django - 將urls.py拆分成多個

Question

我知道，這實際上是django的基礎知識但是

我試圖將我的巨大的urls.py拆分爲3個部分。所有3個urls_xxx.py都在一個文件夾下：mysite。

然後在主urls.py我所做的：

#urls.py 
urlpatterns += patterns('', 

    url(r'^', include('urls_search')), 
    url(r'^', include('urls_ajax')), 
    url(r'^', include('urls_sitemaps')), 

) 

但他說：

ImportError at/
No module named urls_search 

這是urls_search.py​​：

urlpatterns = patterns('myapp.views', 
    url(r'^advancedsearch/$', 'advsearch', name="advsearch"), 
    url(r'^advancedsearch/tc/$', 'advtcsearch', name="advtcsearch"), 
) 

我的URL配置：

ROOT_URLCONF = 'mysite.urls' 

我做錯了什麼？我使用Django 1.4

Answer 1

刪除.py這樣的：

url(r'^', include('mysite.urls_search')), 

Answer 2

可以在主 'urls.py' 這樣做：

from mysites.url_1 import urlpatterns as url1 
from mysites.url_2 import urlpatterns as url2 
from mysites.url_3 import urlpatterns as url3 

urlpatterns = url1 
urlpatterns += url2 
urlpatterns += url3 

Answer 3

在主要url.py

from mysites import url1 
from mysites import url2 
from mysites import url3 

urlpatterns = url1.urlpatterns 
urlpatterns += url2.urlpatterns 
urlpatterns += url3.urlpatterns 

等