I would like an easy way to override the Django error page views so that the request g

You should read about <a href="https://docs.djangoproject.com/en/4.0/ref/urls/#handler

Here's what worked for me New config/error_views.py: <div class=

I'll leave that up to <a class="user-mention notranslate" data-hovercard-type="user" d

Agree with <a class="user-mention notranslate" data-hovercard-type="user" data-hoverca

Provide error page views that add the guid to the HTML output about django-guid HOT 5 CLOSED

andrew-cybsafe commented on September 27, 2024

Provide error page views that add the guid to the HTML output

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Comments (5)

sondrelg commented on September 27, 2024

You should read about handler500. You can very simply add a callable like this:

def handler500(request):
    return TemplateResponse(request, '500.html', {'correlation_id': correlation_id.get()}, status=500)

where the 500.html contains the HTML you wish to render 👍

Please feel free to report back with a solution when you're done. Could be cool to add a generalised example to the docs 🙏

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andrew-cybsafe commented on September 27, 2024

Here's what worked for me

New config/error_views.py:

from django.http import (
    HttpResponseBadRequest,
    HttpResponseForbidden,
    HttpResponseNotFound,
    HttpResponseServerError,
)
from django_guid import get_guid

ERROR_PAGE_TEMPLATE = """
<!doctype html>
<html lang="en">
<head>
  <title>%(title)s</title>
</head>
<body>
  <h1>%(title)s</h1>
  <p>%(details)s</p>
  <p>Request ID: %(guid)s</p>
</body>
</html>
"""


def server_error(request):
    return HttpResponseServerError(
        ERROR_PAGE_TEMPLATE
        % {
            "title": "Server Error (500)",
            "details": "An unexpected error occurred whilst processing the request",
            "guid": get_guid(),
        }
    )


def permission_denied(request, exception, template_name="403.html"):
    return HttpResponseForbidden(
        ERROR_PAGE_TEMPLATE
        % {
            "title": "Permission Denied (403)",
            "details": "No access to requested resource",
            "guid": get_guid(),
        }
    )


def bad_request(request, exception, template_name="400.html"):
    return HttpResponseBadRequest(
        ERROR_PAGE_TEMPLATE
        % {
            "title": "Bad Request (400)",
            "details": "",
            "guid": get_guid(),
        }
    )

In config/urls.py add:

handler500 = "config.error_views.server_error"
handler403 = "config.error_views.permission_denied"
handler400 = "config.error_views.bad_request"

Would you be amenable to including the "default with guid" error handlers to the package?

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sondrelg commented on September 27, 2024

I'll leave that up to @JonasKs to decide, but I think adding template code to the library is a little out-of-scope. Think the styling etc. should be up to the user to define. Think it's a nice thing to describe how to set up in the docs though 👏

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JonasKs commented on September 27, 2024

Agree with @sondrelg 😊 Documentation PR very welcome, but I'd prefer not to maintain templates - styling is so different between everyone.

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sondrelg commented on September 27, 2024

I'll close this for now. If anyone else runs into similar issues, we now at least have your example to reference @andrew-cybsafe 👏

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Provide error page views that add the guid to the HTML output about django-guid HOT 5 CLOSED

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