search

Home  >  Q&A  >  body text

Get employees who have not clocked in for 11 consecutive days

<p>I am trying to get employees from database who have not marked attendance for 11 consecutive days, For this I have employee table and attendance table but the problem I have with this is that there is no record in the attendance table so how can I get</p> <p>I tried many queries, some of them are as follows: </p> <pre class="brush:php;toolbar:false;">SELECT e.name, e.full_name FROM tabEmployee e LEFT JOIN ( SELECT employee, employee MIN(attendance_date) AS first_attendance_date FROM tabAttendance GROUP BY employee ) ar ON e.name = ar.employee WHERE ar.first_attendance_date IS NULL OR NOT EXISTS ( SELECT 1 FROM tabAttendance WHERE employee = e.name AND attendance_date >= ar.first_attendance_date AND attendance_date < DATE_ADD(ar.first_attendance_date, INTERVAL 11 DAY) )</pre> <p>Another query:</p> <pre class="brush:php;toolbar:false;">SELECT e.name, COUNT(a.`attendance_date`), COUNT(e.`name`) from tabEmployee e LEFT JOIN tabAttendance a ON e.name = a.`employee` WHERE a.`employee` IS NULL GROUP BY e.`name`</pre> <p><br /></p>
P粉103739566P粉103739566475 days ago491

reply all(1)I'll reply

  • P粉231079976

    P粉2310799762023-08-19 07:58:16

    Using LAG() analysis function, we can try:

    WITH cte AS (
        SELECT e.name, e.full_name,
               LAG(a.attendance_date) OVER (PARTITION BY a.employee
                                            ORDER BY a.attendance_date) AS lag_attendance_date
        FROM tabAttendance a
        INNER JOIN tabEmployee e ON e.name = a.employee
    )
    
    SELECT DISTINCT name, full_name
    FROM cte
    WHERE DATEDIFF(attendance_date, lag_attendance_date) > 11;
    

    The basic strategy here is to generate the lag (previous consecutive value) of the attendance date in the CTE. We then report only employees with a gap of 11 days or longer.

    reply
    0
  • Cancelreply