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Home/ Questions/Q 9143075
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T09:56:02+00:00 2026-06-17T09:56:02+00:00

I have a table that is going to have several time stamp entries added

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I have a table that is going to have several time stamp entries added throughout the day with a specific employee ID tied to each entry. I am curious how I would get the first timestamp of the day and the last time stamp of the day to calculate amount of time worked for that specific employee on the specific date. My table is below:

+----+------------+----------+---------+---------------------+-----------+------------+-----------+---------+
| id | employeeID | date     | timeIn  | jobDescription      | equipType | unitNumber | unitHours | timeOut |
+----+------------+----------+---------+---------------------+-----------+------------+-----------+---------+
|  1 |          1 | 01/13/13 | 8:17 pm | Worked in Hubbard   | Dozer     | 2D         |     11931 | 8:17 pm |
|  2 |          1 | 01/13/13 | 8:17 pm | Worked in Jefferson | Excavator | 01E        |      8341 | 8:18 pm |
+----+------------+----------+---------+---------------------+-----------+------------+-----------+---------+

so far I have a query like this to retrieve the time values:

$stmt = $conn->prepare('SELECT * FROM `timeRecords` WHERE `date`= :dateToday AND `employeeID` = :employeeID ORDER BY employeeID ASC');
    $stmt->execute(array(':employeeID' => $_SESSION['employeeID'], ':dateToday' => $dateToday));

But I am unsure of how to obtain the greatest value in the timeOut column

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T09:56:03+00:00Added an answer on June 17, 2026 at 9:56 am

    Really, you just need the aggregate MAX() and MIN() grouped by employeeID. Use the TIMEDIFF() function to calculate the difference in time between the two.

    SELECT 
      `employeeID`,
      MIN(`timeIn`) AS `timeIn`,
      MAX(`timeOut`) AS `timeOut`,
      TIMEDIFF(MAX(`timeOut`), MIN(`timeIn`)) AS `totalTime`
    FROM `timeRecords`
    WHERE
      `date` = :dateToday
      AND `employeeID` = :employeeID
    /* Selecting only one employeeID you don't actually need the GROUP BY */
    GROUP BY `employeeID`
    

    However, this won’t report the total time worked if an employee clocks in and out several times during one day. In that case, you would need to SUM() the result of the TIMEDIFF() for each of the in/out pairs.

    Something like:

    SELECT
      `employeeID`,
      /* Assumes no times overlap across rows */
      SEC_TO_TIME(SUM(TIME_TO_SEC(TIMEDIFF(`timeOut`, `timeIn`)))) AS `totalTime`
    FROM `timeRecords`
    WHERE
      `date` = :dateToday
      AND `employeeID` = :employeeID
    GROUP BY `employeeID`
    
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