python - Django:将此行的 urls.py 中的使用 'url' 更改为 'path' (?P.+)

Question

我正在使用 django2.0 并希望将 urls.py 的格式从这里转换为

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

urlpatterns = [
        url(r'^$',  views.post_list , name='post_list'),
        url(r'^(?P<slug>[\w-]+)/$', views.post_detail, name="detail"),
        url(r'^category/(?P<hierarchy>.+)/$', views.show_category, name='category'),
        ]

类似的东西:

from django.urls import path
from . import views

app_name='home'

urlpatterns = [
    path('', views.post_list, name='post_list'),
    path(<slug:slug>/, views.post_detail, name="detail"),
    path(???),  
]

我应该如何以新格式编写最后一行代码。感谢您提供的任何帮助。

Answer 1

documentation指定五个路径转换器:int、path、str、uuid、slug。

正则表达式模式.+表示“匹配一个或多个字符”，字符可以是任何内容。因此，完全等效的路径转换器是 path:

path - Matches any non-empty string, including the path separator, '/'. This allows you to match against a complete URL path rather than just a segment of a URL path as with str.

所以你可以这样写:

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

urlpatterns = [
    path('', views.post_list, name='post_list'),
    path('<slug:slug>/', views.post_detail, name="detail"),
    path('category/<b><path:hierarchy></b>/', views.show_category, name='category'),
]

如果这种层次结构不需要必须匹配斜杠，那么您可以使用str来代替:

str - Matches any non-empty string, excluding the path separator, '/'. This is the default if a converter isn’t included in the expression.