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gmarsi
Asked: 2020-07-03 01:30:28 +0800 CST 2020-07-03 01:30:28 +0800 CST

Access links created in a <textarea>

  • 772

From a text created in a <textarea>, I need to be able to access those links that have been introduced.

For example, I enter:

Lorem ipsum dolor, sit amet consectetur adipisicing elit. https://www.stackoverflow.com Beatae eum nobis distinctiveo sapiente maxime ex dolorum, fuga voluptas architecto

https://www.stackoverflow.com

https://www.stackoverflow.com

Asperiores voluptatibus cupiditate sequi incident aperiam. https://www.stackoverflow.com Repellat recusandae quas eos error?

(The Stackoverflow editor already detects them automatically, but let's imagine it's plain text)

How could I, from the plain text, change the text link and turn it into a label <a>to display it in a Django template?

I thought maybe you could search the text for the link with a regex and replace it, but I don't know how to do it. Of course, the regex that I have tried and that works is

regex = r'http[s]?://(?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+'
python

1 Answers

  1. Best Answer

    In principle you can use django.utils.html.urlize or django.utils.html.urlizetrunc for this instead of resorting to implementing it yourself via regex:

    text = """Lorem ipsum dolor, sit amet consectetur
adipisicing elit. https://www.stackoverflow.com
Beatae eum nobis distinctio sapiente maxime ex dolorum,
fuga voluptas architecto https://ru.stackoverflow.com
https://es.stackoverflow.com asperiores
voluptatibus cupiditate sequi incidunt aperiam.
https://pt.stackoverflow.com Repellat recusandae
quas eos error?
"""


from django.utils.html import urlize
print(urlize(text))


    Lorem ipsum dolor, sit amet consectetur
adipisicing elit. <a href="https://www.stackoverflow.com">https://www.stackoverflow.com</a>
Beatae eum nobis distinctio sapiente maxime ex dolorum,
fuga voluptas architecto <a href="https://ru.stackoverflow.com">https://ru.stackoverflow.com</a>
<a href="https://es.stackoverflow.com">https://es.stackoverflow.com</a> asperiores
voluptatibus cupiditate sequi incidunt aperiam.
<a href="https://pt.stackoverflow.com">https://pt.stackoverflow.com</a> Repellat recusandae
quas eos error?

    To implement something like this with regex , you can make use of re.subenclosing your expression in a group and build the tag from there. For example, using the expression you provide yourself:

    import re

regex = re.compile(r'(http[s]?://(?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+)')
urlize_text = re.sub(regex, r'<a href="\1">\1</a>', text)

    Extra

    If we only want to extract the links, ignoring the rest of the text, we can use re.finditer:

    import re

regex = re.compile(r'(http[s]?://((?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+))')
urls = [match.expand(r'<a href="\1">\1</a>') for match in re.finditer(regex, text)]
print(urls)

    ['<a href="https://www.stackoverflow.com">https://www.stackoverflow.com</a>',
 '<a href="https://ru.stackoverflow.com">https://ru.stackoverflow.com</a>',
 '<a href="https://es.stackoverflow.com">https://es.stackoverflow.com</a>',
 '<a href="https://pt.stackoverflow.com">https://pt.stackoverflow.com</a>'
]
    • 4