Django:条件URL模式？

Question

我想用另外一种 robots.txt根据我的服务器是生产还是开发文件.

为此,我想以不同的方式路由请求 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'}))

但是,这不起作用,因为我没有patterns正确使用该对象:我明白了AttributeError at /robots.txt - 'tuple' object has no attribute 'resolve'.

我怎样才能在Django中正确执行此操作？

Answer 1

试试这个:

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

由于您希望追加tuple类型,append不起作用.
另外,pattern()打电话urlresolver给你.在你的情况下,你不是,因此错误.