How to print line breaks in Python Django template

Use the linebreaks filter.

For example:

{{ value|linebreaks }}

If value is Joel\nis a slug, the output will be <p>Joel<br />is a slug</p>.

You can also use the linebreaksbr filter to simply convert all newlines to <br> without additional <p>.

Example:

{{ value|linebreaksbr }}

If value is Joel\nis a slug, the output will be Joel<br>is a slug.

The difference from Ignacio's answer (linebreaks filter) is that linebreaks tries to guess the paragraphs in a text and wrap every paragraph in <p> where linebreaksbr simply substitutes newlines with <br>.

Here's a demo:

>>> from django.template.defaultfilters import linebreaks>>> from django.template.defaultfilters import linebreaksbr>>> text = 'One\nbreak\n\nTwo breaks\n\n\nThree breaks'>>> linebreaks(text)'<p>One<br />break</p>\n\n<p>Two breaks</p>\n\n<p>Three breaks</p>'>>> linebreaksbr(text)'One<br />break<br /><br />Two breaks<br /><br /><br />Three breaks'

All of these answers don't explicitly mention that {{ value|linebreaksbr }} should be used in the display template ie the get request not the post request.

So if you had a template to display say post.details, it would be

<h4>Below are the post details</h4>{{ post.details|linebreaksbr }}

And not in the post form

<form method="post">    {% csrf_token %}    {{ form|linebreaksbr }}    <button>Save post</button></form>

Had this same issue and after figuring it our decided that someone outthere might find it handy.Happy coding!