在Django框架中,我们一个项目可能有多个应用,如果所有的url放在一个模块里,当应用多的时候,那就先得很臃肿,这时候就需要按应用把url分离
cmdb
cmdb
settings.py
....
cms
__init__.py
models.py
urls.py
views.py
...
lend
__init__.py
views.py
urls.py
这里根目录下cms/urls.py文件如何引用 cms/lend/urls.py 呢?
cms/urls.py:
from django.conf.urls import patterns,include
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns(('cms.views'),
url(r'^$','login'),
url(r'^lend/', include('cms.lend.urls'))
)
cms/lend/urls.py:
from django.conf.urls import patterns,include,url
from cms.lend import views ##引用cms.lend.views
urlpatterns = [
url(r'^list/$', views.list_lend), ##通过http://127.0.0.1:5000/cms/lend/list访问cms/lend/views.py:list_lend方法
]
##注意这里的中括号
cms/lend/views.py:
def list_lend(request):
return render_to_response("lend/list.html", locals())
这样就能正确访问cms/lend/views.py 模块的内容了