[英]How to have separate urls for the same app in Django 1.9
我有一個應用程序,將有2種網址,一個將包含在其他應用程序中,一個將包含在設置應用程序中。 我希望有一種方法只包含部分網址而不為其創建單獨的文件。
# records app -- urls.py
urlpatterns = [
url(r'^create/$', RecordCreate.as_view(), name="record-create"),
url(r'^(?P<pk>\d+)/update/$', RecordUpdate.as_view(), name="record-update"),
url(r'^(?P<pk>\d+)/delete/$', RecordDelete.as_view(), name="record-delete"),
]
urlpatterns_types = [
url(r'^$', RecordTypeList.as_view(), name="record-type-list"),
url(r'^(?P<pk>\d+)/$', RecordTypeDetail.as_view(), name="record-type-detail"),
url(r'^create/$', RecordTypeCreate.as_view(), name="record-type-create"),
url(r'^(?P<pk>\d+)/update/$', RecordTypeUpdate.as_view(), name="record-type-update"),
url(r'^(?P<pk>\d+)/delete/$', RecordTypeDelete.as_view(), name="record-type-delete"),
]
現在在設置應用程序中,我想只包含urlpatterns_types
網址。 但是我試着把它們包括在內但我做不到
我發現創建單獨文件然后將它們作為模塊包含的唯一方法
以下是預期結果的示例
# player app -- urls.py
from django.conf.urls import patterns, include, url
from .views import *
urlpatterns = [
# Records App Urls
url(r'^(?P<player_id>\d+)/records/', include('records.urls')),
]
# settings app -- urls.py
from django.conf.urls import patterns, include, url
from .views import *
urlpatterns = [
# Records App Urls
url(r'^(?P<player_id>\d+)/records/', include('records.urls.urlpatterns_types')),
]
項目樹
-- soccer_game
-- soccer_game
-- settings.py
-- urls.py
-- players
-- models.py
-- urls.py
-- views.py
-- main_settings
-- models.py
-- urls.py
-- views.py
您可以導入模塊並傳遞URL本身列表:
# settings app -- urls.py
from django.conf.urls import patterns, include, url
from records import urls as records_urls
from .views import *
urlpatterns = [
# Records App Urls
url(r'^(?P<player_id>\d+)/records/', include(records_urls.urlpatterns_types)),
]
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