python - Django REST 框架 JSONParser().parse(request) 引发错误

Question

在 Django View 中，我正在尝试这个 -

@csrf_exempt
def customer_list(request):
    """
    List all customers, or create a new customer.
    """
    if request.method == 'GET':
        snippets = Customer.objects.all()
        serializer = CustomerSerializer(snippets, many=True)
        return JsonResponse(serializer.data, safe=False)

    elif request.method == 'POST':
        data = JSONParser().parse(request)
        serializer = CustomerSerializer(data=data)
        if serializer.is_valid():
            serializer.save()
            return JsonResponse(serializer.data, status=201)
        return JsonResponse(serializer.errors, status=400)

在从 postman 发布数据时，我得到

JSON parse error - Expecting value: line 1 column 1 (char 0)

Answer 1

问题在于您将整个请求对象传递给 JSON 解析器，而不是包含实际 JSON 内容的正文。

但无论如何你不应该自己这样做。让 DRF 为您做这件事。

from rest_framework.decorators import api_view
from rest_framework.response import Response

@api_view()
def customer_list(request):
    """
    List all customers, or create a new customer.
    """
    if request.method == 'GET':
        snippets = Customer.objects.all()
        serializer = CustomerSerializer(snippets, many=True)
        return Response(serializer.data)

    elif request.method == 'POST':
        serializer = CustomerSerializer(data=request.data)
        if serializer.is_valid():
            serializer.save()
            return Response(serializer.data, status=201)
        return Response(serializer.errors, status=400)

或者更好的是，使用基于类的 View ，它已经为您包含了所有这些功能:

from rest_framework import generics

class CustomerList(generics.ListCreateAPIView):
     model = Customer
     serializer_class = CustomerSerializer

这就是您所需要的。