Python 如何调用函数inside view.py

这是当前的项目结构

R6Scorextractor


        R6Scoreex
                 migrations
                 templates
                          R6Scoreex
                                   header.html
                                   home.html


                 __Init__.py
                 settings.py
                 urls.py 
                 views.py
                 models.py
                 apps.py
                 admin.py
                 tests.py
        R6Scorextractor
                 __Init__.py
                 settings.py
                 urls.py 
        manage.py

R6Scorextractor/R6scoreex/url.py

from django.conf.urls import url
from . import views

urlpatterns = [
    url(r'^$', views.index, name='index'),
    url(r'^simple_upload/', views.simple_upload, name='simple_upload'),

]

R6Scorextractor/R6scoreex/views.py

R6Scorextractor/R6Scorextractor/url.py

from django.conf.urls import url
from django.contrib import admin
from django.conf.urls import include

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    url(r'^/', include('R6scoreex.urls')),
]

我只是想知道如何调用R6scoreex模块的simple_upload inside views.py。如何为它编写URL，当我执行以下操作时，服务器会给我404错误

url(r'^/simple_upload/$', views.simple_upload, name='simple_upload'),

那么，是什么原因导致我在添加上述代码后出现404错误？我在这里做错了什么

---

您很可能有正则表达式问题

这个模式适合我：（R6Scorextractor/R6Scorextractor/url.py）

和（R6Scorextractor/R6scoreex/url.py）

---

你好您是否编写了此url（r'^/simple\u upload$/'，views.simple\u upload，name='simple\u upload'），其中的url.py？正确的一个是R6Scorextractor/r6scorex/url.py尝试删除regex部分的$and/以检查它是否有效。你也不需要在那里导入include。是的，我在r6scorex/url.py中写了这个url（r'^simple\u upload/$，views.simple\u upload，name='simple\u upload'），它给了我404。老实说，这对我来说不起作用，我也不知道为什么会发生这种情况，这就是我自己发布这个问题的原因。你能做一件事吗？如果我能制作一个压缩文件并发送给你，你能在你的系统上运行它并告诉我吗

url(r'^/simple_upload/$', views.simple_upload, name='simple_upload'),

from django.conf.urls import url
from django.contrib import admin
from django.conf.urls import include

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    url(r'^/', include('R6scoreex.urls')),
]

from django.conf.urls import url
from . import views

urlpatterns = [
    url(r'^$', views.index, name='index'),
    url(r'^simple_upload/', views.simple_upload, name='simple_upload'),

]