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!

What's wrong with the answer to the other post? It finds URL's in text using a regex. What doesn't work? What's broken? Why repeat that question? What's wrong with the answer to http://stackoverflow.com/questions/720113/find-hyperlinks-in-text-using-python-twitter-related? — S.Lott, May 08 '09 at 14:45

score 91 · Accepted Answer · edited Dec 09 '18 at 21:32

91

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"))

edited Dec 09 '18 at 21:32

Iulian Onofrei

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answered May 08 '09 at 14:39

Andrew Hare

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I get an "invalid syntax" with the last line. – Kyle Hayes May 08 '09 at 14:52
Ok, got it to work without the print statement for some reason – Kyle Hayes May 08 '09 at 14:53
Good point - I simply copy/pasted the original regex. I fixed it to be a bit more robust and included your suggestion - thanks! – Andrew Hare May 08 '09 at 17:51
3

If you get a syntax error on the print statement, you're probably using Python 3.0, which removes the print statement and instead simply provides a print("Hello, world.") function instead. – Brandon Rhodes May 08 '09 at 17:55
6

Modify the above to take into account trailing quote around most URLs, especially when parsing HTML: re.search("(?Phttps?://[^\s'\"]+)", myString).group("url") – Paul Kenjora Mar 20 '17 at 18:48
Need to use `re.findall("(?Phttps?://[^\s]+)", myString)` in order to get list of all the URLs present in the string. – Animesh Kumar Nov 06 '19 at 05:15
What is this P syntax? I've never see that in regex. – rosstex Jul 25 '20 at 23:06
Why is this accepted answer? It doesn't work if there is " inside the string. – Raymond Dec 12 '21 at 12:32
@rosstex Just found its something related to re module. See example under match.group – Greg Iven Apr 11 '22 at 21:57

SilentGhost · Answer 2 · 2010-06-03T12:12:45.220

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']

I like this solution the most, since that it allows to extract multiple urls — pedromendessk, Nov 06 '19 at 10:16

score 31 · Answer 3 · answered Jun 11 '18 at 03:20

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']

Please don't post identical answers to multiple questions. Post one good answer, then vote/flag to close the other questions as duplicates. If the question is not a duplicate, *tailor your answers to the question.* — Machavity, Jun 11 '18 at 03:30
It gives me an `invalid syntax` error on the second character of this part `(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9]))`. — Carlos Oliveira, Mar 04 '20 at 18:11
@CarlosOliveira `regex=ur"..."` should be `regex = r"..."`, at least in Python 3. — supermitch, Dec 01 '21 at 02:04

score 16 · Answer 4 · answered Jul 04 '17 at 22:21

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.")

score 12 · Answer 5 · answered Feb 17 '15 at 00:26

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?

Useful regex, but quite murky. For example, imagine I wanted to drop support for the TLD `.ni`. I see two instances of `.ni` in the regex (I was expecting just one instance). Why the repetition? And should I remove both or just the first occurrence? Would be useful for all of us to get minor instructions on editing it to our needs. — Hassan Baig, Apr 30 '18 at 05:59

score 5 · Answer 6 · edited May 23 '17 at 12:10

5

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")

edited May 23 '17 at 12:10

Community

1
1

answered Jun 06 '12 at 03:09

Adam

461
4
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score 3 · Answer 7 · edited Jun 03 '10 at 11:59

3

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

edited Jun 03 '10 at 11:59

SilentGhost

307,395
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answered Jun 03 '10 at 11:55

bogdan

338
6
10

I fixed your post, stop messing it please. – SilentGhost Jun 03 '10 at 12:00
5

or you could jsut do: print re.findall("(?Phttps?://[^\s]+)", myString) – bogdan Jun 03 '10 at 13:02

score 2 · Answer 8 · answered Jan 22 '21 at 08:35

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)

score -1 · Answer 9 · answered Jan 03 '15 at 10:10

-1

[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 Jan 03 '15 at 10:10

kyrenia

5,431
9
63
93

Mohammad Monfared · Answer 10 · 2020-05-05T12:37:13.340

-1

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/']

edited May 05 '20 at 12:37

answered May 05 '20 at 12:22

Mohammad Monfared

578
3
19

duplicate answer, the autor of `URLExtract` posted a similar answer in 2017. – jpoppe Aug 14 '23 at 12:05

Extracting a URL in Python

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