前言:本文以学习记录的形式发表出来,前段时间苦于照模型聚合中group by 找了很久,官方文章中没有很明确的说出group by,但在文档中有提到!!!

正文(最后编辑于2016-11-12):

聚合:

LOrder.objects.values('com_chnl_name').annotate(Count('serv_id'))

#相当于select count(serv_id) from LOrder group by com_chnl_name

模型高级运用—条件表达式+数据库函数:

条件表达式:

from django.db.models import IntegerField,Sum,When,Q,Case #模型执行查询集

LOrder.objects.values('com_chnl_name').annotate(

cdma=Sum(

Case(When(Q(prod_offer_name__contains='语音CDMA') & Q(service_offer_name='订购'),then=1),

output_field=IntegerField())

),

)

#相当于select sum(case when prod_offer_name like’%.语音CDMA %’ and service_offer_name='订购' then 1 end) , com_chnl_name from LOrder group by com_chnl_name,该SQL语句也算是SQL中的中级部分. 其中&相当于and 对应的|相当于or

问题:聚合函数得到的结果如果不存在,那么会返回None,这对于结果全是数字的话,进行数字加减乘除很不方便,所以运用数据库函数将None改为0,以下例子:

数据库函数和条件表达式一起运用:

from django.db.models import IntegerField,Sum,When,Q,Case,Value as V #模型执行查询集

from django.db.models.functions import Coalesce #模型部分:数据库函数

LOrder.objects.values('com_chnl_name').annotate(

cdma_dg=Coalesce(Sum(

Case(When(Q(prod_offer_name__contains='语音CDMA') & Q(service_offer_name='订购'),then=1),

output_field=IntegerField())

),V(0)),

>>> # Get a screen name from least to most public

>>> from django.db.models import Sum, Value as V

>>> from django.db.models.functions import Coalesce

>>> Author.objects.create(name='Margaret Smith', goes_by='Maggie')

>>> author = Author.objects.annotate(

...    screen_name=Coalesce('alias', 'goes_by', 'name')).get()

>>> print(author.screen_name)

Maggie

>>> # Prevent an aggregate Sum() from returning None

>>> aggregated = Author.objects.aggregate(

...    combined_age=Coalesce(Sum('age'), V(0)),

...    combined_age_default=Sum('age'))

>>> print(aggregated['combined_age'])

0

>>> print(aggregated['combined_age_default'])

None