Passing request to custom Django template loader

Question

I want to write custom template loader for my Django app which looks for a specific folder based on a key that is part of the request.

Let me get into more details to be clear. Assume that I will be getting a key on every request(which I populate using a middleware).

Example: request.key could be 'india' or 'usa' or 'uk'.

I want my template loader to look for the template "templates/<key>/<template.html>". So when I say {% include "home.html" %}, I want the template loader to load "templates/india/home.html" or "templates/usa/home.html" or "templates/uk/home.html" based on the request.

Is there a way to pass the request object to a custom template loader?

score 5 · Answer 1 · answered Dec 29 '14 at 19:33

I've been searching for the same solution and, after a couple days of searching, decided to use threading.local(). Simply make the request object global for the duration of the HTTP request processing! Commence rotten tomato throwing from the gallery.

Let me explain:

As of Django 1.8 (according to the development version docs) the "dirs" argument for all template finding functions will be deprecated. (ref)

This means that there are no arguments passed into a custom template loader other than the template name being requested and the list of template directories. If you want to access paramters in the request URL (or even the session information) you'll have to "reach out" into some other storage mechanism.

import threading
_local = threading.local()

class CustomMiddleware:

    def process_request(self, request):
         _local.request = request

def load_template_source(template_name, template_dirs=None):
    if _local.request:
        # Get the request URL and work your magic here!
        pass

In my case it wasn't the request object (directly) I was after but rather what site (I'm developing a SaaS solution) the template should be rendered for.

Wolph · Answer 2 · 2014-04-12T19:27:08.247

2

To find the template to render Django uses the get_template method which only gets the template_name and optional dirs argument. So you cannot really pass the request there.

However, if you customize your render_to_response function to pass along a dirs argument you should be able to do it.

For example (assuming you are using a RequestContext as most people would):

from django import shortcuts
from django.conf import settings

def render_to_response(template_name, dictionary=None, context_instance=None, content_type=None, dirs):
    assert context_instance, 'This method requires a `RequestContext` instance to function'
    if not dirs:
        dirs = []
    dirs.append(os.path.join(settings.BASE_TEMPLATE_DIR, context_instance['request'].key)
    return shortcuts.render_to_response(template_name, dictionary, context_instance, content_type, dirs)

edited Apr 12 '14 at 19:27

answered Apr 12 '14 at 13:06

Wolph

78,177
11
137
148

thanks for this. This is really useful. Let's say I write this custom render_to_response, how do I ensure django uses this function? For example, when I write `{% include "home.html" %}` how will Django know it has to use the custom render_to_response? – UnTechie Apr 12 '14 at 13:31
1

Django won't use the custom `render_to_response` but it should pass along the `dirs` variable to the template parser so the includes get this parameter as well. – Wolph Apr 12 '14 at 19:26
I don't quite follow. If Django won't use this custom `render_to_response`, then how will it pass along the dirs that will be used by an `include` statement? – UnTechie Apr 16 '14 at 07:45
The `render_to_response` method is just a shortcut to creating a template loader, rendering the template to a string and creating a httpresponse from the string. Since the `dirs` argument is passed on to the loader the template object will get it and the renderer will know about your `dirs` parameter. – Wolph Apr 16 '14 at 15:35

Passing request to custom Django template loader

2 Answers2

Linked