Question

我正在尝试从我拥有的一些基本数据中将DATEDIFF的累积总和转换为百分比，这是一个小快照：

ID     IIn                       IOut
AB123  2015-11-06 15:24:44.057   2015-11-14 01:00:00.000
QA565  2015-10-27 20:12:19.753   2015-11-06 03:00:00.000
UN555  2015-12-29 06:29:23.417   2016-01-03 08:00:00.000
LG602  2015-08-07 16:52:13.573   2015-08-11 03:00:00.000

ETC ETC

然后，我使用DATEDIFF获得几天的时间：

SELECT ID, DATEDIFF(hour, IIn, IOut)/24.0 IDays
FROM TimeTable

哪个给我：

ID     IDays
AB123  7.416666
QA565  9.291666
UN555  5.083333
LG602  3.458333

我想要的是将ID的数目除以其IDay的数目（四舍五入），从最低的IDay到最高的累积百分比，如下所示：< / p>

ID     IDays  IDaysPer
LG602  3      12.5
UN555  5      33.33
AB123  7      62.49
QA565  9      100

Answer 1

您可以使用几个窗口聚合来完成此操作，将原始查询放在CTE中以方便使用（子查询也可以）：

declare @timeTable table (ID char(5) not null, IIn datetime not null,
                          IOut datetime not null)
insert into @timeTable(ID,IIn,IOut) values
('AB123','2015-11-06T15:24:44.057','2015-11-14T01:00:00.000'),
('QA565','2015-10-27T20:12:19.753','2015-11-06T03:00:00.000'),
('UN555','2015-12-29T06:29:23.417','2016-01-03T08:00:00.000'),
('LG602','2015-08-07T16:52:13.573','2015-08-11T03:00:00.000')

;With Diffs as (
    SELECT ID, DATEDIFF(hour, IIn, IOut)/24.0 IDays
    FROM @timeTable
)
select
    *,
    (
      SUM(IDays) OVER (ORDER BY IDays, ID)
      /
      SUM(IDays) OVER ()
    ) * 100 as IDaysPer
from
    Diffs
order by IDays

请注意，我不太了解您的“四舍五入”要求，但您应该能够使用围绕适当的计算方法包装的任何通用四舍五入技术。所以我的输出与您的输出不太匹配：

ID    IDays                                   IDaysPer
----- --------------------------------------- ---------------------------------------
LG602 3.458333                                13.696300
UN555 5.083333                                33.828300
AB123 7.416666                                63.201300
QA565 9.291666                                100.000000

Answer 2

在这里，输出与您的匹配...

        create table #TEMp
(ID VARCHAR(100)
,IIn datetime
,IOut datetime
)


insert into #temp(ID,IIn,IOut) values
('AB123','2015-11-06T15:24:44.057','2015-11-14T01:00:00.000'),
('QA565','2015-10-27T20:12:19.753','2015-11-06T03:00:00.000'),
('UN555','2015-12-29T06:29:23.417','2016-01-03T08:00:00.000'),
('LG602','2015-08-07T16:52:13.573','2015-08-11T03:00:00.000')

select ID,IDays AS Idays,ROUND(CAST(SUM(IDays) OVER(ORDER BY IDays) AS FLOAT)/CAST(SUM(IDays)OVER() AS FLOAT) * 100,2) AS IdaysPer
from
(
select *,ROUND(DATEDIFF(hour, IIn, IOut)/24,0) IDays
from #TEMP
)T

Answer 3

考虑时间表已经有了数据

WITH t1 (ID, IDays)
AS (
    SELECT ID, DATEDIFF(hour, IIn, IOut) / 24.0 AS IDays
    FROM TimeTable
)
SELECT 
    ID, FLOOR(IDays), 
    (FLOOR(IDays) / (SELECT SUM(FLOOR(IDays)) FROM t1 t2 WHERE t1.IDays <= t2.IDays)) * 100.0 AS IDaysPer
FROM t1
ORDER BY 2 ASC

SQL Server DATEDIFF的累积总和成百分比

3 个答案: