ee|eg|eh|er|es|et|eu|fi|fj|fk|fm|fo|fr|ga|gb|gd|ge|gf|gg|gh|gi|gl|gm|gn|gp|gq|gr|gs|gt|gu|gw|gy|hk|hm|hn|hr|ht|hu|id|ie|il|im|in|io|iq|ir|is|it|je|jm|jo|jp|ke|kg|kh|ki|km|kn|kp|kr|kw|ky|kz|la|lb|lc|li|lk|lr|ls|lt|lu|lv|ly|ma|mc|md|me|mg|mh|mk|ml|mm|mn|mo|mp|mq|mr|ms|mt|mu|mv|mw|mx|my|mz|na|nc|ne|nf|ng|ni|nl|no|np|nr|nu|nz|om|pa|pe|pf|pg|ph|pk|pl|pm|pn|pr|ps|pt|pw|py|qa|re|ro|rs|ru|rw|sa|sb|sc|sd|se|sg|sh|si|sj|Ja|sk|sl|sm|sn|so|sr|ss|st|su|sv|sx|sy|sz|tc|td|tf|tg|th|tj|tk|tl|tm|tn|to|tp|tr|tt|tv|tw|tz|ua|ug|uk|us|uy|uz|va|vc|ve|vg|vi|vn|vu|wf|ws|ye|yt|yu|za|zm|zw)b/?(?!@)))"""

I call that file urlmarker.py and when I need it I just import it, eg.

import urlmarker
import re
re.findall(urlmarker.URL_REGEX,'some text news.yahoo.com more text')

cf. http://daringfireball.net/2010/07/improved_regex_for_matching_urls and What's the cleanest way to extract URLs from a string using Python?

Answered By: dranxo

If you want to extract URLs from any text you can use my urlextract. It finds URL based on TLD found in text. It expands to both side from TLD position an gets whole URL. Its easy to use. Check it: https://github.com/lipoja/URLExtract

    from urlextract import URLExtract

    extractor = URLExtract()
    urls = extractor.find_urls("Text with URLs: stackoverflow.com.")
Answered By: Jan Lipovský

You can use the following monstrous regex:

b((?:https?://)?(?:(?:www.)?(?:[da-z.-]+).(?:[a-z]{2,6})|(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?).){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)|(?:(?:[0-9a-fA-F]{1,4}:){7,7}[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,7}:|(?:[0-9a-fA-F]{1,4}:){1,6}:[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,5}(?::[0-9a-fA-F]{1,4}){1,2}|(?:[0-9a-fA-F]{1,4}:){1,4}(?::[0-9a-fA-F]{1,4}){1,3}|(?:[0-9a-fA-F]{1,4}:){1,3}(?::[0-9a-fA-F]{1,4}){1,4}|(?:[0-9a-fA-F]{1,4}:){1,2}(?::[0-9a-fA-F]{1,4}){1,5}|[0-9a-fA-F]{1,4}:(?:(?::[0-9a-fA-F]{1,4}){1,6})|:(?:(?::[0-9a-fA-F]{1,4}){1,7}|:)|fe80:(?::[0-9a-fA-F]{0,4}){0,4}%[0-9a-zA-Z]{1,}|::(?:ffff(?::0{1,4}){0,1}:){0,1}(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9]).){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])|(?:[0-9a-fA-F]{1,4}:){1,4}:(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9]).){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])))(?::[0-9]{1,4}|[1-5][0-9]{4}|6[0-4][0-9]{3}|65[0-4][0-9]{2}|655[0-2][0-9]|6553[0-5])?(?:/[w.-]*)*/?)b

Demo regex101

This regex will accept urls in the following format:

INPUT:

add1 http://mit.edu.com abc
add2 https://facebook.jp.com.2. abc
add3 www.google.be. uvw
add4 https://www.google.be. 123
add5 www.website.gov.us test2
Hey bob on www.test.com. 
another test with ipv4 http://192.168.1.1/test.jpg. toto2
website with different port number www.test.com:8080/test.jpg not port 80
www.website.gov.us/login.html
test with ipv4 192.168.1.1/test.jpg.
search at google.co.jp/maps.
test with ipv6 2001:0db8:0000:85a3:0000:0000:ac1f:8001/test.jpg.

OUTPUT:

http://mit.edu.com
https://facebook.jp.com
www.google.be
https://www.google.be
www.website.gov.us
www.test.com
http://192.168.1.1/test.jpg
www.test.com:8080/test.jpg
www.website.gov.us/login.html
192.168.1.1/test.jpg
google.co.jp/maps
2001:0db8:0000:85a3:0000:0000:ac1f:8001/test.jpg

Explanations:

  • b is used for word boundary to delimit the URL and the rest of the text
  • (?:https?://)? to match http:// or https// if present
  • (?:(?:www.)?(?:[da-z.-]+).(?:[a-z]{2,6}) to match standard url (that might start with www. (lets call it STANDARD_URL)
  • (?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?).){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?) to match standard Ipv4 (lets call it IPv4)
  • to match the IPv6 URLs: (?:(?:[0-9a-fA-F]{1,4}:){7,7}[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,7}:|(?:[0-9a-fA-F]{1,4}:){1,6}:[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,5}(?::[0-9a-fA-F]{1,4}){1,2}|(?:[0-9a-fA-F]{1,4}:){1,4}(?::[0-9a-fA-F]{1,4}){1,3}|(?:[0-9a-fA-F]{1,4}:){1,3}(?::[0-9a-fA-F]{1,4}){1,4}|(?:[0-9a-fA-F]{1,4}:){1,2}(?::[0-9a-fA-F]{1,4}){1,5}|[0-9a-fA-F]{1,4}:(?:(?::[0-9a-fA-F]{1,4}){1,6})|:(?:(?::[0-9a-fA-F]{1,4}){1,7}|:)|fe80:(?::[0-9a-fA-F]{0,4}){0,4}%[0-9a-zA-Z]{1,}|::(?:ffff(?::0{1,4}){0,1}:){0,1}(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9]).){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])|(?:[0-9a-fA-F]{1,4}:){1,4}:(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9]).){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])) (lets call it IPv6)
  • to match the port part (lets call it PORT) if present: (?::[0-9]{1,4}|[1-5][0-9]{4}|6[0-4][0-9]{3}|65[0-4][0-9]{2}|655[0-2][0-9]|6553[0-5])
  • to match the (?:/[w.-]*)*/?) target object part of the url (html file, jpg,…) (lets call it RESSOURCE_PATH)

This gives the following regex: 

b((?:https?://)?(?:STANDARD_URL|IPv4|IPv6)(?:PORT)?(?:RESSOURCE_PATH)b

Sources:

IPv6: Regular expression that matches valid IPv6 addresses

IPv4: https://www.safaribooksonline.com/library/view/regular-expressions-cookbook/9780596802837/ch07s16.html

PORT: https://stackoverflow.com/a/12968117/8794221

Other sources:
https://code.tutsplus.com/tutorials/8-regular-expressions-you-should-know–net-6149

$ more url.py

import re

inputString = """add1 http://mit.edu.com abc
add2 https://facebook.jp.com.2. abc
add3 www.google.be. uvw
add4 https://www.google.be. 123
add5 www.website.gov.us test2
Hey bob on www.test.com. 
another test with ipv4 http://192.168.1.1/test.jpg. toto2
website with different port number www.test.com:8080/test.jpg not port 80
www.website.gov.us/login.html
test with ipv4 (192.168.1.1/test.jpg).
search at google.co.jp/maps.
test with ipv6 2001:0db8:0000:85a3:0000:0000:ac1f:8001/test.jpg."""

regex=ur"b((?:https?://)?(?:(?:www.)?(?:[da-z.-]+).(?:[a-z]{2,6})|(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?).){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)|(?:(?:[0-9a-fA-F]{1,4}:){7,7}[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,7}:|(?:[0-9a-fA-F]{1,4}:){1,6}:[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,5}(?::[0-9a-fA-F]{1,4}){1,2}|(?:[0-9a-fA-F]{1,4}:){1,4}(?::[0-9a-fA-F]{1,4}){1,3}|(?:[0-9a-fA-F]{1,4}:){1,3}(?::[0-9a-fA-F]{1,4}){1,4}|(?:[0-9a-fA-F]{1,4}:){1,2}(?::[0-9a-fA-F]{1,4}){1,5}|[0-9a-fA-F]{1,4}:(?:(?::[0-9a-fA-F]{1,4}){1,6})|:(?:(?::[0-9a-fA-F]{1,4}){1,7}|:)|fe80:(?::[0-9a-fA-F]{0,4}){0,4}%[0-9a-zA-Z]{1,}|::(?:ffff(?::0{1,4}){0,1}:){0,1}(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9]).){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])|(?:[0-9a-fA-F]{1,4}:){1,4}:(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9]).){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])))(?::[0-9]{1,4}|[1-5][0-9]{4}|6[0-4][0-9]{3}|65[0-4][0-9]{2}|655[0-2][0-9]|6553[0-5])?(?:/[w.-]*)*/?)b"

matches = re.findall(regex, inputString)
print(matches)

OUTPUT:

$ python url.py 
['http://mit.edu.com', 'https://facebook.jp.com', 'www.google.be', 'https://www.google.be', 'www.website.gov.us', 'www.test.com', 'http://192.168.1.1/test.jpg', 'www.test.com:8080/test.jpg', 'www.website.gov.us/login.html', '192.168.1.1/test.jpg', 'google.co.jp/maps', '2001:0db8:0000:85a3:0000:0000:ac1f:8001/test.jpg']
Answered By: Allan

In case of extracting from HTML source:

from urlextract import URLExtract
from requests import get

url = "sample.com/samplepage/"
req = requests.get(url)
text = req.text
# or if you already have the html source:
# text2 = "This is html for ex <a href='http://google.com/'>Google</a> <a href='http://yahoo.com/'>Yahoo</a>"
text = text.replace(' ', '').replace('=','')
extractor = URLExtract()
print(extractor.find_urls(text))

output (text2):

['http://google.com/', 'http://yahoo.com/']

Answered By: Mohammad Monfared

Simply follow below code and enjoy….!!!!

import requests
from bs4 import BeautifulSoup
url = "your url"//Any url that you want to fetch.
r = requests.get(url)
htmlContent = r.content
soup = BeautifulSoup(htmlContent, 'html.parser')

anchors = soup.find_all('a')
all_links = set()
for link in anchors:
    if(link.get('href') != '#'): 
        linkText = url+str(link.get('href'))
        all_links.add(link)
        print(linkText)
Answered By: PRIYA SHARMA
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