sql - 获取给定日期之间的所有日期

Question

我有这个表格和示例数据。我想要获取整个月或特定日期的出勤情况以及他的工作时间或缺勤天数等信息。

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，可以显示员工工作了多少小时，但它仅显示表中存在数据的天数。我想显示提供的日期之间的所有日期，如果没有数据，则列中应为 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; 

只需确保在 SELECT 语句中使用 t 中的日期而不是 Attendance 表即可。

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