Question

输入表

Start time     End Time
8/12/14 17:00  8/14/14 12:00

我需要将时间拆分并显示输出如下

11:00 to 23:00    23:00 to 11:00
      19                24

计算工作

Date     11:00 to 23:00    23:00 to 11:00
8/12/14        6                1
8/13/14        12               12
8/14/14        1                11

Total          19               24

Answer 1

您可以使用递归公用表表达式生成开始日期和结束日期之间的小时序列，然后使用case语句来评估它所属的范围，然后求和汇总：

DECLARE @T TABLE (StartTime DATETIME, EndTime DATETIME)
INSERT @T VALUES ('8/12/14 17:00', '8/14/14 12:00')

;WITH cte AS 
    (
    SELECT StartTime AS dt, EndTime
    FROM @T
    UNION  ALL
    SELECT DATEADD(MINUTE, 60, dt) AS dt, EndTime
    FROM cte
    WHERE dt < DATEADD(HOUR, -1, EndTime)
    )

SELECT 
  SUM(
    CASE 
      WHEN CAST(dt AS TIME) >= '11:00' AND CAST(dt AS TIME) < '23:00' THEN 1 
      ELSE 0 
    END) AS '11:00 to 23:00',
  SUM(
    CASE 
      WHEN CAST(dt AS TIME) >= '11:00' AND CAST(dt AS TIME) < '23:00' THEN 0 
      ELSE 1 
    END) AS '23:00 to 11:00'
FROM cte

我确信它可以改进但它应该能给你想要的结果。

Sample SQL Fiddle

示例输出：

11:00 to 23:00 23:00 to 11:00
-------------- --------------
19             24

此SQL Fiddle以30分钟为间隔显示处理时间，输出为：

11:00 to 23:00 23:00 to 11:00
-------------- --------------
23.5           25.5

两个日期之间的班次

1 个答案: