Django的urls.py查询

2013-08-04 135 views 1 likes

我希望从我的仪表板页面浏览链接时，该网址应该是/dashboard/link。当我写url(r'dashboard/inbox/','apps.dashboard.views.inbox', name = 'grabhalo_inbox'),它确实读取我提供的views参数。Django的urls.py查询

而是当我写url(r'inbox/','apps.dashboard.views.inbox', name = 'grabhalo_inbox'),时，我得到预期的输出，但URL是/inbox。我想要的是/dashboard/inbox。

我哪里错了？

这里是我的根urls.py文件

这里是我的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'), 
    )

来源

2013-08-04 PythonEnthusiast

回答

只需添加^和$（开始和结束字符串）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查看，http://mydomain.com/dashboard/inbox/处理 - 由apps.dashboard.views.inbox处理。

来源

2013-08-04 04:50:40 alecxe

Whollaaa ...谢谢:) – PythonEnthusiast

Django的urls.py查询

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