从2个日历天的工作时间计算工作日 - SQL Server

Question

请参阅下面的数据图片：

表格结构：

CREATE TABLE [dbo].[tblData](
    [agentName] [nvarchar](255) NULL,
    [DateTime] [datetime] NULL,
    [loggedMins] [float] NULL,
    [activeMinutes] [float] NULL,
    [holdMinutes] [float] NULL,
    [inactiveMinutes] [float] NULL
) ON [PRIMARY]

示例数据：

Insert Into TblData Values ('Doe, John', '01/21/2018 23:30:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/22/2018 00:00:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/22/2018 00:30:00', 30, 0.018, 0, 29.982) 
Insert Into TblData Values ('Doe, John', '01/22/2018 01:00:00', 30, 0, 0, 29.982) 
Insert Into TblData Values ('Doe, John', '01/22/2018 01:30:00', 30, 0, 0, 29.016) 
Insert Into TblData Values ('Doe, John', '01/22/2018 02:00:00', 30, 0, 0, 0) 
Insert Into TblData Values ('Doe, John', '01/22/2018 02:30:00', 30, 0, 0, 18.132) 
Insert Into TblData Values ('Doe, John', '01/22/2018 03:00:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/22/2018 03:30:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/22/2018 04:00:00', 27.55, 0, 0, 27.55) 
Insert Into TblData Values ('Doe, John', '01/22/2018 18:00:00', 19.1166666666667, 1.86578666666667, 0, 16.9660416666667) 
Insert Into TblData Values ('Doe, John', '01/22/2018 18:30:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/22/2018 19:00:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/22/2018 19:30:00', 30, 0, 0, 26.532) 
Insert Into TblData Values ('Doe, John', '01/22/2018 20:00:00', 30, 0, 0, 14.568) 
Insert Into TblData Values ('Doe, John', '01/22/2018 20:30:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/22/2018 21:00:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/22/2018 21:30:00', 30, 0, 0, 11.232) 
Insert Into TblData Values ('Doe, John', '01/22/2018 22:00:00', 30, 0, 0, 13.266) 
Insert Into TblData Values ('Doe, John', '01/22/2018 22:30:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/22/2018 23:00:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/22/2018 23:30:00', 30, 0, 0, 16.551) 
Insert Into TblData Values ('Doe, John', '01/23/2018 00:00:00', 30, 4.416, 0, 25.584) 
Insert Into TblData Values ('Doe, John', '01/23/2018 00:30:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/23/2018 01:00:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/23/2018 01:30:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/23/2018 18:00:00', 30, 0, 0, 30) 
Insert Into TblData Values ('Doe, John', '01/23/2018 18:30:00', 30, 0, 0, 30) 

我的问题是我需要根据＆＃39; DateTime＆＃39;来计算工作日期。柱。 John Doe工作时间为下午6点至凌晨4点。需要将工作日期添加到新列，该列应等于6PM日期。以下是我的逻辑：

1. 将当前列的DateTime与上一列以及agentName进行比较。
2. 如果差异是半小时且同事相同，请更新上一行日期的最后一栏。

我很难翻译成代码:(帮助！

Answer 1

Matther Baker的回答非常好！我按agentName和time排序行;然后将agentName添加到代码中，以便每个代理的时间分别计算：

--CREATING A TEMP TABLE
Select 
    row_number() over(order by agentName, [DateTime] asc) as [Rnk], * 
Into ##TmpData
from TblData
Order by agentName, [DateTime] asc

--ACTUAL LOGIC

SELECT  *,
       (SELECT MIN(DateTime) FROM   tblData S
        WHERE  S.DateTime <= M.DateTime
        AND S.DateTime > DATEADD(HOUR, -15, M.DateTime) and S.agentName = M.agentName
       ) ShiftStart
FROM    ##TmpData M


--DROPPING TEMP TABLE
Drop Table ##TmpData

我看到没有人做过超过13个小时的转变，因此增加了缓冲区并将时间间隔改为15个。

非常感谢你们！

Answer 2

看看这个。我将日期时间仅限于时间，并设置中午的支点，如果时间是中午之后，那么今天。如果它在中午之前那么它的昨天。如果是早期开始/晚期结束等，我选择正午。

SELECT  *
,       CASE WHEN CAST(DateTime AS TIME) > '12:00:00' THEN CAST(DateTime AS DATE)
             ELSE DATEADD(DAY, -1, CAST(DateTime AS DATE))
        END AS WorkDate
FROM    @tblData

这将为您提供一个可以分组的列。也许把它包装成cte来帮助。

Answer 3

第二个回答：

SELECT  *
,       (SELECT MIN(DateTime)
         FROM   tblData S
         WHERE  S.DateTime <= M.DateTime
                AND S.DateTime > DATEADD(HOUR, -12, M.DateTime)
        ) ShiftStart
FROM    tblData M

可能是你最好的替代方案 - 这可以解释为一个班次不会超过12小时，但你可以调整一下。也不是最有效的方法。