刚解决。
from django.urls import path,re_path
from . import views
urlpatterns = [
path('',views.myblog,name='first-url'),
re_path(r'^article//(?P<article_id>\d+)/$', views.article_page),
]
浏览器输入127.0.0.1:8000/app/article//1/
from django.urls import path,re_path
from . import views
urlpatterns = [
path('',views.myblog,name='first-url'),
re_path(r'^article//(?P<article_id>\d+)/$', views.article_page),
]
浏览器输入127.0.0.1:8000/app/article//1/
2019-05-19
最新回答 / Xiaoyu_Wu
你好!你是否在浏览器访问 index 之前,在命令行工具上运行 python manage.py runserver ?要先运行这个命令,得到服务地址和端口,才能在浏览器上访问。
2019-05-06
最赞回答 / 热粥
浏览器地址应该是 127.0.0.1/blog/article/1/另外,path里面加个参数
urlpatterns = [ path('index/', views.index), path('article/<int:article_id>/', views.article_page, name='article_page'), ]
2019-04-27