Find min and max data column in Table

I have a table that specifies exactly what date and time each employee was in a particular office.

EmployeeTable looks like this:

id	EmployeeID	DateP	TimeP
1	11111	1397/01/02	01:30
2	11111	1398/05/09	05:30
3	11111	1398/06/07	05:10
4	22222	1398/08/09	06:12
5	22222	1399/02/01	07:15
6	11111	1399/07/02	08:51
7	11111	1399/08/06	12:20
8	33333	1399/09/04	20:01
9	33333	1399/12/08	22:05
10	33333	1400/01/01	23:11
11	33333	1400/02/05	14:10
12	22222	1400/04/05	16:25

I want exactly select Min and Max date and time for each Employee when present in a office:

id	EmployeeID	MinDateP	TimeMinDateP	MaxDateP	TimeMaxDateP
1	11111	1397/01/02	01:30	1398/06/07	05:10
2	22222	1398/08/09	06:12	1399/02/01	07:15
3	11111	1399/07/02	08:51	1399/08/06	12:20
4	33333	1399/09/04	20:01	1400/02/05	14:10
5	22222	1400/04/05	16:25	1400/04/05	16:25

My SQL code is:

with tab1 as 
(
    select * 
    from EmployeeTable
), tab2 as 
(
    select 
        t1.*,
        case when lag(t1.EmployeeID) over(order by t1.id) is null then 1
            when lag(t1.EmployeeID) over(order by t1.id) = t1.EmployeeID then 0
            else 1
       end lg
  from tab1 t1
)
, tab3 as (
select t1.*,
       sum(t1.lg) over(order by t1.id) grp
  from tab2 t1
)
select t1.EmployeeID,
       min(t1.DateP) as min,
       TimeP,
       max(t1.DateP)as max,
       TimeP
  from tab3 t1
 group by t1.EmployeeID, t1.grp

But above codes has error. Can every body help me?

Answer

This is a gaps and islands problem. One approach to solve this uses the difference in row numbers method:

WITH cte AS (
    SELECT *, ROW_NUMBER() OVER (ORDER BY DateP, TimeP) rn1,
              ROW_NUMBER() OVER (PARTITION BY EmployeeID ORDER BY DateP, TimeP) rn2
    FROM EmployeeTable
),
cte2 AS (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY EmployeeID, rn1 - rn2
                                 ORDER BY DateP, TimeP) rn_first,
              ROW_NUMBER() OVER (PARTITION BY EmployeeID, rn1 - rn2
                                 ORDER BY DateP DESC, TimeP DESC) rn_last
    FROM cte
)

SELECT
    EmployeeID,
    MAX(CASE WHEN rn_first = 1 THEN DateP END) AS MinDateP,
    MAX(CASE WHEN rn_first = 1 THEN TimeP END) AS TimeMinDateP,
    MAX(CASE WHEN rn_last = 1  THEN DateP END) AS MaxDateP,
    MAX(CASE WHEN rn_last = 1  THEN TimeP END ) AS TimeMaxDateP
FROM cte2
GROUP BY
    EmployeeID,
    rn1 - rn2
ORDER BY
    MIN(DateP),
    MIN(TimeP);

Note that the logic in the second CTE would be totally unnecessary if you were using a single datetime column to represent the date and time. It is usually not beneficial to separate date and time as you are currently doing.