基于自动分配给员工的上班时间
表事件:
emp_reader_id EVENTID DT
3 1 2019-07-14 17:00:00.000
3 0 2019-07-14 10:00:00.000
3 1 2019-07-13 17:50:00.000
3 0 2019-07-13 10:05:00.000
3 1 2019-07-12 16:00:00.000
3 0 2019-07-12 08:55:00.000
declare
@start_date date='2019-07-12'
,@end_date date ='2019-07-14'
;WITH ByDays AS
( -- Number the entry register in each day
SELECT
emp_reader_id,
dt AS T,
CONVERT(VARCHAR(10),dt,102) AS Day,
FLOOR(CONVERT(FLOAT,dt)) DayNumber,
ROW_NUMBER() OVER(PARTITION BY FLOOR(CONVERT(FLOAT,dt)) ORDER BY dt) InDay
FROM trnevents
where
(
CONVERT(VARCHAR(26), dt, 23) >= CONVERT(VARCHAR(26), @start_date, 23)
and CONVERT(VARCHAR(26), dt, 23) <=CONVERT(VARCHAR(26), @end_date, 23)
)
)
,Diffs AS
(
SELECT
E.Day,
E.emp_Reader_id,
E.T ET,
O.T OT,
O.T-E.T Diff,
DATEDIFF(S,E.T,O.T) DiffSeconds -- difference in seconds
FROM
(
SELECT
BE.emp_Reader_id,
BE.T,
BE.Day,
BE.InDay
FROM ByDays BE
WHERE BE.InDay % 2 = 1
) E -- Even rows
INNER JOIN
(
SELECT
BO.emp_reader_id,
BO.T,
BO.Day,
BO.InDay
FROM ByDays BO
WHERE BO.InDay % 2 = 0
) O -- Odd rows
ON E.InDay + 1 = O.InDay -- Join rows (1,2), (3,4) and so on
AND E.Day = O.Day -- in the same day
)
SELECT * FROM Diffs
DECLARE @start TIME(0) = '9:00 AM', @end TIME(0) = '18:00 PM';
WITH x(n) AS
(
SELECT TOP (DATEDIFF(HOUR, @start, @end) + 1)
rn = ROW_NUMBER() OVER (ORDER BY [object_id])
FROM sys.all_columns
ORDER BY [object_id]
)
SELECT
t = DATEADD(HOUR, n-1, @start)
,cast(DATEADD(HOUR, n-1, @start) as varchar(50))+' shift'
FROM x
ORDER BY t;
如果员工的上班时间在8.30到9.30 am之间,则分配为9.00班次 如果是9.30到10.30。它分配了10.00班次
预期输出:
Day emp_Reader_id ET OT Diff DiffSeconds Shift
2019.07.12 3 2019-07-12 08:55:00.000 2019-07-12 16:00:00.000 1900-01-01 07:05:00.000 25500 09:00:00 shift
2019.07.13 3 2019-07-13 10:05:00.000 2019-07-13 17:50:00.000 1900-01-01 07:45:00.000 27900 10:00:00 shift
2019.07.14 3 2019-07-14 12:00:00.000 2019-07-14 21:00:00.000 1900-01-01 07:00:00.000 25200 12:00:00 shift
答案 0 :(得分:1)
如果员工的上班时间在8.30到9.30 am之间,则从9.30到10.30分配到9.00班次。它分配了10.00班次
如果我理解正确,则可以使用case
表达式:
select e.*,
(case when dt >= '08:30:00' and dt < '09:30:00'
then 'Shift 09:00'
when dt >= '09:30:00' and dt < '10:30:00'
then 'Shift 10:00'
end) as shift
from Trnevents e
如果您想要一个更通用的解决方案,每天休息时间间隔为30分钟,则减去30分钟并提取小时数:
select e.*,
datepart(hour, dateadd(minute, -30, dt)) as shift
from e;
答案 1 :(得分:1)
两种解决方案,一种采用LEAD。 首先是没有铅:
select
CAST(t1.DT as date) AS "Day",
t1.emp_reader_id AS emp_Reader_id,
t1.DT AS ET,
t2.DT AS OT,
t1.DT - t2.DT As Diff,
DATEDIFF(s, t1.DT, t2.DT) As DiffSeconds,
cast(dateadd(HOUR,datepart(HH,t1.DT)+ round(datepart(MINUTE,t1.dt)/60.0,0),0) as time) as Shift
from trnevents t1
inner join trnevents t2 on t2.emp_reader_id=t1.emp_reader_id and t2.EVENTID=1 and CAST(t2.DT as date)= CAST(t1.DT as date)
where t1.eventID=0
order by t1.DT
或:
SELECT
Day,
emp_reader_id,
ET,
OT,
ET-OT AS Diff ,
DATEDIFF(s,ET,OT) as DiffSeconds,
cast(dateadd(HOUR,datepart(HH,ET)+ round(datepart(MINUTE,ET)/60.0,0),0) as time) as Shift
FROM (
select
CAST(t1.DT as date) AS "Day",
t1.emp_reader_id AS emp_Reader_id,
t1.DT AS ET,
LEAD(t1.DT) over (order by emp_reader_id,dt) AS OT,
eventid,
--t1.DT - t2.DT As Diff,
--DATEDIFF(s, t1.DT, t2.DT) As DiffSeconds,
cast(dateadd(HOUR,datepart(HH,t1.DT)+ round(datepart(MINUTE,t1.dt)/60.0,0),0) as time) as Shift
from trnevents t1) x
where x.EVENTID=0
两个查询都产生相同的结果(第二个查询可能更快)