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I have a table with time-in/time-out data in a structure as shown below.

id date timein timeout duration
3 2024-01-02 01:00 02:00 01:00
3 2024-01-02 02:30 07:00 04:30
3 2024-01-02 09:00 09:10 00:10
3 2024-01-02 09:10 09:30 00:20
3 2024-01-02 12:00 12:25 00:25

Time-in positions with a duration less than 30 minutes should not be considered for calculation. however, if two consecutive time-in for a day is more than or equal to 30 minutes then it should be considered. like in the above example, row 3 time-out is same as row 4 time-in 09:00-09:10/09:1--09:30. they are considered as single time-in and should be added if their total time is greater than 30 minutes. I came up with the following query to get the day wise total duration excluding durations less than 30 minutes but not including the time-in duration as explained above.

SELECT
    DATE,  TIME_FORMAT(SEC_TO_TIME(SUM(TIME_TO_SEC(duration))),'%H:%i' ) AS total,
    SUM(TIME_TO_SEC(duration)) AS seconds
FROM
    `timeintable`
WHERE
    id = '3' AND( DATE BETWEEN '2024-01-01' AND '2024-01-10') AND(TIME_TO_SEC(duration) >= 1800)
GROUP BY
    DATE
ORDER BY
    DATE ASC

Is it possible to achieve the same through SQL query rather than php array loop? My attempt using LEAD/LAG functions were unsuccessful.

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  • however, if two consecutive time-in for a day is more than or equal to 30 minutes then it should be considered. Does these timeranges must be adjacent? I.e. if in 4th row the times are 9:11-9:31 then this row should not be combined with row 3 and the combined value should not be taken into account? Also - if more than 2 consecutive timeranges gives 30 min or more whereas any 2 of them are less (9:00-9:10, 9:10-9:20, 9:20-9:30) does these rows should be counted?
    – Akina
    Commented Jan 17 at 5:16
  • What is MySQL version precisely? SELECT VERSION();
    – Akina
    Commented Jan 17 at 5:19
  • @Akina it should be considered only when they are adjacent. If the time-out time is equal to the time-in time of next row on same day. More than two consecutive rows satisfying above condition should also be considered for checking if greater than 30 minutes. MySQL version is 10.6.16-MariaDB-cll-lve. Commented Jan 17 at 5:31
  • MySQL version is 10.6.16-MariaDB-cll-lve. MariaDB <> MySQL !!!
    – Akina
    Commented Jan 17 at 5:39

1 Answer 1

1
WITH 
cte1 AS (
  SELECT *, CASE WHEN timein = LAG(timeout) OVER (PARTITION BY id, `date` ORDER BY timein)
                 THEN 0 
                 ELSE 1
                 END not_adjacent  
  FROM test
-- WHERE id IN ( {needed ids list} )
--   AND `date` IN ( {needed dates list} )
  ),
cte2 AS (
  SELECT *,
         SUM(not_adjacent) OVER (PARTITION BY id, `date` ORDER BY timein) group_no
  FROM cte1
  ),
cte3 AS (
  SELECT id, `date`, TIMESTAMPDIFF(SECOND, MIN(timein), MAX(timeout)) duration
  FROM cte2
  GROUP BY id, `date`, group_no
  HAVING duration >= TIME_TO_SEC('00:30:00')
  )
SELECT id, `date`, SEC_TO_TIME(SUM(duration)) daily_duration
FROM cte3
GROUP BY id, `date`

step-by-step fiddle with some comments.

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  • Thank you @Akina. When I test it , I get the error like 'Warning: #1292 Incorrect datetime value: '18:30'" probably because time-in and timeout are saved as VARCHAR in my table. Commented Jan 17 at 6:45
  • @ShahabMohd time-in and timeout are saved as VARCHAR Adjust cte1 and perform correct convertion (CAST) from VARCHAR to TIME (also - replace an asterisk with definite columns names and perform the convertion for them too). PS. Next time - always provide a sample of your table as CREATE TABLE + INSERT INTO, not as table-formatted text.
    – Akina
    Commented Jan 17 at 7:15
  • Thank you. I will do that Commented Jan 18 at 11:18

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