Question

Django 2.1，python 3.6，djangorestframework。

当我进入以下网址时，我可以看到我的数据（很棒！）http://127.0.0.1:8000/api/cards/1这就是我在api页面上看到的-

HTTP 200 OK
Allow: GET, PUT, PATCH, DELETE, HEAD, OPTIONS
Content-Type: application/json
Vary: Accept

[
    {
        "id": "1",
        "card_title": "Hello"
    },
]

我希望能够转到该网址以获取相同的数据-http://127.0.0.1:8000/api/cards/title/Hello

我该如何更新我的视图和URL？

基本网址

urlpatterns = [
    ...
    path('api/cards/', include('cards.api.urls')),
]

cards.api.urls.py

urlpatterns = [
    path('', CardListView.as_view()),
    path('<str:pk>/', CardDetailView.as_view()),
]

urlpatterns = format_suffix_patterns(urlpatterns)

cards.api.views.py

class CardList(generics.ListCreateAPIView):
    permission_classes = ()

    queryset = Card.objects.all()
    serializer_class = CardSerializer


class CardDetail(generics.RetrieveUpdateDestroyAPIView):
    #permisssion_classes = (UserPermission,) # set the permission class
    permission_classes = ()

    queryset = Card.objects.all()
    serializer_class = CardSerializer

我尝试将其添加到cards.api.urls.py path('api/cards/title/<str:pk>/', CardDetail.as_view()),中，但它仍在查看id变量而不是card_title变量。

Answer 1

我认为这会对您有所帮助，

# base urls.py
urlpatterns = [
    ...
    path('api/', include('cards.api.urls')), # remove "cards/" from url
]

和创建如下的新视图类 CardTitleDetail 并添加 lookup_field 属性

class CardTitleDetail(generics.RetrieveUpdateDestroyAPIView):
    lookup_field = 'card_title'
    permission_classes = ()
    queryset = Card.objects.all()
    serializer_class = CardSerializer

# cards.api.urls.py
urlpatterns = [
    path('cards/', CardListView.as_view()),  # add "cards/" to the url
    path('cards/<str:pk>/', CardDetailView.as_view()),  # add "cards/" to the url
    path('cards/title/<str:card_title>/', CardTitleDetail.as_view()),  # this is the new url
]

urlpatterns = format_suffix_patterns(urlpatterns)

注意
在整个数据库中， card_title 属性应为 unique ，否则将引发异常！

Answer 2

您可以在您的cards.api.urls.py文件中添加另一个网址，以重定向到相同的api以获得相同的响应。

将此添加到您的urls.py文件urlpatterns中。

path('title/<str:pk>/', CardDetailView.as_view()),

然后尝试点击此http://127.0.0.1:8000/api/cards/title/1

如何制作不同的Django Api网址路径

2 个答案:

将此添加到您的urls.py文件urlpatterns中。