How To find attendance Mispunch in sql - sql

Question

I have a table tbl_attendance in SQL Server with data is in this format
card_no adate time
-----------------------------------------
13 2016-08-01 2016-08-01 09:30:00
13 2016-08-01 2016-08-01 11:00:00
13 2016-08-01 2016-08-01 12:00:00
13 2016-08-01 2016-08-01 15:00:00
13 2016-08-01 2016-08-01 16:00:00
13 2016-08-02 2016-08-02 09:30:00
but when I execute my query, I want to get results in this format
card_no adate time_in time_in
----------------------------------------------------------------
13 2016-08-01 2016-08-01 09:30:00 2016-08-01 11:00:00
13 2016-08-01 2016-08-01 12:00:00 2016-08-01 15:00:00
13 2016-08-01 2016-08-01 16:00:00 null
13 2016-08-02 2016-08-02 09:30:00 null
Please help as soon as possible

Answer 1

Try this trick
;WITH cte
AS (SELECT *,
((Row_number()OVER(partition BY [adate] ORDER BY [time])-1)%2) + 1 AS seq_no,
((Row_number()OVER(partition BY [adate] ORDER BY [time])-1)/2) + 1 rang_no
FROM Yourtable)
SELECT [card_no],
[adate],
time_in = Min(CASE WHEN seq_no = 1 THEN [time] END),
time_in = Min(CASE WHEN seq_no = 2 THEN [time] END)
FROM cte
GROUP BY [card_no],
[adate],
rang_no
Demo
Schema Setup
IF Object_id('tempdb.dbo.#Table1') is not null
DROP TABLE #Table1
CREATE TABLE #Table1
([card_no] int, [adate] date, [time] datetime);
Sample Data
INSERT INTO #Table1
([card_no], [adate], [time])
VALUES
(13, '2016-08-01 00:00:00', '2016-08-01 09:30:00'),
(13, '2016-08-01 00:00:00', '2016-08-01 11:00:00'),
(13, '2016-08-01 00:00:00', '2016-08-01 12:00:00'),
(13, '2016-08-01 00:00:00', '2016-08-01 15:00:00'),
(13, '2016-08-01 00:00:00', '2016-08-01 16:00:00'),
(13, '2016-08-02 00:00:00', '2016-08-02 09:30:00');
Query
;WITH cte
AS (SELECT *,
((Row_number()OVER(partition BY [adate] ORDER BY [time])-1)%2) + 1 AS seq_no,
((Row_number()OVER(partition BY [adate] ORDER BY [time])-1)/2) + 1 rang_no
FROM #Table1)
SELECT [card_no],
[adate],
time_in = Min(CASE WHEN seq_no = 1 THEN [time] END),
time_in = Min(CASE WHEN seq_no = 2 THEN [time] END)
FROM cte
GROUP BY [card_no],
[adate],
rang_no
Result:
card_no adate time_in time_in
------ ---------- ----------------------- -----------------------
13 2016-08-01 2016-08-01 09:30:00.000 2016-08-01 11:00:00.000
13 2016-08-01 2016-08-01 12:00:00.000 2016-08-01 15:00:00.000
13 2016-08-01 2016-08-01 16:00:00.000 NULL
13 2016-08-02 2016-08-02 09:30:00.000 NULL

Answer 2

Another way with OUTER APPLY (is used to get next row with time > than in current row) and ROW_NUMBER (with %2 to get only odd rows):
SELECT t.card_no,
t.adate,
t.[time],
t.time_in
FROM (
SELECT y.*,
p.[time] as time_in,
ROW_NUMBER() OVER (PARTITION BY y.card_no, y.adate ORDER BY y.[time])%2 as seq
FROM YourTable y
OUTER APPLY (
SELECT TOP 1 *
FROM YourTable
WHERE [time] > y.[time] and [adate] = y.adate
ORDER BY [time] ASC
) p
) as t
WHERE t.seq = 1
ORDER BY [time]
Output:
card_no adate time time_in
13 2016-08-01 2016-08-01 09:30:00 2016-08-01 11:00:00
13 2016-08-01 2016-08-01 12:00:00 2016-08-01 15:00:00
13 2016-08-01 2016-08-01 16:00:00 NULL
13 2016-08-02 2016-08-02 09:30:00 NULL