Python “Django”；主页「；app url.py问题

我是Django的新手，当我布置我的网站时，我是用以下方式做的：

这个项目是一个“门户网站”

• app1“home”是一个包含主页、登录和注册的应用程序 以及其他“站点范围”功能
• App 2“存货”是一种基本资产存货
• App 3“仪表板”是一组基于库存资产的状态仪表板

现在我正在尝试添加登录功能，我有一个主项目urls.py，如下所示：

文件：/portal/url.py

from django.conf import settings
from django.conf.urls import patterns, include, url

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    url(r'^$', include('home.urls'), name='home'),
    url(r'^admin/', include(admin.site.urls), name='admin'),
    url(r'^inventory/', include('inventory.urls'), name='inventory'),
    url(r'^dashboard/', include('dashboard.urls'), name='dashboard'),
    url(r'^capacity/', include('capacity.urls'), name='capacity'),
)

from django.conf.urls import patterns, url
from home import views

urlpatterns = patterns('',
    url(r'^test/$', views.index, name='home_test'),
    url(r'^ajax_login/$', views.ajax_login, name='ajax_login'),
    url(r'^$', views.index, name='home_index'),
)

from django.conf import settings
from django.conf.urls import patterns, include, url

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    url(r'^admin/', include(admin.site.urls), name='admin'),
    url(r'^inventory/', include('inventory.urls'), name='inventory'),
    url(r'^dashboard/', include('dashboard.urls'), name='dashboard'),
    url(r'^capacity/', include('capacity.urls'), name='capacity'),
    url(r'^', include('home.urls'), name='home'),
)

我的计划是使用…/home/url.py来管理任何站点范围的功能，如登录、注册等

现在home index视图显示的很好，但是…/home/url.py中的任何其他内容都会给我一个404

文件：/home/url.py

from django.conf import settings
from django.conf.urls import patterns, include, url

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    url(r'^$', include('home.urls'), name='home'),
    url(r'^admin/', include(admin.site.urls), name='admin'),
    url(r'^inventory/', include('inventory.urls'), name='inventory'),
    url(r'^dashboard/', include('dashboard.urls'), name='dashboard'),
    url(r'^capacity/', include('capacity.urls'), name='capacity'),
)

from django.conf.urls import patterns, url
from home import views

urlpatterns = patterns('',
    url(r'^test/$', views.index, name='home_test'),
    url(r'^ajax_login/$', views.ajax_login, name='ajax_login'),
    url(r'^$', views.index, name='home_index'),
)

from django.conf import settings
from django.conf.urls import patterns, include, url

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    url(r'^admin/', include(admin.site.urls), name='admin'),
    url(r'^inventory/', include('inventory.urls'), name='inventory'),
    url(r'^dashboard/', include('dashboard.urls'), name='dashboard'),
    url(r'^capacity/', include('capacity.urls'), name='capacity'),
    url(r'^', include('home.urls'), name='home'),
)

在这一点上，我想我的问题是双重的：我的做法正确吗？如果是，如何获得所需的功能？如果没有，我应该如何改变事情的设置/布置方式，以正确的“最佳实践”方式进行

提前谢谢

编辑

多亏了dt0xff和holdenweb，它才得以工作，接受了holdenweb的答案，因为它更准确，包括需要将主url包含放在其他url的下面

这是我的新门户/url.py文件，供参考

文件：/portal/url.py

from django.conf import settings
from django.conf.urls import patterns, include, url

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    url(r'^$', include('home.urls'), name='home'),
    url(r'^admin/', include(admin.site.urls), name='admin'),
    url(r'^inventory/', include('inventory.urls'), name='inventory'),
    url(r'^dashboard/', include('dashboard.urls'), name='dashboard'),
    url(r'^capacity/', include('capacity.urls'), name='capacity'),
)

from django.conf.urls import patterns, url
from home import views

urlpatterns = patterns('',
    url(r'^test/$', views.index, name='home_test'),
    url(r'^ajax_login/$', views.ajax_login, name='ajax_login'),
    url(r'^$', views.index, name='home_index'),
)

from django.conf import settings
from django.conf.urls import patterns, include, url

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    url(r'^admin/', include(admin.site.urls), name='admin'),
    url(r'^inventory/', include('inventory.urls'), name='inventory'),
    url(r'^dashboard/', include('dashboard.urls'), name='dashboard'),
    url(r'^capacity/', include('capacity.urls'), name='capacity'),
    url(r'^', include('home.urls'), name='home'),
)

---

Django在URL中的外观如下：

获取根URL列表-门户/URL

首先查找与当前url匹配的

如果是include，则转到includeurl，删除匹配的部分

你的问题就在这里

url(r'^$', include('home.urls'), name='home'),

我是说，这里

'^$'

您希望url匹配“url的开始和结束”。它对root（

dunno.com/

）很好，但对其他人却不行，因为url将包含更多内容

所以，只要删除

$

，就可以了

url(r'^', include('home.urls'), name='home'),

---

问题在于您的第一个模式，它将只匹配空URL。因此，任何空URL都会导致分析

home/urls.py

URL，但即使在这种情况下，也只有空URL会匹配

不幸的是，如果没有公共前缀，那么模式

“^”

将匹配每个URL（因为它们都从开头开始…）

---

因此，您应该为所有页面使用一个公共前缀，或者在测试之前将home URL规范放在末尾，以使其他URL有机会匹配。

您永远不需要在project/URLs.py添加正则表达式（$）

，尽管即使空字符串有一个开头，此模式也会匹配所有字符串。因此，顶级列表中的任何其他URL都不可能匹配，因此我建议对列表重新排序。请在回答中添加解释。