Webdevelop Docs

Appearance

Doing ajax with django admin ​

Problem ​

Some times we want to make an ajax request to django admin default views But you cannot override templates and return data in the format you need.

Idea ​

  • we don't want to rewrite django admin code
  • we don't want to copy/paste any existing functions so we don't break compatability with new django version
  • we prefer to write as little code as possible

Solution ​

We can easily replace view template. and return data in a format we need

  1. Create new TemplateResponse class which will check if its an ajax request then change template file. Example, update history template for ajax request:
python
if 'x-requested-with' in request.headers and \
  request.headers['x-requested-with'] == 'XMLHttpRequest':
  # handle object history request correctly
  if "admin/object_history.html" in template:
      template = ["admin/ajax/object_history.html"] + template

Full example

python
from django.template.response import SimpleTemplateResponse, TemplateResponse


class AjaxTemplateResponse(TemplateResponse):
    rendering_attrs = SimpleTemplateResponse.rendering_attrs + ["_request"]

    def __init__(
        self,
        request,
        template,
        context=None,
        content_type=None,
        status=None,
        charset=None,
        using=None,
        headers=None,
    ):
        if 'x-requested-with' in request.headers and \
            request.headers['x-requested-with'] == 'XMLHttpRequest':
            # handle object history request correctly
            if "admin/object_history.html" in template:
                template = ["admin/ajax/object_history.html"] + template
        super(TemplateResponse, self).__init__(
            template, context, content_type, status, charset, using, headers=headers
        )
        self._request = request

TemplateResponse.__init__ = AjaxTemplateResponse.__init__