Turn URL into HTML link

Question

In Python, if I have a URL, what is the simplest way to turn something like:

http://stackoverflow.com

into:

<a href="http://stackoverflow.com">http://stackoverflow.com</a>

So far I have tied a lot with Regular Expressions, but nothing works at all.

I am downvoting because the example information you provided does not match what you actually wanted for an answer. — SethMMorton, Feb 21 '15 at 02:13
possible duplicate of [replace URLs in text with links to URLs](http://stackoverflow.com/questions/1727535/replace-urls-in-text-with-links-to-urls) — Waylan, Feb 21 '15 at 03:42

score 3 · Accepted Answer · answered Feb 21 '15 at 02:03

3

You can use str.format:

>>> link = 'http://stackoverflow.com'
>>> print('<a href="{0}">{0}</a>'.format(link))
<a href="http://stackoverflow.com">http://stackoverflow.com</a>
>>>

Note however that you need to number the format fields since you are repeating an argument.

answered Feb 21 '15 at 02:03

What if I have a string containing multiple urls and non-urls? – Progo Feb 21 '15 at 02:07
I'm not sure what you mean by that. Could you give an example input/output? – Feb 21 '15 at 02:11
@Progo This answer meets the criteria as given in the original question. If your criteria is more complicated then you need to give that information as well. – SethMMorton Feb 21 '15 at 02:12
input: "Hey bob, go to http://stackexchange.com, they have cool sites like http://stackoverflow.com". The two links would be formatted. – Progo Feb 21 '15 at 02:12
@SethMMorton This is true. I will accept the answer. I have to wait 2 minutes though – Progo Feb 21 '15 at 02:13
@Progo - That's significantly different than what you said in your question. You might be able to use a Regex: `re.sub('(\w+\.com)', r'\1', data)` where `data` is your input string. Of course, this is a very simple solution. You will probably want to make the pattern a lot more robust. That, or look into libraries designed to work with HTML or URLs. – Feb 21 '15 at 02:18
@iCodez Sorry... I didn't phrase the question as well as I would like to have. Your solution works very well. Thanks! – Progo Feb 21 '15 at 02:19
@iCodez this will not work with links that do not end with `.com` or that specify a sub resource: `www.stackoverflow.com/questions/28641210` EDIT: looks like i read over the second statement of your comment. – syntonym Feb 21 '15 at 02:27
not working for urls like `http://my?param1&param2` – em2er Mar 03 '21 at 16:27

score 3 · Answer 2 · answered Feb 21 '15 at 02:19

3

You could use regex.

>>> import re
>>> s = "http://stackoverflow.com www.foo.com"
>>> re.sub(r'\b((?:https?:\/\/)?(?:www\.)?(?:[^\s.]+\.)+\w{2,4})\b', r'<a href="\1">\1</a>', s)
'<a href="http://stackoverflow.com">http://stackoverflow.com</a> <a href="www.foo.com">www.foo.com</a>'

answered Feb 21 '15 at 02:19

Avinash Raj

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Thanks for the answer. It works, but iCodez deserves the excepted answer. Thanks though. – Progo Feb 21 '15 at 02:21
seems like you're trying to search and replace links in an input string within a tag. – Avinash Raj Feb 21 '15 at 02:23
Correct. As I said, your answer works, but since I didn't phrase my original question very well and iCodez answered the original question **and** answered this one in a comment, he deserves the accept. +1 though. – Progo Feb 21 '15 at 02:25

score 1 · Answer 3 · answered Mar 03 '21 at 17:56

You could use a library to convert URL to HTML tag, e.g. bleach:

from bleach import linkify
url = 'https://stackoverflow.com?with=param1&other=param2'
atag = linkify(url)
print(atag)

will output

<a href="https://stackoverflow.com?with=param1&amp;other=param2" rel="nofollow">https://stackoverflow.com?with=param1&amp;other=param2</a>

score 0 · Answer 4 · answered Feb 21 '15 at 02:04

0

url = 'http://stackoverflow.com'
reference = '<a href=\"'+ url + '\">' + url + '</a>'

answered Feb 21 '15 at 02:04

user14241

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No, I don't know what the link is. I string and each url needs to be formatted. – Progo Feb 21 '15 at 02:06
iCodez answer is better – user14241 Feb 21 '15 at 02:06

score 0 · Answer 5 · answered Feb 21 '15 at 02:05

Assuming you already have that string in a variable, or can get it, I'd suggest you use string.format.

link = '<a href="{0}">{0}</a>'
stackOverflowLink = link.format("http://stackoverflow.com")

stackOverflowLink will then contain
<a href="http://stackoverflow.com">http://stackoverflow.com</a>

Turn URL into HTML link

5 Answers5