Question

我有一个模特：

class Scenario(models.Model):
    tasks = models.ManyToManyField(Task, blank=True)

它的序列化器：

class ScenarioSerializer(serializers.ModelSerializer):

    class Meta:
        model = Scenario
        fields = '__all__'

以及创建方案的视图：

@api_view(['GET', 'POST'])
def scenarios_list(request):
    """
    List all scenarios, or create a new.
    """
    if request.method == 'GET':
        scenarios = Scenario.objects.all()
        serializer = ScenarioSerializer(scenarios, many=True)
        return Response(serializer.data)

    elif request.method == 'POST':
        serializer = ScenarioSerializer(data=request.data)
        if serializer.is_valid():
            serializer.save()
            return Response(serializer.data, status=status.HTTP_201_CREATED)
        else:
            return Response(
                serializer.errors, status=status.HTTP_400_BAD_REQUEST)

我从我的前端发送以下数据：

[
    {id: 3, title: "Three", how_often: "DS", how_important_task: "EI", role: "Lorem", …},

    {id: 1, title: "One", how_often: "MO", how_important_task: "RI", role: "Lorem", …},

    {id: 6, title: "Six", how_often: "WO", how_important_task: "EI", role: "", …},

    {id: 4, title: "Four", how_often: "", how_important_task: "", role: "", …}
]

这引发了我BAD REQUEST错误。我做错了什么？

Answer 1

将您的请求有效负载更改为dict类似对象，如下所示（问题为list），然后发送

{title: "Three", how_often: "DS", how_important_task: "EI", role: "Lorem", …}

我从请求有效负载中删除了id密钥，因为根据您的序列化程序和模型，您不必在创建新对象时指定id，

Django休息错误请求错误

1 个答案: