序列化器上的Django Rest Framework条件字段 [英] Django Rest Framework Conditional Field on Serializer
本文介绍了序列化器上的Django Rest Framework条件字段的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个引用通用关系我想详细序列化。
I have a model that references a Generic Relation that I want to serialize in a detailed manner.
class AType(models.Model):
foo = CharField()
class BType(models.Model):
bar = PositiveIntegerField()
class ToSerialize(models.Model):
scope_limit = models.Q(app_label="app", model="atype") | \
models.Q(app_label="app", model="btype")
content_type = models.ForeignKey(ContentType, limit_choices_to=scope_limit)
object_id = models.PositiveIntegerField()
content_object = generic.GenericForeignKey('content_type', 'object_id')
我想要ToSerialize视图集的list方法的JSON如下所示:
I'd like the JSON for the list method of the ToSerialize viewset to look like:
[
{
"atype": { "id": 1, "foo": "a" }
},
{
"atype": { "id": 2, "foo": "b" }
},
{
"btype": { "id": 1, "bar": "1" }
},
{
"btype": { "id": 2, "bar": "2" }
}
]
有没有办法让ToSerialize对象的视图集的序列化程序根据将实现此效果的content_type / object_id生成条件字段?
Is there a way I can have the serializer for the ToSerialize object's viewset produce "conditional fields" based on the content_type/object_id that will achieve this effect?
推荐答案
使用 SerializeMethodField :
class YourSerializer(serializers.ModelSerializer):
your_conditional_field = serializers.SerializerMethodField()
class Meta:
model = ToSerialize
def get_your_conditional_field(self, obj):
# do your conditional logic here
# and return appropriate result
return obj.content_type > obj.object_id
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