bde*_*dex 3 python django django-rest-framework
我在序列化中间“枢轴”模型并附加到 Django Rest Framework 中多对多关系中的每个项目时遇到困难。
例子:
模型.py:
class Member(models.Model):
name = models.CharField(max_length = 20)
groups = models.ManyToManyField('Group', through='Membership')
class Group(models.Model):
name = models.CharField(max_length = 20)
class Membership(models.Model):
member = models.ForeignKey('Member')
group = models.ForeignKey('Group')
join_date = models.DateTimeField()
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序列化器.py:
class MemberSerializer(ModelSerializer):
class Meta:
model = Member
class GroupSerializer(ModelSerializer):
class Meta:
model = Group
class MembershipSerializer(ModelSerializer):
class Meta:
model = Membership
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我尝试遵循答案: Inclusion intermediary (through model) in returns in Django Rest Framework
但这不完全是我需要的
我需要生成以下输出
{
"id": 1,
"name": "Paul McCartney",
"groups": [
{
"id": 3,
"name": "Beatles",
"membership": {
"id": 2,
"member_id": 1,
"group_id": 3,
"join_date": "2018-08-08T13:43:45-0300"
}
}
]
}
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在此输出中,我返回组中每个项目的相关“通过模型”。
如何以这种方式生成序列化模型?
根据您想要显示输出的方式,我建议您将模型更改为:
class Group(models.Model):
name = models.CharField(max_length=20)
members = models.ManyToManyField(
'Membership',
related_name='groups',
related_query_name='groups',
)
class Member(models.Model):
name = models.CharField(max_length=20)
class Membership(models.Model):
group = models.ForeignKey(
'Group',
related_name='membership',
related_query_name='memberships',
)
join_date = models.DateTimeField()
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Group模型和Member模型如何ManytoMany,没有问题你让Group模型中的关系。在序列化中输出它是最简单的。related_name和related_query_name用于进行序列化和指向嵌套关系。
最后,您的序列化可能是这样的(我用 create 方法举例说明):
class MembershipSerializer(ModelSerializer):
class Meta:
fields = ("id", "join_date",)
class GroupSerializer(ModelSerializer):
memberships = MembershipSerializer(many=True)
class Meta:
model = Group
fields = ("id", "name", "memberships",)
class MemberSerializer(ModelSerializer):
groups = GroupSerializer(many=True)
class Meta:
model = Member
fields = ("id", "name", "groups")
def create(self):
groups_data = validated_data.pop('groups')
member = Member.objects.create(**validated_data)
for group in groups_data:
memberships_data = group.pop('memberships')
Group.objects.create(member=member, **group)
for memberhip in memberships:
Membership.objects.create(group=group, **memberships)
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输出将是:
{
"id": 1,
"name": "Paul McCartney",
"groups": [
{
"id": 3,
"name": "Beatles",
"memberships": [
{
"id": 2,
"join_date": "2018-08-08T13:43:45-0300"
}
]
}
]
}
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在此输出中,我没有“嵌套”父 ID,但您也可以这样做,只需在 fields 属性中声明即可。
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