如何按日期分组订单

Question

我有一个订单明细表，我想按每个日期查找价格总和，并将其显示为

对于前

   count   sum            date 
    5      619.95        2015-11-19
    3      334.97        2015-11-18
    4      734.96        2015-11-18
    5      1129.95       2015-11-18

我写了查询以获取count和sum为

select count(id), sum([price])
from [OrderDetails]
where [date]between '2015-10-29 05:15:00' and '2015-11-09 00:01:00'
group by datepart(day,[date])

但无法实现与日期。 如何做呢？

Answer 1

您需要在查询的SELECT部分中包括要分组的内容：

select count(id), sum([price]), datepart(day,[date]) as [date]
from [OrderDetails]
where [date] between '2015-10-29 05:15:00' and '2015-11-09 00:01:00'
group by datepart(day,[date]);

Answer 2

您的名为date的列似乎同时具有日期和时间部分。 我建议将其转换为日期，对于select和group by ：

select count(id), sum(price), cast([date] as date) as thedate
from OrderDetails
where [date] between '2015-10-29 05:15:00' and '2015-11-09 00:01:00'
group by cast([date] as date) 
order by thedate;

注意： date是列的较差名称，因为它是内置类型的名称。