Question

我的项目结构如下，

..
├── project
│    ├── app1
│    └── app2
│         ├── app2_views
│              ├── main_view.py
│              ├── view1.py
│              ├── view2.py

我想在我的view1.py中调用main_view.py中定义的类

我尝试了以下代码，

main.py

from rest_framework import viewsets
from app2.app2_views.view1.py import class1

class mainclass(viewsets.ModelViewSet):

    class1.view1_data()

view1.py

from app2.models.py import table1 
from rest_framework import viewsets
from app2.serializers.py import seri1
from django.db.models import Sum

class class1(viewsets.ModelViewSet):

    def view1_data():
        queryset = table1.objects.values('column')
        serializer_class = seri1

seri1.py

from rest_framework import serializers
from app2.models import table1
class seri1(serializers.HyperlinkedModelSerializer):
    class Meta:
        model = table1
        fields = ['column']

但是我在main.py说

时遇到错误

should either include a `serializer_class` attribute, or override the `get_serializer_class()` method

因为我已经在view1.py中加入了序列化程序类，所以我不想在main_view.py中再做一次。我是python的新手，我在这里做错了什么？我也想保持view1.py>class1和main.py>mainclass都可以通过url（其余框架管理门户）访问吗？

当前我的urls.py是，

from django.urls import include, path
from rest_framework import routers
from . import views
from .app2 import view1, main_view

router = routers.DefaultRouter(trailing_slash=False)
router.register(r'^view1', main_view.mainclass, basename='mis_trx_data')
router.register(r'^main_view', view1.class1, basename='mis_trx_data')

urlpatterns = [
    path('', include(router.urls)),
    path('api-auth/', include('rest_framework.urls', namespace='rest_framework'))
]

谢谢您的建议

Answer 1

您需要在serializer_class中指定一个名为ModelViewSet的类级别变量

class class1(viewsets.ModelViewSet):
    serializer_class = MySerializer
    def view1_data():
        queryset = table1.objects.values('column')
        serializer = self.serializer_class(request.data)

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