Question

我们如何在两次之间制作商店程序？

喜欢，

Total Hr=(( outtime - intime) - (breakin time - breakout time) )

Answer 1

create procedure sp_time_outtime

@id bigint

as

begin

declare

 @totalhour as nvarchar(50),

@tth as bigint,

@ttm as bigint,

@inhour as bigint,

@outhour as bigint,

@breakinhour as bigint select @tth = datediff(mi, intime, (case when  breakouttime is null then isnull(breakintime,GETDATE()) else GETDATE() end) ) from time where empid=@id   and CONVERT(date,GETDATE()) = CONVERT(date,[date])

 select @ttm = datediff(mi, isnull(breakintime,'1/1/1990'),ISNULL(breakouttime,isnull (breakintime,'1/1/1990'))) from time where empid=@id  and CONVERT(date,GETDATE()) = CONVERT(date,[date])


 select @totalhour = (@tth -@ttm)

 SET @totalhour = CASE WHEN (@tth -@ttm) >= 60 THEN (SELECT CAST(((@tth -@ttm) / 60) AS VARCHAR(2)) + ' :H :' +  
                CASE WHEN ((@tth -@ttm) % 60) > 0 THEN
                    CAST(((@tth -@ttm) % 60) AS VARCHAR(2)) + ' M'
                ELSE
                    ''
                END)  ELSE 
        CAST(((@tth -@ttm) % 60) AS VARCHAR(2)) + 'M'
    END
 if ((select outtime from time where empid=@id and convert(date,GETDATE())=CONVERT(date,[date])) IS NULL)
begin
 update time set outtime=GETDATE(),totalhour=@totalhour where 
empid=@id and CONVERT(date,GETDATE()) = CONVERT(date,[date])
end END

计算2个时间戳之间的总小时数？

1 个答案: