如何在 get 方法 (drf) 中传递 json 参数？答案

【问题标题】：How can i pass a json param in get method (drf)?如何在 get 方法 (drf) 中传递 json 参数？
【发布时间】：2023-01-27 00:39:49
【问题描述】：

class getAlarmlogsFilterByDistincts(APIView):
    permission_classes = (IsAuthenticated,)

    def get(self, request, *args, **kwargs): 
        column_name = request.data['column_name'] #just an idea, dont works
        distincts_list = request.data['distincts_list'] #just an idea, dont works
        column_values = [Alarmlog.objects.filter(**{column_name: x}) for x in distincts_list ]
        
        serialized_response = AlarmlogSerializer(column_values)

        return Response(serialized_response.data, status=status.HTTP_200_OK)

我的看法或多或少是这样的。我需要在 column_name 中接收一个字符串，在 distincts_list 中接收一个列表（可能是 json 格式）。

我尝试了什么：

class AlarmlogsFilterByDistinctsSerializer(serializers.Serializer): 
    distincts_list = serializers.ListField()
    column_name = serializers.CharField()  


class getAlarmlogsFilterByDistincts(CreateAPIView):
    permission_classes = (IsAuthenticated,)
    serializer_class = AlarmlogsFilterByDistinctsSerializer

    def post(self, request, *args, **kwargs): 

        column_values = columnFilter(request.data['column_name'], request.data['distincts_list'])
        
        response_serialized = serializers.serialize('json', column_values)

        print(response_serialized)
        return Response(response_serialized, status=status.HTTP_200_OK)

我试过使用 post 方法，但我的回复是一个字符串： "[{\"model\": \"alarms.alarmlog\", \"pk\": 26, \"fields\": {\"created_at\": \"2023-01-12T17:06:19.016Z\", \"updated_at\": \"2023-01-12T17:06:19.016Z\", \"name\": \"string43\", \"description\": \"string\", \"severity\": \"low\", \"type\": \"string\", \"period\": \"string\", \"value\": \"string\", \"reference\": \"string\", \"source\": 2147483647, \"third_party_id\": \"string\", \"device\": \"string\", \"extra_content\": {}, \"occurred_at\": \"2023-01-12T16:29:39.055Z\", \"equipment\": 7}}, {\"model\": \"alarms.alarmlog\", \"pk\": 27,...

附言。对不起我的英语

【问题讨论】：

标签： python django django-rest-framework

【解决方案1】：

你可以在下面返回数据

import json


return Response(json.dumps(response_serialized), status=status.HTTP_200_OK)

【讨论】：