django - 使用API​​-View在Django-Rest-Framework中进行分页

Question

我目前有一个API View 设置，如下所示:

class CartView(APIView):
    authentication_classes = [SessionAuthentication, TokenAuthentication]
    permission_classes = [IsAuthenticated, ]
    api_view = ['GET', 'POST']

    def get(self, request, format=None):
        try:
            cart = request.user.cart
        except Cart.DoesNotExist:
            cart = Cart.objects.create(user=request.user)
        cart_details = cart.cart_details.all()
        serializer = CartDetailSerializer(cart_details, many=True, fields=['id', 'item', 'quantity', 'product_type'])
        return Response(serializer.data)

这里CartDetailSerializer是普通的ModelSerializer。

我想对这个API进行分页。但是，在DRF的文档中，我发现了这一点:

If you're using a regular APIView, you'll need to call into the pagination API yourself to ensure you return a paginated response.

没有提供有关如何对常规APIView API进行分页的示例。

任何人都可以发表我可以在上述情况下使用的示例。

谢谢。

Answer 1

使用常规APIView时，您需要使用Django自己的Paginator类。

Django Pagination in Views

在您的情况下，您可以在将查询集发送给序列化程序之前对其进行分页。

像这样:

def get(self, request, format=None):
    try:
        cart = request.user.cart
    except Cart.DoesNotExist:
        cart = Cart.objects.create(user=request.user)
    cart_details = cart.cart_details.all()

    paginator = Paginator(cart_details, 10)
    page = request.GET.get('page')

    try:
        cart_details = paginator.page(page)
    except PageNotAnInteger:
        # If page is not an integer, deliver first page.
        cart_details = paginator.page(1)
    except EmptyPage:
        # If page is out of range (e.g. 9999), deliver last page of results.
        cart_details = paginator.page(paginator.num_pages)
    serializer = CartDetailSerializer(cart_details, many=True, fields=['id', 'item', 'quantity', 'product_type'])
    return Response(serializer.data)

希望这可以帮助。