在python url模式中找不到url的反向

Question

我进行了URL配置

urlpatterns = [
path('', lambda request: redirect('stocks', permanent=False)),
path('stocks', views.stocks.index, name='stocks.index'),
path('stocks/<slug:slug>', views.stocks.show, name='stocks.show'),
path('stocks/<slug:slug>/load', views.stocks.load, name='stocks.load')
]

这些URL是可以的，它们可以工作，但不是第一个

path('', lambda request: redirect('stocks', permanent=False)),

当我尝试时，它会给出http://127.0.0.1:8000

NoReverseMatch at /
Reverse for 'stocks' not found. 'stocks' is not a valid view function or pattern name.

但如果我尝试http://127.0.0.1:8000/stocks一切正常

这里怎么了？

Answer 1

您可能应该使用stocks.index而不是stocks

path('', lambda request: redirect('stocks.index', permanent=False)),