如何选择每天记录的最后一个条目?在此示例中,我需要过去5天中每天的最后订购商品编号和最后DateOrdered条目。这是我的桌子:
ItemNumber | DateOrdered
1 2020-04-01 08:00:00.000
3 2020-04-01 09:00:00.000
5 2020-04-01 10:00:00.000
4 2020-04-02 09:00:00.000
6 2020-04-02 10:00:00.000
7 2020-04-03 08:00:00.000
3 2020-04-03 09:00:00.000
2 2020-04-03 10:00:00.000
5 2020-04-04 10:00:00.000
8 2020-04-05 08:00:00.000
2 2020-04-05 09:00:00.000
8 2020-04-05 10:00:00.000
这是我需要的结果:
ItemNumber | DateOrdered
5 2020-04-01 10:00:00.000
6 2020-04-02 10:00:00.000
2 2020-04-03 10:00:00.000
5 2020-04-04 10:00:00.000
8 2020-04-05 10:00:00.000
这是我所能接近的:
with tempTable as
(
select
*,
row_number() over(partition by datediff(d, 0, DateOrdered) order by DateOrdered desc) as rn
from myTable
)
select *
from tempTable
where rn = 1
答案 0 :(得分:0)
您快到了。您只需要修复分区的定义,即可将属于同一天的所有行汇总在一起。
这应该做到:
with tempTable as
(
select
*,
row_number() over(partition by cast(DateOrdered as date) order by DateOrdered desc) as rn
from myTable
)
select *
from tempTable
where rn = 1