获取所提供日期之间的所有日期

Question

我有这个表和样本数据.我希望得到整个月或特定日期的出勤率和信息,例如他工作的时间或他不在的日子.

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,告诉员工工作了多少小时,但它只显示表中存在数据的天数.我想显示提供日期之间的所有日期,如果没有数据,它应该在列中为NULL.

这是查询:

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，您需要使用递归方式生成可能的日期：

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; 

只需确保在语句中使用日期而t不是表即可。AttendanceSELECT

注意：如果您的日期周期很大，那么不要忘记使用查询提示OPTION (MAXRECURSION 0)，默认情况下它具有100递归级别。