获取连续11天没有打卡的员工
P粉103739566
P粉103739566 2023-08-18 09:31:19
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<p>我正在尝试从数据库中获取连续11天没有标记出勤的员工, 为此我有员工表和出勤表,但是我在这方面遇到的问题是出勤表中没有记录,所以我该如何获取</p> <p>我尝试了很多查询,其中一些如下:</p> <pre class="brush:php;toolbar:false;">SELECT e.name, e.full_name FROM tabEmployee e LEFT JOIN ( SELECT 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>另一个查询:</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>
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P粉103739566

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P粉231079976

使用 LAG() 分析函数,我们可以尝试:

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;

这里的基本策略是在 CTE 中生成出勤日期的 lag(前一个连续值)。然后我们只报告有 11 天或更长间隔的员工。

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