Question

我有这张桌子和样本数据。我想获得整个月或特定日期的出勤率以及诸如他工作的小时数或缺勤的天数之类的信息。

CREATE TABLE Attendance
(
    [EmpCode] int,
    [TimeIn] datetime,
    [TimeOut] datetime
)

INSERT INTO Attendance VALUES (12, '2018-08-01 09:00:00', '2018-08-01 17:36:00');
INSERT INTO Attendance VALUES (12, '2018-08-02 09:00:00', '2018-08-02 18:10:00');
INSERT INTO Attendance VALUES (12, '2018-08-03 09:25:00', '2018-08-03 16:56:00');
INSERT INTO Attendance VALUES (12, '2018-08-04 09:13:00', '2018-08-05 18:09:00');
INSERT INTO Attendance VALUES (12, '2018-08-06 09:00:00', '2018-08-07 18:15:00');
INSERT INTO Attendance VALUES (12, '2018-08-07 09:27:00', '2018-08-08 17:36:00');
INSERT INTO Attendance VALUES (12, '2018-08-08 09:35:00', '2018-08-09 17:21:00');
INSERT INTO Attendance VALUES (12, '2018-08-10 09:00:00', '2018-08-10 17:45:00');
INSERT INTO Attendance VALUES (12, '2018-08-11 09:50:00', '2018-08-11 17:31:00');
INSERT INTO Attendance VALUES (12, '2018-08-13 09:23:00', '2018-08-13 17:19:00');
INSERT INTO Attendance VALUES (12, '2018-08-15 09:21:00', '2018-08-15 17:36:00');
INSERT INTO Attendance VALUES (12, '2018-08-16 09:00:00', '2018-08-16 17:09:00');
INSERT INTO Attendance VALUES (12, '2018-08-17 09:34:00', '2018-08-17 17:29:00');
INSERT INTO Attendance VALUES (12, '2018-08-18 09:00:00', '2018-08-18 17:10:00');
INSERT INTO Attendance VALUES (12, '2018-08-20 09:34:00', '2018-08-20 17:12:00');
INSERT INTO Attendance VALUES (12, '2018-08-21 09:20:00', '2018-08-21 17:15:00');
INSERT INTO Attendance VALUES (12, '2018-08-22 09:12:00', '2018-08-22 17:19:00');
INSERT INTO Attendance VALUES (12, '2018-08-23 09:05:00', '2018-08-23 17:21:00');
INSERT INTO Attendance VALUES (12, '2018-08-24 09:07:00', '2018-08-24 17:09:00');
INSERT INTO Attendance VALUES (12, '2018-08-25 09:12:00', '2018-08-25 17:05:00');
INSERT INTO Attendance VALUES (12, '2018-08-27 09:21:00', '2018-08-27 17:46:00');
INSERT INTO Attendance VALUES (12, '2018-08-28 09:17:00', '2018-08-28 17:12:00');
INSERT INTO Attendance VALUES (12, '2018-08-29 09:00:00', '2018-08-29 17:36:00');
INSERT INTO Attendance VALUES (12, '2018-08-30 09:12:00', '2018-08-30 17:24:00');

我有一个查询，可以告诉员工工作了多少小时，但它仅显示表中数据存在的天数。我想显示提供的日期之间的所有日期，以防万一没有数据，列中应该为NULL。

以下是查询：

SELECT
    [EmpCode],
    FirstIN    = CAST(MIN([TimeIn]) AS TIME), 
    LastOUT    = CAST(MAX([TimeOut]) AS TIME), 
    CONVERT(VARCHAR(6), Datediff(second, CAST(MIN([TimeIn]) AS TIME), CAST(MAX([TimeOut]) AS TIME))/3600) 
       + ':'
       + RIGHT('0' + CONVERT(VARCHAR(2), (Datediff(second, CAST(MIN([TimeIn]) AS TIME), CAST(MAX([TimeOut]) AS TIME)) % 3600) / 60), 2) 
       + ':' 
       + RIGHT('0' + CONVERT(VARCHAR(2), Datediff(second, CAST(MIN([TimeIn]) AS TIME), CAST(MAX([TimeOut]) AS TIME)) % 60) , 2 ) AS HoursSpent,
    CAST(COALESCE(TimeIn, TimeOut) AS DATE) [Date]    
FROM Attendance
WHERE CAST(COALESCE(TimeIn, TimeOut) AS DATE) BETWEEN '2018-08-01' AND '2018-08-25'    
GROUP BY EmpCode, TimeIn, TimeOut

Answer 1

为此，您需要使用递归方法来生成可能的日期：

with t as (
     select '2018-08-01' as startdt
     union all
     select dateadd(day, 1, startdt)
     from t
     where startdt < '2018-08-25'
)

select . . . 
from t left join
     Attendance at
     on cast(coalesce(at.TimeIn, at.TimeOut) as date) = t.startdt;

只需确保使用t语句中的Attendance中的日期而不是SELECT表中的日期。

注意：如果您没有太多的日期范围，那么不要忘记使用查询提示OPTION (MAXRECURSION 0)。通过取消默认它具有100递归级别。

Answer 2

您可以尝试递归CTE来填充日期，然后与之结合以获取时间间隔

DECLARE @From DATETIME = '2018-08-01' ,@To DATETIME= '2018-08-25'

;WITH CTE
AS
(
    SELECT
        [EmpCode] EmpId,
        MyDate = @From
        FROM Attendance A
    UNION ALL
    SELECT
        EmpId,
        MyDate = DATEADD(DAY,1,MyDate)
        FROM CTE
            WHERE MyDate < @To
)
SELECT
[EmpCode] = CTE.EmpId,
CTE.MyDate,
FirstIN    = CAST(MIN([TimeIn]) AS TIME), 
LastOUT    = CAST(MAX([TimeOut]) AS TIME), 
CONVERT(VARCHAR(6), Datediff(second, CAST(MIN([TimeIn]) AS TIME), CAST(MAX([TimeOut]) AS TIME))/3600) 
       + ':'
       + RIGHT('0' + CONVERT(VARCHAR(2), (Datediff(second, CAST(MIN([TimeIn]) AS TIME), CAST(MAX([TimeOut]) AS TIME)) % 3600) / 60), 2) 
       + ':' 
       + RIGHT('0' + CONVERT(VARCHAR(2), Datediff(second, CAST(MIN([TimeIn]) AS TIME), CAST(MAX([TimeOut]) AS TIME)) % 60) , 2 ) 

       AS HoursSpent,
CAST(CTE.MyDate AS DATE) [Date]

FROM CTE
    LEFT JOIN Attendance A
        ON A.EmpCode = CTE.EmpId
            AND CAST(CTE.MyDate AS DATE) = CAST(COALESCE(TimeIn, TimeOut) AS DATE)
GROUP BY CTE.EmpId, TimeIn, TimeOut,CTE.MyDate
    ORDER BY 6

Answer 3

使用计数表的另一种方法。这样做的好处是，rCTE是RBAR的一种形式。 Tally表的想法并不那么明显，但是更快，并且，如果您的天数超过100天，则不需要添加OPTION (MAXRECURSION 0)。实际上，此示例最多可以处理10,000天，而这应该绰绰有余：

DECLARE @EmpCode int = 12;

WITH N AS(
    SELECT N
    FROM (VALUES(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL)) N(N)),
Tally AS(
    SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) -1 AS I
    FROM N N1 --10
         CROSS JOIN N N2 --100
         CROSS JOIN N N3 --1000
         CROSS JOIN N N4 --10000
),
Dates AS(
    SELECT DATEADD(DAY, T.I, TT.MinTimeIn) AS CalendarDate,
           @EmpCode AS EmpCode
    FROM Tally T
         CROSS APPLY (SELECT MIN(CONVERT(date,TimeIn)) AS MinTimeIn,
                             MAX(CONVERT(date,TimeOut)) AS MaxTimeOut
                      FROM Attendance
                      WHERE EmpCode = @EmpCode) TT
     WHERE DATEADD(DAY, T.I, TT.MinTimeIn) <= CONVERT(date, TT.MaxTimeOut))
SELECT CalendarDate
       EmpCode,
       TimeIn,
       TimeOut
FROM Dates D
     LEFT JOIN Attendance A ON D.CalendarDate = CONVERT(date,A.TimeIn)
                           AND D.EmpCode = A.EmpCode;

获取提供的日期之间的所有日期

3 个答案: