[英]Django-REST: Get User by username instead of primary-key
I am using the basic UserViewSet derived from ModelViewSet.我正在使用从 ModelViewSet 派生的基本 UserViewSet。
Retrieving a user with a primary-key via api/users/<pk>
works fine.通过
api/users/<pk>
使用主键检索用户工作正常。
But I also want to be able to retrieve a User by Username.但我也希望能够通过用户名检索用户。
I have added a new detail route but I always get 404
on my server when I try to get the user with the url /api/users/retrieve_by_username/altoyr
.我添加了一个新的详细路由,但是当我尝试使用 url
/api/users/retrieve_by_username/altoyr
获取用户时,我的服务器上总是得到404
。
class UserViewSet(viewsets.ModelViewSet):
queryset = User.objects.all()
serializer_class = UserSerializer
@detail_route(methods=['get'])
def retrieve_by_username(self, request, username=None):
try:
user = User.objects.get(userName=username)
return Response(user)
except User.DoesNotExist:
return Response("No user with username found!", status=status.HTTP_400_BAD_REQUEST)
The Urls are registered via a router:网址是通过路由器注册的:
router = DefaultRouter()
router.register(r'users', views.UserViewSet)
# The API URLs are now determined automatically by the router.
# Additionally, we include the login URLs for the browsable API.
urlpatterns = [
url(r'^', include(router.urls)),
url(r'^api-auth/', include('rest_framework.urls', namespace='rest_framework'))
]
I think I am missing an important part of building rest urls.我想我错过了构建休息网址的重要部分。
You can do this by adding a list route like:您可以通过添加一个列表路由来做到这一点,如:
@list_route(methods=['get'], url_path='retrieve_by_username/(?P<username>\w+)')
def getByUsername(self, request, username ):
user = get_object_or_404(User, username=username)
return Response(UserSerializer(user).data, status=status.HTTP_200_OK)
and the url will be like:网址将如下所示:
/api/users/retrieve_by_username/altoyr
You can override the retrieve() ViewSet action.您可以覆盖retrieve() ViewSet 操作。 You'll find more detail here: https://www.django-rest-framework.org/api-guide/viewsets/
您可以在此处找到更多详细信息: https : //www.django-rest-framework.org/api-guide/viewsets/
from django.shortcuts import get_object_or_404
from rest_framework.response import Response
class UserViewSet(viewsets.ModelViewSet):
queryset = User.objects.all()
serializer_class = UserSerializer
def retrieve(self, request, pk=None):
queryset = User.objects.filter(username=pk)
contact = get_object_or_404(queryset, pk=1)
serializer = ContactSerializer(contact)
return Response(serializer.data)
The answer from Anush Devendra is right but need a little update due to deprecations on v3.9 . Anush Devendra的答案是正确的,但由于v3.9 的弃用需要进行一些更新。
action decorator replaces
list_route
anddetail_route
Bothlist_route
anddetail_route
are now deprecated in favour of the single action decorator.action 装饰器替换
list_route
和detail_route
现在不推荐使用list_route
和detail_route
以支持单个动作装饰器。 They will be removed entirely in 3.10.它们将在 3.10 中完全删除。 The action decorator takes a boolean detail argument.
动作装饰器接受一个布尔细节参数。
- Replace detail_route uses with @action(detail=True).
用@action(detail=True) 替换 detail_route 使用。
- Replace list_route uses with @action(detail=False).
用@action(detail=False) 替换 list_route 使用。
...
from rest_framework.decorators import action
from rest_framework.response import Response
from django.shortcuts import get_object_or_404
from rest_framework import status
class UserViewSet(viewsets.ModelViewSet):
...
@action(methods=['get'], detail=False,
url_path='username/(?P<username>\w+)')
def getByUsername(self, request, username):
user = get_object_or_404(User, username=username)
data = UserSerializer(user, context={'request': request}).data
return Response(data, status=status.HTTP_200_OK)
I add context={'request': request}
because I have url
as HyperlinkedIdentityField in my serializer.我添加
context={'request': request}
因为我的序列化程序中有url
作为 HyperlinkedIdentityField 。 If you don't have it, you probably don't need it.如果你没有它,你可能不需要它。
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