T-SQL 解法
DECLARE @Temp TABLE (EmpNO char(10) , [WorkDate] datetime ,WorkType char(4))
INSERT INTO @Temp VALUES('001','2013/01/10','早班')
INSERT INTO @Temp VALUES('001','2013/01/11','早班')
INSERT INTO @Temp VALUES('001','2013/01/12','早班')
INSERT INTO @Temp VALUES('001','2013/01/13','早班')
INSERT INTO @Temp VALUES('001','2013/01/14','休息')
INSERT INTO @Temp VALUES('001','2013/01/15','休息')
INSERT INTO @Temp VALUES('001','2013/01/16','休息')
INSERT INTO @Temp VALUES('001','2013/01/17','晚班')
INSERT INTO @Temp VALUES('001','2013/01/18','早班')
INSERT INTO @Temp VALUES('001','2013/01/21','早班')
INSERT INTO @Temp VALUES('001','2013/01/22','早班')
INSERT INTO @Temp VALUES('001','2013/01/23','早班')
INSERT INTO @Temp VALUES('001','2013/01/24','早班')
SELECT
T.EmpNO ,
T.WorkType ,
MIN(WorkDate) AS StartDate , -- 找出該班別起始日期
COUNT(*) AS [Counts] -- 該班別連續天數
FROM
(
SELECT
EmpNO ,
WorkType ,
WorkDate ,
-- 重點邏輯
DATEADD(DD , -1 * ROW_NUMBER() OVER (PARTITION BY EmpNO , WorkType ORDER BY WorkDate) , WorkDate) AS Grp
FROM @Temp
) AS T
GROUP BY T.EmpNO , T.WorkType , T.Grp
ORDER BY EmpNO , StartDate
邏輯分析
以 001 早班為例說明
SELECT
EmpNO ,
WorkType ,
WorkDate ,
ROW_NUMBER() OVER (PARTITION BY EmpNO , WorkType ORDER BY WorkDate) AS ROWNO ,
DATEADD(DD , -1 * ROW_NUMBER() OVER (PARTITION BY EmpNO , WorkType ORDER BY WorkDate) , WorkDate) AS Grp
FROM @Temp
WHERE EmpNO = '001'
AND WorkType = '早班'
判斷連續日期為此問題重點,利用 ROW_NUMBER() 產生 ROWNO 欄位後,把此 ROWNO 資料去減掉 WorkDate,從上圖的 Grp 欄位可以看出,假如該班別為連續日期,Grp 欄位內的日期資料會是相同日期,依此來判斷是否為連續日期。