django2.2 配置urls

    xiaoxiao2022-07-06  204

    项目名下:

    urls.py

    from django.contrib import admin from django.urls import path,include

    urlpatterns = [     path('admin/', admin.site.urls),     path('',include('booktest.urls')), ]

    APP下:

    urls.py

    from django.contrib import admin from django.urls import path from . import views

    urlpatterns = [     # path('admin/', admin.site.urls),     path('index/', views.index) ]

    然后开启服务:访问以下地址

    http://127.0.0.1:8000/index/

     

    最新回复(0)