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我希望在从我的仪表板页面浏览链接时，该 url 应该是/dashboard/link. 当我写url(r'dashboard/inbox/','apps.dashboard.views.inbox', name = 'grabhalo_inbox'),它时，它确实会读取我views提供的论点。

相反，当我编写时，url(r'inbox/','apps.dashboard.views.inbox', name = 'grabhalo_inbox'),我得到了预期的输出，但 url 是/inbox. 我想要的是/dashboard/inbox。

我哪里错了？

这是我的根urls.py文件

urlpatterns = patterns('',
    (r'^',include('apps.auth.urls')),
    (r'^',include('apps.dashboard.urls')),
    (r'^static/(?P<path>.*)$', 'django.views.static.serve', {'document_root':settings.MEDIA_ROOT}),
    url(r'^admin/', include(admin.site.urls)),
)

这是我的dashboard.urls.py

urlpatterns = patterns('',
    url(r'dashboard/','apps.dashboard.views.dashboard', name = 'grabhalo_dashboard'),
    url(r'sent/','apps.dashboard.views.sent', name = 'grabhalo_sent'),
    url(r'inbox/','apps.dashboard.views.inbox', name = 'grabhalo_inbox'),
    )
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1 回答 1

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只需将^and $（字符串的开头和结尾）添加到您的dashboard/url 正则表达式：

urlpatterns = patterns('',
    url(r'^dashboard/$','apps.dashboard.views.dashboard', name = 'grabhalo_dashboard'),
    url(r'^sent/$','apps.dashboard.views.sent', name = 'grabhalo_sent'),
    url(r'^dashboard/inbox/$','apps.dashboard.views.inbox', name = 'grabhalo_inbox'),
    )

然后，http://mydomain.com/dashboard/将由apps.dashboard.views.dashboard视图处理， http://mydomain.com/dashboard/inbox/- 由apps.dashboard.views.inbox.

于 2013-08-04T04:50:40.343 回答