How do you extract a url from a string using python?

For example:

string = "This is a link http://www.google.com"

How could I extract ‘http://www.google.com’ ?

(Each link will be of the same format i.e ‘http://’)

Answers:

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Method 1

There may be few ways to do this but the cleanest would be to use regex

>>> myString = "This is a link http://www.google.com"
>>> print re.search("(?P<url>https?://[^s]+)", myString).group("url")
http://www.google.com

If there can be multiple links you can use something similar to below

>>> myString = "These are the links http://www.google.com  and http://stackoverflow.com/questions/839994/extracting-a-url-in-python"
>>> print re.findall(r'(https?://[^s]+)', myString)
['http://www.google.com', 'http://stackoverflow.com/questions/839994/extracting-a-url-in-python']
>>>

Method 2

In order to find a web URL in a generic string, you can use a regular expression (regex).

A simple regex for URL matching like the following should fit your case.

    regex = r'('

    # Scheme (HTTP, HTTPS, FTP and SFTP):
    regex += r'(?:(https?|s?ftp)://)?'

    # www:
    regex += r'(?:www.)?'

    regex += r'('

    # Host and domain (including ccSLD):
    regex += r'(?:(?:[A-Z0-9][A-Z0-9-]{0,61}[A-Z0-9].)+)'

    # TLD:
    regex += r'([A-Z]{2,6})'

    # IP Address:
    regex += r'|(?:d{1,3}.d{1,3}.d{1,3}.d{1,3})'

    regex += r')'

    # Port:
    regex += r'(?::(d{1,5}))?'

    # Query path:
    regex += r'(?:(/S+)*)'

    regex += r')'

If you want to be even more precise, in the TLD section, you should ensure that the TLD is a valid TLD (see the entire list of valid TLDs here: https://data.iana.org/TLD/tlds-alpha-by-domain.txt):

    # TLD:
    regex += r'(com|net|org|eu|...)'

Then, you can simply compile the former regex and use it to find possible matches:

    import re

    string = "This is a link http://www.google.com"

    find_urls_in_string = re.compile(regex, re.IGNORECASE)
    url = find_urls_in_string.search(string)

    if url is not None and url.group(0) is not None:
        print("URL parts: " + str(url.groups()))
        print("URL" + url.group(0).strip())

Which, in case of the string “This is a link http://www.google.com“ will output:

    URL parts: ('http://www.google.com', 'http', 'google.com', 'com', None, None)
    URL: http://www.google.com

If you change the input with a more complex URL, for example “This is also a URL https://www.host.domain.com:80/path/page.php?query=value&a2=v2#foo but this is not anymore” the output will be:

    URL parts: ('https://www.host.domain.com:80/path/page.php?query=value&a2=v2#foo', 'https', 'host.domain.com', 'com', '80', '/path/page.php?query=value&a2=v2#foo')
    URL: https://www.host.domain.com:80/path/page.php?query=value&a2=v2#foo

NOTE: If you are looking for more URLs in a single string, you can still use the same regex, but just use findall() instead of search().

Method 3

There is another way how to extract URLs from text easily. You can use urlextract to do it for you, just install it via pip:

pip install urlextract

and then you can use it like this:

from urlextract import URLExtract

extractor = URLExtract()
urls = extractor.find_urls("Let's have URL stackoverflow.com as an example.")
print(urls) # prints: ['stackoverflow.com']

You can find more info on my github page: https://github.com/lipoja/URLExtract

NOTE: It downloads a list of TLDs from iana.org to keep you up to date. But if the program does not have internet access then it’s not for you.

Method 4

This extracts all urls with parameters, somehow all above examples haven’t worked for me

import re

data = 'https://net2333.us3.list-some.com/subscribe/confirm?u=f3cca8a1ffdee924a6a413ae9&id=6c03fa85f8&e=6bbacccc5b'

WEB_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 href="https://getridbug.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="765736">[email protected]</a>)[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/?(<a href="https://getridbug.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="edd2ccad">[email protected]</a>)))"""
re.findall(WEB_URL_REGEX, text)

Method 5

You can extract any URL from a string using the following patterns,

1.

>>> import re
>>> string = "This is a link http://www.google.com"
>>> pattern = r'[(http://)|w]*?[w]*.[-/w]*.w*[(/{1})]?[#-./w]*[(/{1,})]?'
>>> re.search(pattern, string)
http://www.google.com

>>> TWEET = ('New Pybites article: Module of the Week - Requests-cache '
         'for Repeated API Calls - http://pybit.es/requests-cache.html '
         '#python #APIs')
>>> re.search(pattern, TWEET)
http://pybit.es/requests-cache.html

>>> tweet = ('Pybites My Reading List | 12 Rules for Life - #books '
             'that expand the mind! '
             'http://pbreadinglist.herokuapp.com/books/'
             'TvEqDAAAQBAJ#.XVOriU5z2tA.twitter'
             ' #psychology #philosophy')
>>> re.findall(pattern, TWEET)
['http://pbreadinglist.herokuapp.com/books/TvEqDAAAQBAJ#.XVOriU5z2tA.twitter']

to take the above pattern to the next level, we can also detect hashtags including URL the following ways

2.

>>> pattern = r'[(http://)|w]*?[w]*.[-/w]*.w*[(/{1})]?[#-./w]*[(/{1,})]?|#[.w]*'
>>> re.findall(pattern, tweet)
['#books', http://pbreadinglist.herokuapp.com/books/TvEqDAAAQBAJ#.XVOriU5z2tA.twitter', '#psychology', '#philosophy']

The above example for taking URL and hashtags can be shortened to

>>> pattern = r'((?:#|http)S+)'
>>> re.findall(pattern, tweet)
['#books', http://pbreadinglist.herokuapp.com/books/TvEqDAAAQBAJ#.XVOriU5z2tA.twitter', '#psychology', '#philosophy']

The pattern below can matches two alphanumeric separated by “.” as URL

>>> pattern = pattern =  r'(?:http://)?w+.S*[^.s]'

>>> tweet = ('PyBites My Reading List | 12 Rules for Life - #books '
             'that expand the mind! '
             'www.google.com/telephone/wire....  '
             'http://pbreadinglist.herokuapp.com/books/'
             'TvEqDAAAQBAJ#.XVOriU5z2tA.twitter '
             "http://-www.pip.org "
             "google.com "
             "twitter.com "
             "facebook.com"
             ' #psychology #philosophy')
>>> re.findall(pattern, tweet)
['www.google.com/telephone/wire', 'http://pbreadinglist.herokuapp.com/books/TvEqDAAAQBAJ#.XVOriU5z2tA.twitter', 'www.pip.org', 'google.com', 'twitter.com', 'facebook.com']

You can try any complicated URL with the number 1 & 2 pattern.
To learn more about re module in python, do check this out
REGEXES IN PYTHON by Real Python.

Cheers!

Method 6

I’ve used a slight variation from @Abhijit’s accepted answer.

This one uses S instead of [^s], which is equivalent but more concise. It also doesn’t use a named group, because there is just one and we can ommit the name for simplicity reasons:

import re

my_string = "This is my tweet check it out http://example.com/blah"
print(re.search(r'(https?://S+)', my_string).group())

Of course, if there are multiple links to extract, just use .findall():

print(re.findall(r'(https?://S+)', my_string))

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0