Get static image URL in Python code

Question

I've got a Python function in my Django project where I need to return some HTML code, including the static URL where my images reside.

In a template, I can do <img src="{% static 'img/my_img.png' %}">. Any idea how I can get this in Python code as part of a return statement?

Obviously I could include the whole URL in my return statement (as per the below) but this would violate Django's DRY principle (and would be problematic in case my image URL would change).

if my.condition:
    return "<img src='/img/my_img.png' />" + obj.name

Using Django 1.8.

Augusto Destrero · Accepted Answer · 2018-11-05 11:34:05Z

2

You can use the template tag in python code, like this:

from django.templatetags.static import static

full_path = static('img/my_img.png')

This will also take into account use cases where the full static path is not simply the concatenation of settings.STATIC_URL with the relative path. See for instance this example in Django documentation.

answered Nov 5, 2018 at 11:34

Augusto Destrero

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rjv · Accepted Answer · 2015-10-31 12:25:05Z

1

You can always import the settings file

from django.conf import settings
...

if my.condition:
    return ("<img src='%s/img/my_img.png' />" % (settings.STATIC_URL)) + obj.nam

answered Oct 31, 2015 at 12:25

rjv

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1 Comment

Augusto Destrero Over a year ago

This answer do not take into account use cases where the full static path is not simply the concatenation of settings.STATIC_URL with the relative path. See for instance this example in Django documentation.

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