sql server - SQL : Calculating employee utilization hours for each day -


sql : have table employee working hours in starttime , endtime column. want calculate working hours each employee each day employee starts he's shift on 1 day , ends on next day.

|employee |starttime         |endtime         | |a        | 01/01/2001 23:00 |02/01/2001 10:00| |b        | 01/01/2001 21:00 |01/01/2001 22:00|  output: |employee |date         |hoursworked               |a        | 01/01/2001  | 1         | |a        | 02/01/2001  | 10        | |b        | 01/01/2001  | 1         |

here approach using recursive ctes:

declare @t table( employee nvarchar(10) ,starttime datetime ,endtime datetime )  insert @t values ('a', '2001-01-01 23:00:00', '2001-01-03 10:00:00')                      ,('a', '2001-01-05 21:00:00', '2001-01-06 22:00:00')                      ,('a', '2001-01-07 21:00:00', '2001-01-08 22:00:00')                      ,('b', '2001-01-01 21:00:00', '2001-01-01 22:00:00')                      ,('b', '2001-01-02 21:00:00', '2001-01-03 02:00:00')                      ,('c', '2001-01-03 02:00:00', '2001-01-04 00:00:00');  cte as(   select 1 lvl, employee, convert(date, starttime) starttime_date, starttime, endtime     @t t   union   select lvl + 1 lvl, c.employee, dateadd(d, 1, c.starttime_date) starttime_date, c.starttime, c.endtime     cte c     dateadd(d, 1, c.starttime_date) < c.endtime ), ctecalc as(   select *         ,convert(date, starttime) startdate         ,case when lvl > 1 convert(datetime,convert(date, dateadd(d, datediff(d, starttime, starttime_date), starttime))) else dateadd(d, datediff(d, starttime, starttime_date), starttime) end starttimenew         ,isnull(convert(datetime, lead(starttime_date) on (partition employee, convert(date, starttime) order starttime_date)), endtime) endtimenew     cte  ) select employee, starttime_date startdate, datediff(minute, starttimenew, endtimenew)/60.0 workhours   ctecalc   order employee, starttime_date   option (maxrecursion 0)

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