認為
class ABC(models.Models):
...
id = models.SlugField(...)
user = models.ForeignKey(settings.AUTH_USER_MODEL)
...
現在我想要類似的東西
[{id: adfsdd, no_of_user_objects: 5}, {id: gdfvsdf, no_of_user_objects: 0}, {id: ergthf, no_of_user_objects: 2}]
我如何實作這一目標？
uj5u.com熱心網友回復：
獲取具有相同 slug id 的用戶數量的模型方法：
class ABC(models.Models):
...
id = models.SlugField(...)
user = models.ForeignKey(settings.AUTH_USER_MODEL)
...
@property # for using in serializer
def no_of_user_objects(self):
return ABC.objects.filter(id=self.id).count()
并且在您的序列化程式中有一個用于查詢集序列化程式的自定義序列化程式：
class ABDSerializer(serializers.ModelSerializer):
class Meta:
model = ABC
fields = ['id', 'no_of_user_objects']
# in your view
queryset = ABC.objects.distinct('id')
serializer = ABDSerializer(queryset, many=True)
print(serializer.data)
# get you [{id: adfsdd, no_of_user_objects: 5}, {id: gdfvsdf, #no_of_user_objects: 0}, {id: ergthf, no_of_user_objects: 2}]
轉載請註明出處，本文鏈接：https://www.uj5u.com/qiye/346374.html
標籤：姜戈 django-models Django 休息框架 django-views django 序列化器
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