Python django rest farmework嵌套关系以使用筛选器
使用django rest farmework实现API,有一个问题一直无法解决:如何过滤关联表?具体代码如下: models.pyPython django rest farmework嵌套关系以使用筛选器,python,django,django-rest-framework,serialization,Python,Django,Django Rest Framework,Serialization,使用django rest farmework实现API,有一个问题一直无法解决:如何过滤关联表?具体代码如下: models.py class Category(models.Model): name = models.CharField(max_length=30) amount = models.IntegerField() class Source(models.Model): name = models.CharField(max_length=50)
class Category(models.Model):
name = models.CharField(max_length=30)
amount = models.IntegerField()
class Source(models.Model):
name = models.CharField(max_length=50)
rss_link = models.URLField()
amount = models.IntegerField()
# ForeignKey
category = models.ForeignKey(Category)
views.py
class CategoryListView(APIView):
def get(self, request):
category = Category.objects.all()
serializers = CategorySerializers(category, many=True)
return Response(serializers.data)
序列化程序.py
class SourceSerializers(serializers.ModelSerializer):
class Meta:
model = Source
fields = ("id","name","amount")
class CategorySerializers(serializers.ModelSerializer):
source_set = SourceSerializers(many=True, read_only=True)
class Meta:
model = Category
fields = ("id","name","amount","source")
class SourceSerializers(serializers.ModelSerializer):
class Meta:
model = Source
fields = ("id","name","amount")
class CategorySerializers(serializers.ModelSerializer):
#Modified the following code
source_set = SourceSerializers(Source.objects.filter(amount=0),many=True, read_only=True)
class Meta:
model = Category
fields = ("id","name","amount","source")
程序运行结果:
[
{
"id": 1,
"name": "study",
"amount": "0",
"source": [
{
"id": 34,
"name": "java",
"amount": "0"
},
{
"id": 35,
"name": "python",
"amount": "0"
},
{
"id": 36,
"name": "css",
"amount": "2"
}
]
}
]
现在我只需要查询0数据“数量”的“来源”,我尝试编写以下代码:
序列化程序.py
class SourceSerializers(serializers.ModelSerializer):
class Meta:
model = Source
fields = ("id","name","amount")
class CategorySerializers(serializers.ModelSerializer):
source_set = SourceSerializers(many=True, read_only=True)
class Meta:
model = Category
fields = ("id","name","amount","source")
class SourceSerializers(serializers.ModelSerializer):
class Meta:
model = Source
fields = ("id","name","amount")
class CategorySerializers(serializers.ModelSerializer):
#Modified the following code
source_set = SourceSerializers(Source.objects.filter(amount=0),many=True, read_only=True)
class Meta:
model = Category
fields = ("id","name","amount","source")
修改了以下代码:
source_set = SourceSerializers(many=True, read_only=True)
到
但结果是一样的,我希望结果是:
[
{
"id": 1,
"name": "study",
"amount": "0",
"source": [
{
"id": 34,
"name": "java",
"amount": "0"
},
{
"id": 35,
"name": "python",
"amount": "0"
}
]
}
]
提前感谢。稍微更改一下序列化程序。可以使用SourceWithCategorySerializer序列化相关的类别对象。然后,您可以基于数量查询源对象,并将该查询集传递给序列化程序 你的看法:
class CategoryListView(APIView):
def get(self, request):
sources = Source.objects.filter(amount=0)
serializers = SourceWithCategorySerializers(sources, many=True)
return Response(serializers.data)
序列化程序:
class SourceWithCategorySerializers(serializers.ModelSerializer):
category = CategorySerializer(read_only=True)
class Meta:
model = Source
fields = ("id","name","amount", "category")
class CategorySerializers(serializers.ModelSerializer):
class Meta:
model = Category
fields = ("id","name","amount")
稍微更改一下序列化程序。可以使用SourceWithCategorySerializer序列化相关的类别对象。然后,您可以基于数量查询源对象,并将该查询集传递给序列化程序 你的看法:
class CategoryListView(APIView):
def get(self, request):
sources = Source.objects.filter(amount=0)
serializers = SourceWithCategorySerializers(sources, many=True)
return Response(serializers.data)
序列化程序:
class SourceWithCategorySerializers(serializers.ModelSerializer):
category = CategorySerializer(read_only=True)
class Meta:
model = Source
fields = ("id","name","amount", "category")
class CategorySerializers(serializers.ModelSerializer):
class Meta:
model = Category
fields = ("id","name","amount")
您可以使用:
我把你的班名的复数“s”去掉了,因为他们让我很痒;) 您可以使用:
我把你的班名的复数“s”去掉了,因为他们让我很痒;)
sources=category.source\u set.filter(amount=0)
,如果“amount”是参数,该参数来自“CategoryListView”,如何将其传递给“get\u sources”?sources=category.source\u set.filter(amount=0)
,如果“amount”是参数,该参数来自“CategoryListView”,如何将其传递给“get\u sources” ?