Find consecutive days absent

Question

I have a MySql table attendance with empno and timeindate columns. I'm having trouble getting the data of an employee with 3 consecutive days absent:

Here is my sample data:

==========
empno:     | timeindate

empno 1    | 2013-10-01 00:00:00
empno 2    | 2013-10-01 00:00:00
empno 1    | 2013-10-02 00:00:00
empno 2    | 2013-10-02 00:00:00
empno 2    | 2013-10-03 00:00:00
empno 2    | 2013-10-04 00:00:00
empno 2    | 2013-10-05 00:00:00
empno 1    | 2013-10-06 00:00:00
empno 2    | 2013-10-06 00:00:00
----------

Please help, I have read several posts but it didn't give me the data I need and I just got confused. I would very much appreciate any answer.

Answer 1

I would try something like this.

WITH T AS
(SELECT empno, dateintime, @row:=@row+1 rownum FROM
attendance JOIN (SELECT @row:=0) init
ORDER BY empno, timeindate)

SELECT DISTINCT t1.empno
FROM T AS t1 INNER JOIN T AS t2
ON t1.empno = t2.empno AND t1.rownum = t2.rownum + 1
WHERE TIMESTAMPDIFF(DAY, t2.dateintime, t1.dateintime) >= 3

First query orders the table by datetime and enumerates rows. Then, you join it with itself, with a shift in row number, getting neighbor datetimes in one row. And then, you just look for entries with a needed difference.

I'm not sure it works, but it should work like that: first query:

empno:     | timeindate          | row

empno 1    | 2013-10-01 00:00:00 | 0
empno 1    | 2013-10-02 00:00:00 | 1
empno 1    | 2013-10-06 00:00:00 | 2
empno 2    | 2013-10-01 00:00:00 | 3
empno 2    | 2013-10-02 00:00:00 | 4
empno 2    | 2013-10-03 00:00:00 | 5
empno 2    | 2013-10-04 00:00:00 | 6
empno 2    | 2013-10-05 00:00:00 | 7
empno 2    | 2013-10-06 00:00:00 | 8

Second query without WHERE clause:

empno 1    | 2013-10-02 00:00:00 | 2013-10-02 00:00:00
empno 1    | 2013-10-02 00:00:00 | 2013-10-06 00:00:00
empno 2    | 2013-10-01 00:00:00 | 2013-10-02 00:00:00
empno 2    | 2013-10-02 00:00:00 | 2013-10-03 00:00:00
etc.

As you see, the WHERE will filter the second row from this result.