MySQL

Question

我有一个 table access 存储员工访问系统的时间以及他注销的时间。

access_id int(11) -- PK, auto-generated
employee_id (11)
in_time bigint(20) 
out_time bigint(20)

我需要的是员工外出的平均时间(in_time - out_time员工上次访问的时间) - 所有员工.

到目前为止我管理的内容：

我可以使用这个（相当复杂的）查询设法计算单个员工的平均值

SELECT AVG (b.in_time - a.out_time) / 60000 AS avginminutes
FROM access a CROSS JOIN access b
WHERE 
    b.access_id = 
    (SELECT MIN(c.access_id)
    FROM access c
    WHERE c.access_id > a.access_id
    and c.employee_id = 1765708 )
AND a.employee_id = 1765708
AND  b.in_time - a.out_time != 0
ORDER BY a.access_id ASC;

我的主要问题是，如何修改此查询以计算所有员工的平均值？在花了很多时间之后，我没有得到任何结果。

其次（但不重要），有没有办法简化查询？

示例数据：

access_id|employee_id|in_time      |out_time
|1       |1765708    |1643720400000|1643727600000
|2       |1765708    |1643728200000|1643734800000
|3       |1765708    |1643735100000|1643738400000
|4       |4344524    |1646125200000|1646128800000
|5       |4344524    |1646129100000|1646134200000
|6       |4344524    |1646134800000|1646142000000
|7       |4344524    |1646149200000|1646156400000

MySQL版本：5.5

Answer 1

(The average time when the employee was out) 
 = 
(MAX(out_time) - MIN(in_time) - SUM(out_time - in_time)) / (COUNT(*) - 1)

即

SELECT employee_id,
       (MAX(out_time) - MIN(in_time) - SUM(out_time - in_time)) / (COUNT(*) - 1) / 60000 avginminutes
FROM access 
GROUP BY 1;

如果某个员工只有一行，当然会失败或产生 NULL（取决于 SQL 模式）。在严格模式下 - 根据 CASE 进行调整。

Answer 2

如果您使用的是 MySQL 8，则可以使用 LAG 函数查找每一行的上一个结束时间：

WITH cte AS (
    SELECT
        employee_id,
        LAG(out_time) OVER (PARTITION BY employee_id ORDER BY in_time) AS prev_out,
        in_time
    FROM t
)
SELECT employee_id, AVG(in_time - prev_out)
FROM cte
GROUP BY employee_id

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