如何在 Django REST 框架中实现过滤器 API答案

【问题标题】：How to implement filter API in Django rest framework如何在 Django REST 框架中实现过滤器 API
【发布时间】：2016-08-02 19:36:44
【问题描述】：

我正在尝试在 django rest 中实现过滤器 api。有点像 -

localhost:8000/api/v1/users/?email=abc.xzy@gmail.com/

所以它应该使用通过的过滤器搜索用户并返回结果。但目前它正在返回所有用户。

URLS.py

    url(r'^api/v1/users/$',   UserViews.UserList.as_view(), name='userlist_view'),
    url(r'^api/v1/users/(?P<email>.+)/$', UserViews.UserList.as_view(), name='userList_view'),
    url(r'^api/v1/users/(?P<pk>[0-9]+)/$', UserViews.UserDetail.as_view(), name='userdetail_view'),

UserViews.py

class UserList(generics.ListAPIView):
    """
    List all users, or create a new user.
    """
    lookup_url_kwarg = "email"

    def get(self, request, format=None):
        if request.user.is_authenticated():
            users = User.objects.all()
            serializer = UserSerializer(users, many=True)
            return Response(serializer.data)
        return Response("User is not authenticated.", status=status.HTTP_400_BAD_REQUEST)

    def post(self, request, format=None):

        valid_paylaod, msg = UserListRepository.validations_create_user(request.data)
        if not valid_paylaod:
            return Response(msg, status=status.HTTP_400_BAD_REQUEST)

        result = UserListRepository.create_user_repo(request)

        if not result.success:
            return Response(str(result.msg), status=result.status )

        return Response(UserSerializer(result.data).data, status=result.status)

   def get_queryset(self):
    # It restricts the userlist by retunning users having emails passed in uery param
    user = self.request.user
    if user.is_authenticated():
         if 'email' in self.request.query_params:
            email = self.request.query_params.get('email', None)
            users = User.objects.get(email= email)
            if not users:
                return Response( "User Not found", status=status.HTTP_404_NOT_FOUND)
            else:
                return Response(UserSerializer(User.objects.all()).data, status.HTTP_200_OK, users)
        else:
            return Response(UserSerializer().data, status=result.status)
    else:
        return Response("User is not authenticated.", status=status.HTTP_400_BAD_REQUEST)

有人能说出为什么请求不去 get_queryset() 方法和去 UserList 方法的 get() 吗？在删除 get 方法时，请求转到 get_queryset(self) 方法。在调试时，我发现我在 return 语句之前得到了有效的响应 -

(Pdb) UserSerializer(result.data).data
{'parent_id': 2, 'id': 31, 'group_id': '4', 'last_name': 'user',         'email': 'agency22_user@agency22.com', 'organization_id': 0, 'first_name':  'agency22'}

但仍然在 APi 响应中，我收到以下错误：

Internal Server Error: /api/v1/users/
 Traceback (most recent call last):
 File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/django/core/handlers/base.py", line 149, in get_response
response = self.process_exception_by_middleware(e, request)
 File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/django/core/handlers/base.py", line 147, in get_response
response = wrapped_callback(request, *callback_args, **callback_kwargs)
Fi e "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/django/views/decorators/csrf.py", line 58, in wrapped_view
return view_func(*args, **kwargs)
File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/django/views/generic/base.py", line 68, in view
return self.dispatch(request, *args, **kwargs)
 File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/rest_framework/views.py", line 466, in dispatch
response = self.handle_exception(exc)
File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/rest_framework/views.py", line 463, in dispatch
response = handler(request, *args, **kwargs)
File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/rest_framework/generics.py", line 201, in get
return self.list(request, *args, **kwargs)
File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/rest_framework/mixins.py", line 43, in list
if page is not None:
File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/rest_framework/serializers.py", line 674, in data
ret = super(ListSerializer, self).data
File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/rest_framework/serializers.py", line 239, in data
self._data = self.to_representation(self.instance)
File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-    packages/rest_framework/serializers.py", line 614, in to_representation
self.child.to_representation(item) for item in iterable
 File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-  packages/django/template/response.py", line 173, in __iter__
raise ContentNotRenderedError('The response content must be '
 django.template.response.ContentNotRenderedError: The response content          must be rendered before it can be iterated over.

我不清楚为什么。我点击的 API URL 是：localhost:8000/api/v1/users?email=agency22_user@agency22.com

【问题讨论】：

self.lookup_url_kwarg，为什么在这里self？
我们甚至可以尝试以下操作 :def get_queryset(self,request): # 它通过返回具有在 uery 中传递的电子邮件的用户来限制用户列表 param email = self.request.query_params.get('email',无）返回 User.objects.filter(email=email) 问题是，请求不去那个方法。

标签： django django-models django-views django-rest-framework

【解决方案1】：

在 Django REST 框架中，有一个关于如何使用过滤器的适当规范。所以你的用户类看起来像。

class UserList(generics.ListAPIView):
    """
    List all users, or create a new user.
    """
    lookup_url_kwarg = "email"
    serializer_class = UserSerializer

    def get(self, request, format=None):
        if request.user.is_authenticated():
            users = User.objects.all()
            serializer = UserSerializer(users, many=True)
            return Response(serializer.data)
        return Response("User is not authenticated.", status=status.HTTP_400_BAD_REQUEST)

    def post(self, request, format=None):

        valid_paylaod, msg = UserListRepository.validations_create_user(request.data)
        if not valid_paylaod:
            return Response(msg, status=status.HTTP_400_BAD_REQUEST)

        result = UserListRepository.create_user_repo(request)

        if not result.success:
            return Response(str(result.msg), status=result.status )

        return Response(UserSerializer(result.data).data, status=result.status)

    def get_queryset(self):
        email = self.request.query_params.get('email', None)
        return User.objects.filter(email=email)

或者您可以尝试不使用 get_queryset 。链接会改变，更新的链接是：

localhost:8000/api/v1/users/abc.xzy@gmail.com/

更新后的代码是

class UserList(generics.ListCreateAPIView):
        """
        List all users, or create a new user.
        """
        lookup_url_kwarg = "email"
        serializer_class = UserSerializer

        def get(self, request, email,format=None):
            if request.user.is_authenticated():
                user_details = User.objects.filter(email=email)
                serializer = UserSerializer(user_details, many=True)
                return Response(serializer.data, status=status.status.HTTP_200_OK)
            return Response("User is not authenticated.", status=status.HTTP_400_BAD_REQUEST)

        def post(self, request, format=None):

            valid_paylaod, msg = UserListRepository.validations_create_user(request.data)
            if not valid_paylaod:
                return Response(msg, status=status.HTTP_400_BAD_REQUEST)

            result = UserListRepository.create_user_repo(request)

            if not result.success:
                return Response(str(result.msg), status=result.status )

            return Response(UserSerializer(result.data).data, status=result.status)

【讨论】：

嗯，有两个api-
@RichaGupta 我已经更新了代码，因为您正在使用两个 API，一个是 GET 另一个是 POST ，所以您需要使用 ListCreateAPIView 那个用户 ListModelMixin （具有可以在现场使用的列表方法get 的方法，这将更有意义）、CreateModelMixin（具有可用于代替 post 方法并且更有意义的 create 方法）和 GenericAPIView（具有当前正在继承的 get 和 post 方法）。

【解决方案2】：

看起来您正在使用来自 django 的 Response 而不是 django-rest-framework。

你必须从 rest_framework.response 导入响应

from rest_framework.response import Response

【讨论】：

【解决方案3】：

我对 UrlConf 感到困惑：

url(r'^api/v1/users/$', UserViews.UserList.as_view(), name='userlist_view'),
url(r'^api/v1/users/(?P<email>.+)/$', UserViews.UserList.as_view(), name='userList_view'),

第一个是localhost:8000/api/v1/users/

第二个是localhost:8000/api/v1/users/abc.xzy@gmail.com/而不是localhost:8000/api/v1/users/?email=abc.xzy@gmail.com/，这是一个查询参数不是kwargs，如果你想use query parameter就勾选这个

class UserList(generics.ListAPIView):
"""
List all users, or create a new user.
"""
lookup_url_kwarg = "email"
serializer_class = UserSerializer

def post(self, request, format=None):

    valid_paylaod, msg = UserListRepository.validations_create_user(request.data)
    if not valid_paylaod:
        return Response(msg, status=status.HTTP_400_BAD_REQUEST)

    result = UserListRepository.create_user_repo(request)

    if not result.success:
        return Response(str(result.msg), status=result.status )

    return Response(UserSerializer(result.data).data, status=result.status)
def get_queryset(self):
"""
It restricts the userlist by return users having emails passed in query param
"""
    queryset = Users.objects.all()
    email = self.request.query_params.get('email', None)
    if email is not None:
        queryset = queryset.filter(email=email)
    return queryset

【讨论】：

我可以通过删除 get() 方法来调用我的 get_query() 方法。我仍然无法解决问题中提到的错误（上）
因为您要覆盖默认行为，所以您使用的是通用视图，默认情况下会为您提供 get() 方法，停止覆盖 get() 并会得到您想要的，但对 get_queryset( )
并从 View 类中获取授权代码 django-rest-framework.org/api-guide/authentication/…

【解决方案4】：

问题是您覆盖了ListAPIView 的get 方法。此方法由 GET 请求调用。 get 的默认实现调用ListModelMixin 的list 方法，然后get_queryset 被调用：

def list(self, request, *args, **kwargs):
    queryset = self.filter_queryset(self.get_queryset())

    page = self.paginate_queryset(queryset)
    if page is not None:
        serializer = self.get_serializer(page, many=True)
        return self.get_paginated_response(serializer.data)

    serializer = self.get_serializer(queryset, many=True)
    return Response(serializer.data)

如果您覆盖此方法，您必须自己调用get_queryset。如果您只想检查用户是否经过身份验证，然后调用默认列表方法，您可以这样做：

  def get(self, request, format=None):
        if request.user.is_authenticated():
            return super(UserList, self).get(request, format)
        return Response("User is not authenticated.", status=status.HTTP_400_BAD_REQUEST)

【讨论】：

我可以通过删除 get() 方法来调用我的 get_query() 方法。我仍然无法解决问题（上）中提到的错误