How to assign shift based on punch time - sql

Question

Based on punch time shift automatically assigned to employee
Table Trnevents:
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;
If employee punch in time between 8.30 to 9.30 am , it assigned to 9.00 shift
if 9.30 to 10.30. it assigned to 10.00 shift
Expected output:
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

Answer 1

Two solutions, one with LEAD.
First is without LEAD:
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
or:
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
Both query produce same result (second one is probably quicker)

Answer 2

If employee punch in time between 8.30 to 9.30 am , it assigned to 9.00 shift if 9.30 to 10.30. it assigned to 10.00 shift
If I understand this correctly, you can use a case expression:
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
If you want a more general solution where the breaks are at 30 minute intervals throughout the day, then subtract 30 minutes and extract the hour:
select e.*,
datepart(hour, dateadd(minute, -30, dt)) as shift
from e;