Django: conditional URL patterns?

Question

I'd like to use a different robots.txt file depending on whether my server is production or development.

To do this, I would like to route the request differently in urls.py:

urlpatterns = patterns('',
   // usual patterns here
)

if settings.IS_PRODUCTION: 
  urlpatterns.append((r'^robots\.txt$', direct_to_template, {'template': 'robots_production.txt', 'mimetype': 'text/plain'}))
else:
  urlpatterns.append((r'^robots\.txt$', direct_to_template, {'template': 'robots_dev.txt', 'mimetype': 'text/plain'}))

However, this isn't working, because I'm not using the patterns object correctly: I get AttributeError at /robots.txt - 'tuple' object has no attribute 'resolve'.

How can I do this correctly in Django?

Answer 1

Try this:

if settings.IS_PRODUCTION: 
  additional_settings = patterns('',
     (r'^robots\.txt$', direct_to_template, {'template': 'robots_production.txt', 'mimetype': 'text/plain'}),
  )
else:
  additional_settings = patterns('',
      (r'^robots\.txt$', direct_to_template, {'template': 'robots_dev.txt', 'mimetype': 'text/plain'}),
  )

urlpatterns += additional_settings

Since you are looking to append tuple types , append does not work.
Also, pattern() calls the urlresolver for you. In your case you were not, hence the error.