Python 为什么我的django url配置未能包含另一个应用程序url模块

我有一个项目，其根url conf内容为：

 from django.conf.urls import patterns, include, url
import funnytest
urlpatterns = patterns(
url(r'^funnytest/', include('funnytest.urls')),
url(r'^helloworld/', funnytest.views.hello),
)

funnytest是本项目的一个应用程序，在funnytest中我编写了一个模块urls.py来配置此应用程序的请求：

from django.conf.urls import patterns, include, url
from views import *
urlpatterns = patterns(
url(r'^hello/$', hello),
)

当我访问localhost/funnytest/hello/时，返回一个dispath错误，表示没有这样的模式

当我访问localhost/helloworld时，它工作得很好

---

为什么呢，应该如何配置～

如果查看patterns函数的定义：

def patterns(prefix, *args):
    pattern_list = []
    for t in args:
        if isinstance(t, (list, tuple)):
            t = url(prefix=prefix, *t)
        elif isinstance(t, RegexURLPattern):
            t.add_prefix(prefix)
        pattern_list.append(t)
    return pattern_list

您将看到，模式在url模式列表之前有一个参数“prefix”

from django.conf.urls import patterns, include, url
import funnytest
urlpatterns = patterns(
    '',
    url(r'^funnytest/', include('funnytest.urls')),
    url(r'^helloworld/', funnytest.views.hello),
)

from django.conf.urls import patterns, include, url
from views import *
urlpatterns = patterns(
    '',
    url(r'^hello/$', hello),
)

在这两个文件中尝试以下操作：添加一个空字符串作为模式的第一个参数

from django.conf.urls import patterns, include, url
import funnytest
urlpatterns = patterns(
    '',
    url(r'^funnytest/', include('funnytest.urls')),
    url(r'^helloworld/', funnytest.views.hello),
)

from django.conf.urls import patterns, include, url
from views import *
urlpatterns = patterns(
    '',
    url(r'^hello/$', hello),
)

---

下面是关于view prefix参数的更好的讨论：