计算 Pivot SQL 中的总天差

Posted 2023-02-15

【中文标题】计算 Pivot SQL 中的总天差

【英文标题】：Count total day difference in a Pivot SQL

【发布时间】：2021-11-25 17:11:51

【问题描述】：

我有两个表中的数据，日期为 PAL 日期和 HUB 日期。 我希望在一个有 6 列的表格中显示天数差异：

Client	1day	2days	3days	4days	5days

Client 将显示分组的客户端，1 Day 将显示多少 datediff 为 1 的计数。 2 Days 将显示多少个 datediff 2，然后是 3，然后是 4，然后是 5。

PAL 表：

Consignment_Number	eventdatetime
T00A1A0001000001	2021-09-27 12:17:00.000
T00A1A0001000002	2021-09-27 12:27:00.000
T00A1A0001000003	2021-09-28 13:37:00.000
T00A1A0001000004	2021-09-28 13:47:00.000
T00A1A0001000005	2021-09-29 13:57:00.000
T00A1A0001000006	2021-09-39 14:07:00.000
T00A1A0001000007	2021-09-30 14:17:00.000
T00A1A0001000008	2021-09-30 14:27:00.000
T00A1A0001000009	2021-09-30 14:37:00.000

HUB 表：

Consignment_Number	eventdatetime
T00A1A0001000001	2021-09-30 18:39:00.000
T00A1A0001000002	2021-10-01 18:40:00.000
T00A1A0001000003	2021-10-01 18:46:00.000
T00A1A0001000004	2021-10-02 18:46:00.000
T00A1A0001000005	2021-10-02 18:46:00.000
T00A1A0001000006	2021-10-03 18:46:00.000
T00A1A0001000007	2021-10-03 18:46:00.000
T00A1A0001000008	2021-10-04 18:46:00.000
T00A1A0001000009	2021-10-05 18:46:00.000

最终它会看起来像这样：

Clients	1day	2days	3days	4days	5days
Client1	5	2	3	1	0
Client2	2	2	2	1	0
Client3	4	2	5	1	0

【问题讨论】：

您使用的是哪个 dbms？

嗨，J，我正在使用 SQL 服务器

【参考方案1】：

select 
  client 
  , sum(case when abs(datediff(day, Consignment_Number,Consignment_Number)) = 1 then 1 else 0 end) 1day
  , sum(case when abs(datediff(day, Consignment_Number,Consignment_Number)) = 2 then 1 else 0 end) 2days
  , sum(case when abs(datediff(day, Consignment_Number,Consignment_Number)) = 3 then 1 else 0 end) 3days
  , ...
  , sum(case when abs(datediff(day, Consignment_Number,Consignment_Number)) = 1 then 1 else 0 end) * 100.0 / count(*) 1dayperc
  , ...
from PAL p
join HUB h
 on h.Consignment_Number = p.Consignment_Number
group by client

【讨论】：

谢谢 Eshirvana，它以我想要的格式工作。你是个传奇！

如果我想对这些应用百分比怎么办？

@sidekick 查看更新的答案

我不能感谢你，你帮了我兄弟！

@sidekick 欢迎您，如果有帮助，请点赞并接受答案。