Question

我一直试图让函数继续计算从员工的StartTime和EndTime开始的工作班次。这是我到目前为止的代码，似乎计算错误的转移。

Shift 1 from 08:00:00 - 16:30:00
Shift 2 from 16:00:00 - 00:30:00
Shift 3 from 00:00:00 - 08:30:00

此外，大多数小时的转变在轮班之间获胜。

谢谢， ģ

`FUNCTION [dbo].[ShiftDifferential] (
 @StartTime time(0), 
 @EndTime time(0)
 )
RETURNS int
AS
BEGIN
 --DECLARE @StartTime time(0)
 --DECLARE @EndTime time(0)

 -- Declare the return variable here
 DECLARE @Shift1StartTime time(0)
 DECLARE @Shift2StartTime time(0)
 DECLARE @Shift3StartTime time(0)
 DECLARE @Shift1EndTime time(0)
 DECLARE @Shift2EndTime time(0)
 DECLARE @Shift3EndTime time(0)

 DECLARE @HrsShift1 decimal(18,2)
 DECLARE @HrsShift2 decimal(18,2)
 DECLARE @HrsShift3 decimal(18,2)

 DECLARE @ShiftDiff int

 --SET @StartTime = '09:00:00'
 --SET @EndTime = '13:00:00'

 SET @Shift1StartTime = '08:00:00' 
 SET @Shift2StartTime = '16:00:00'
 SET @Shift3StartTime = '00:00:00'
 SET @Shift1EndTime = '16:30:00'
 SET @Shift2EndTime = '00:30:00'
 SET @Shift3EndTime = '08:30:00'



 --SELECT DATEDIFF(HH, @Shift1StartTime, @EndTime)

 -- hours are between shift 3 and shift 1
 if DATEDIFF(HH, @Shift1StartTime, @StartTime) < 0 AND (DATEDIFF(hh, @Shift1StartTime, @EndTime) < 8.0 AND DATEDIFF(hh, @Shift1StartTime, @EndTime) > 0)
  begin
   --PRINT 'Shift 3-1 step1'
   SET @HrsShift3 = DATEDIFF(HH, @StartTime, @Shift1StartTime)
   SET @HrsShift1 = DATEDIFF(HH, @Shift1StartTime, @Endtime)
   --PRINT @HrsShift3
   --PRINT @HrsShift1

   -- get shift with most hours
   if @HrsShift3 > @HrsShift1
    begin
     SET @ShiftDiff = 3
    end
   else
    begin
     SET @ShiftDiff = 1
    end
  end

 -- hours are in shift 1 
 if (DATEDIFF(HH, @Shift1StartTime, @StartTime) = 0 AND DATEDIFF(HH, @Shift1StartTime, @EndTime) <= 8) 
   OR (DATEDIFF(HH, @Shift1StartTime, @StartTime) > 0  AND DATEDIFF(HH, @Shift1StartTime, @EndTime) <= 8)
  begin
   --PRINT 'Shift 1 step2'
   SET @HrsShift3 = 0
   SET @HrsShift1 = DATEDIFF(HH, @StartTime, @EndTime)
   --PRINT @HrsShift3
   --PRINT @HrsShift1

   -- only one shift with hours
   SET @ShiftDiff = 1
  end

 -- hours are between shift 1 and shift 2
 if DATEDIFF(HH, @Shift2StartTime, @StartTime) < 0 and (DATEDIFF(HH, @Shift2StartTime, @EndTime) < 8.0 AND DATEDIFF(HH, @Shift2StartTime, @EndTime) > 0)  
  begin
   --PRINT 'Shift 1-2 step1' 
   SET @HrsShift1 = DATEDIFF(HH, @StartTime, @Shift2StartTime)
   SET @HrsShift2 = DATEDIFF(HH, @Shift2StartTime, @Endtime)
   --PRINT @HrsShift1
   --PRINT @HrsShift2

   -- get the shift with most hours
   if @HrsShift1 > @HrsShift2
    begin
     SET @ShiftDiff = 1
    end
   else
    begin
     SET @ShiftDiff = 2
    end
  end

 -- hours are in shift 2 
 if (DATEDIFF(HH, @Shift2StartTime, @StartTime) = 0 AND DATEDIFF(HH, @Shift2StartTime, @EndTime) <= 8) 
  OR (DATEDIFF(HH, @Shift2StartTime, @StartTime) > 0  AND DATEDIFF(HH, @Shift2StartTime, @EndTime) <= 8)
  begin
   --PRINT 'Shift 2 step2'
   SET @HrsShift3 = 0
   SET @HrsShift1 = DATEDIFF(HH, @StartTime, @EndTime)
   --PRINT @HrsShift3
   --PRINT @HrsShift1   

   -- only one shift with hours
   SET @ShiftDiff = 2
  end

 -- hours are between shift 2 and shift 3 - overnight shift
 if DATEDIFF(HH, @StartTime, @EndTime) < 0  
  begin
   --PRINT 'Shift 2-3 step1' 
   SET @HrsShift2 = DATEDIFF(HH, @StartTime, '23:59:59') + DATEDIFF(HH, '00:00:00', '00:30:00')
   SET @HrsShift3 = DATEDIFF(HH, '00:30:00', @EndTime)
   --PRINT @HrsShift2
   --PRINT @HrsShift3

   -- get the shift with most hours
   if @HrsShift2 > @HrsShift3
    begin
     SET @ShiftDiff = 2
    end
   else
    begin
     SET @ShiftDiff = 3
    end
  end

 -- hours are in shift 3 
 if (DATEDIFF(HH, @Shift3StartTime, @StartTime) = 0  AND DATEDIFF(HH, @Shift3StartTime, @EndTime) <= 8)
  OR (DATEDIFF(HH, @Shift3StartTime, @StartTime) > 0  AND DATEDIFF(HH, @Shift3StartTime, @EndTime) <= 8)
  begin
   --PRINT 'Shift 3 step2'
   SET @HrsShift2 = 0
   SET @HrsShift3 = DATEDIFF(HH, @StartTime, @EndTime)
   --PRINT @HrsShift2
   --PRINT @HrsShift3   

   -- only one shift with hours
   SET @ShiftDiff = 3
  end

 RETURN @ShiftDiff;
END`

Answer 1

感觉就像是在试图采用迭代方法解决基于集合的问题。

听起来我觉得你正试图找到与时间表条目中输入的次数最重叠的班次。

完成这项工作后，您会发现它基于设置，并且可以轻松扩展，因为如果您愿意，可以使用Times表。这假定事情始于使用time类型，但我将其转换为datetime，因为转移可能会在第二天结束。

with 
Shifts as
(
SELECT ShiftID, cast(StartTime as datetime) as StartTime, 
    case 
      when EndTime < StartTime 
      then dateadd(day,1,cast(EndTime as datetime)) 
      else cast(EndTime as datetime) 
    end as EndTime
FROM Shifts
),
Times as 
(select 1 as TimeID,
    cast(@StartTime as datetime) as StartTime, 
    case 
      when @EndTime < @StartTime 
      then dateadd(day,1,cast(@EndTime as datetime)) 
      else cast(@EndTime as datetime) 
    end as EndTime
),
Overlaps as
(
select s.ShiftID, t.TimeID, 
       case when s.StartTime > t.StartTime then s.StartTime else t.StartTime end as StartOverlap,
       case when s.EndTime < t.EndTime then s.EndTime else t.EndTime end as EndOverlap,
from Shifts s
cross join Times t
),
OrderedOverlaps as
(
select *, row_number() over(partition by TimeID order by datediff(min,StartOverlap,EndOverlap) desc) as RowNum
from Overlaps
)
select s.ShiftID, t.TimeID
from OrderedOverlaps
where RowNum = 1;

从StartTime和EndTime计算Shift的SQL函数

1 个答案: