Executing code for a custom Django 404 page

Question

I am getting ready to deploy my first Django application and am hitting a bit of a roadblock. My base template relies on me passing in the session object so that it can read out the currently logged in user's name. This isn't a problem when I control the code that is calling a template.

However, as part of getting this app ready to be deployed, I need to create a 404.html page. I extended my base template just like I've done with my other pages, but I don't see a way to pass in the session object so that I can utilize it. Is there a way to have Django call a custom method to render your 404 rather than just rendering your 404.html for you?

AlbertoPL · Accepted Answer · 2009-07-22 19:05:11Z

13

You need to override the default view handler for the 404 error. Here is the documentation on how to create your own custom 404 view function:

http://docs.djangoproject.com/en/dev/topics/http/views/#customizing-error-views

answered Jul 22, 2009 at 19:05

AlbertoPL

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4 Comments

Chris Lieb Over a year ago

Just a note for other people: You need to define handler404 in the urls.py in the root of your project, not in one of your apps.

thebjorn Over a year ago

It must be in the file settings.ROOT_URLCONF points to (which normally is in the root of your project, but doesn't have to be...)

AlanSE Over a year ago

I found this answer because that tutorial didn't make sense to me. Does the handler404 line go in the urls.py urlpatterns list? How does the syntax of it make any sense?

AlanSE Over a year ago

Answered myself - yes it does go in urls.py, and outside of the urlpaterns. And yes, it is a string that holds the name in the format that you would otherwise use in an import statement. I found these details surprising, but it works.

Alexander Ljungberg · Accepted Answer · 2009-07-22 19:08:17Z

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Define your own 404 handler. See Django URLs, specifically the part about handler404.

answered Jul 22, 2009 at 19:08

Alexander Ljungberg

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