我的模型是:
class Mo(Model):
dur = DurationField(default=timedelta(0))
price_per_minute = FloatField(default=0.0)
minutes_int = IntegerField(default=0) # dublicates data from dur , only for test
一些测试数据:
for i in range(4):
m = Mo(dur=timedelta(minutes=i),
minutes_int=i,
price_per_minute=10.0)
m.save()
我想乘以dur
by price_per_minute
并找到Sum
:
r = Mo.objects.all().aggregate(res=Sum(F('dur')*F('price_per_minute')))
print(r['res']/(10e6 * 60))
but:
Invalid connector for timedelta: *.
说明:在数据库中DurationField
存储为包含微秒的简单 BIGINT,如果我知道如何聚合它,我会将其除以 (10e6 * 60) 并支付分钟数。
如果我使用一个简单的IntegerField
相反,一切正常:
r = Mo.objects.all().aggregate(res=Sum(F('minutes_int')*F('price_per_minute')))
print(r['res']/(10e6*60))
所以我需要在聚合中进行一些转换为整数,是否可以将持续时间转换为一些额外的字段?
r = Mo.objects.all().extra({'mins_int': 'dur'}).aggregate(res=Sum(F('mins_int')*F('price_per_minute')))
print(r['res']),
but
Cannot resolve keyword 'mins_int' into field. Choices are: dur, id, minutes_int, price_per_minute