Python 如何在django中为json响应定义嵌套列表

我是Django的新手，所以我想创建一个代表航运公司运营的基本应用程序。我有一份包含货物的工作单。因此，my models.py包含以下内容：

class WorkOrder (models.Model):
    status = models.CharField(max_length=300,default = "New")
    source = models.CharField(max_length=300)
    destination = models.CharField(max_length=
    material = models.CharField(max_length=300)
    shipmentlist = [] //PROBLEMATIC CODE

class Shipment (models.Model):
    expected_startDate = models.DateTimeField(default=timezone.now)
    expected_endDate = models.DateTimeField(default=timezone.now)
    shipment_status = models.CharField(max_length=300,default = "Not Started")

我在serializers.py中定义了两个序列化程序WorkOrderSerializer和ShipmentSerializer。我想返回包含在工单对象中的装运列表

class WorkOrderSerializer        
    generated_date = models.DateTimeField(default=timezone.now)
    status = models.CharField(max_length=300, default="New")
    source = models.CharField(max_length=300)
    destination = models.CharField(max_length=300)
    material = models.CharField(max_length=300)
    shipmentlist = ShipmentSerializer(many=True)

class ShipmentSerializer
    expected_startDate = models.DateTimeField(default=timezone.now)
    expected_endDate = models.DateTimeField(default=timezone.now)
    shipment_status = models.CharField(max_length=300, default="Not Started")

我遵循这里指定的模型。

在my views.py中，当我收到这样的请求时，我正在调用序列化程序

def workorder_operations(request,workorder_pk):
    workorder = Work_Order.objects.filter(pk=workorder_pk)
    serializer = Work_Order_Serializer(workorder)

这将生成一个json，如下所示

{
    "shipmentlist":[]
}

我对两件事完全感到困惑：

为什么它只显示装运对象而不显示其他对象，即使它无法解析数据或其他内容

为什么货物没有被填充 当我改用ModelSerializer并以这种方式定义序列化程序时，一切都很好：

class ShipmentSerializer(serializers.ModelSerializer):
    class Meta:
        model = Shipment
        fields =('expected_startDate','expected_endDate','shipment_status')
class WorkOrderSerializer(serializers.ModelSerializer):
    class Meta:
        model = WorkOrder
        fields =('request_id','generated_date','status', 'source','destination','material')

我想知道的是如何表示对象的嵌套列表，以便可以正确地序列化它们。 我希望WorkOrder对象的json看起来像这样：（注意：json和模型中的变量名可能不匹配，因此请忽略这一点，因为我已经剥离了一些变量，以避免使示例复杂化。）

---

在

WorkOrder

模型中添加

ManyToManyField

class Shipment (models.Model):
    expected_startDate = models.DateTimeField(default=timezone.now)
    expected_endDate = models.DateTimeField(default=timezone.now)
    shipment_status = models.CharField(max_length=300,default = "Not Started")


class WorkOrder (models.Model):
    status = models.CharField(max_length=300,default = "New")
    source = models.CharField(max_length=300)
    destination = models.CharField(max_length=
    material = models.CharField(max_length=300)
    shipments = models.ManyToManyField(Shipment, related_name='shipments')

序列化程序会是这样的

class ShipmentSerializer(serializers.ModelSerializer):
    class Meta:
        model = Shipment
        fields = [f.name for f in model._meta.fields]

class WorkOrderSerializer(serializers.ModelSerializer):
    shipments = ShipmentSerializer(many=True)
    class Meta:
        model = WorkOrder
        fields = [f.name for f in model._meta.fields] + ['shipments']

---

我尝试了你的建议，当我尝试点击URL时，我得到了以下错误：没有这样的表：workorder\u发货。我的数据库当前没有任何发货，因此我希望它显示一个空的发货列表，其中包含工单的其余数据。另外一点是，我在workorder的数据库中输入了部分内容，其中我没有输入任何装运数据。我必须这么做吗？只是确保我没有在你认为不会出错的地方出错：）你迁移了数据库吗？该死！那是丢失的一环！非常感谢您的帮助！：-）

class ShipmentSerializer(serializers.ModelSerializer):
    class Meta:
        model = Shipment
        fields = [f.name for f in model._meta.fields]

class WorkOrderSerializer(serializers.ModelSerializer):
    shipments = ShipmentSerializer(many=True)
    class Meta:
        model = WorkOrder
        fields = [f.name for f in model._meta.fields] + ['shipments']