404 view -> redirect to index

[galago]

I try to redirect from 404 view to my index page. In url.py i added: 

handler404 = 'index.views.custom404'

in my index i have:

def custom404(request):    

    return HttpResponseRedirect(reverse('index'))

But it doesn't work.

Should I do something more?

[Ivo Brodien]

Are you sure, that you have a url pattern called ‘’view? in reverse you have to pass the name of the rule, not the views method name.

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[galago]

Yes I have a index pattern:

urlpatterns = patterns('',

    url(r'^$', 'index.views.index', name='index'),.......

When I add print to my method:

def custom404(request):

    print 'foo'

    return HttpResponseRedirect(reverse('index'))

It prints out but it doesn't redirect me to index. Instead it displays my 404.html page.

[Ivo Brodien]

I guess that django catches the 404 Error Code somehow somewhere else than in the handler? Maybe you have to set the HTTP Code to 200?

[Thomas]

why just not copy the 'templates/index.html' template to 'templates/404.html' - ready

*not*sure*about*the*correctness*of*this*

[Thomas]

The book "The definitive Guide to Django" says on page 476:

---8<---

... if you want to override the 404 view, you can specify 'handler404' in your URLconf, like so:

from django.conf.urls.defaults import *

urlpatterns = patterns ('',
...
)

handler404 = 'mysite.views.my_custom_404_view'

Behind the scenes ...

---8<---

hope this helps,

good luck,
TR

Am 18.01.2011 um 13:42 schrieb Ivo Brodien:

[natebeacham]

Thomas: That is what he has.

Make sure you turn DEBUG off. Your custom 404 handler will only be
used when you are not in debug mode.

Cheers

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