mysql - SQL 跨项目的不同工作日期，不包括休息日期

Question

考虑以下模式；

CREATE TABLE `Project Assignment`
    (`Employee` varchar(1), `Project Id` int, `Project Assignment Date` date, `Project Relieving Date` date)
;

INSERT INTO `Project Assignment`
    (`Employee`, `Project Id`, `Project Assignment Date`, `Project Relieving Date`)
VALUES
    ('A', 1, '2018-04-01', '2019-12-25'),
    ('A', 2, '2019-06-15', '2020-03-31'),
    ('A', 3, '2019-09-07', '2020-05-20'),
    ('A', 4, '2020-07-14', '2020-12-15')
;


CREATE TABLE `Break`
    (`Break Id` int, `Employee` varchar(1), `Project Id` int, `Break Start Date` date, `Break End Date` date)
;

INSERT INTO `Break`
    (`Break Id`, `Employee`, `Project Id`, `Break Start Date`, `Break End Date`)
VALUES
    (1, 'A', 1, '2018-09-01', '2018-09-30'),
    (2, 'A', 1, '2019-10-05', '2019-11-30'),
    (3, 'A', 2, '2019-10-15', '2019-11-15'),
    (4, 'A', 3, '2019-11-01', '2019-11-10'),
    (5, 'A', 2, '2020-01-01', '2020-01-10'),
    (6, 'A', 3, '2020-01-01', '2020-01-10')
;

在项目期间，员工可以在每个项目中进行一次或多次休息。中断不会在 Project 中重叠，但可以在项目之间重叠。

我们想要员工至少分配一个项目的天数（减去）员工在所有分配项目上的休息天数。

通过使用以下查询，我能够得出员工被分配到项目的不同天数：

SELECT merged.employee,
    SUM(DATEDIFF(merged.EndDate,merged.`Project Assignment Date`)+1) assigned_days
FROM (SELECT
        s1.employee, s1.`Project Assignment Date`,
        MIN(IFNULL(t1.`Project Relieving Date`,CURDATE())) AS EndDate
    FROM `Project Assignment` s1
    INNER JOIN `Project Assignment` t1
        ON t1.employee = s1.employee
        AND s1.`Project Assignment Date` <= IFNULL(t1.`Project Relieving Date`,CURDATE())
        AND NOT EXISTS( SELECT * FROM `Project Assignment` t2
            WHERE t2.employee = s1.employee 
                AND IFNULL(t1.`Project Relieving Date`,CURDATE()) >= t2.`Project Assignment Date` 
                AND IFNULL(t1.`Project Relieving Date`,CURDATE()) < IFNULL(t2.`Project Relieving Date`,CURDATE()))
    WHERE NOT EXISTS( SELECT * FROM `Project Assignment` s2
        WHERE s2.employee = s1.employee
            AND s1.`Project Assignment Date` > s2.`Project Assignment Date` 
            AND s1.`Project Assignment Date` <= IFNULL(s2.`Project Relieving Date`,CURDATE()))
    GROUP BY s1.employee, s1.`Project Assignment Date`
    ORDER BY s1.`Project Assignment Date`) merged
GROUP BY merged.employee

结果：

| employee | assigned_days |
| -------- | ------------- |
| A        | 936           |

但是想不出一种方法来推导出这个人在所有分配的项目上的休息天数。

预期结果：

+----------+---------------+------------+-------------+
| employee | assigned_days | break_days | worked_days |
+==========+===============+============+=============+
| A        | 936           | 50         | 886         |
+----------+---------------+------------+-------------+

Mariadb 10.3.29

锻炼break_days的解释

+----------+---------+-------------+------------------+-----------------+-------------------------------------------------------------------------------------------------------------------+
| Employee | Project | Break Start | Break End        | Days Considered | Remarks                                                                                                           |
+==========+=========+=============+==================+=================+===================================================================================================================+
| A        | 1       |  2018-09-01 |  2018-09-30      | 30              | Only one project assigned so consider whole break                                                                 |
+----------+---------+-------------+------------------+-----------------+-------------------------------------------------------------------------------------------------------------------+
| A        | 1       |  2019-10-05 |  2019-11-30      | 10              | 3 Projects were   assigned during these breaks. The common days of break fall between   2019-11-01 and 2019-11-10 |
+----------+---------+-------------+------------------+                 |                                                                                                                   |
| A        | 2       |  2019-10-15 |  2019-11-15      |                 |                                                                                                                   |
+----------+---------+-------------+------------------+                 |                                                                                                                   |
| A        | 3       |  2019-11-01 |  2019-11-10      |                 |                                                                                                                   |
+----------+---------+-------------+------------------+-----------------+-------------------------------------------------------------------------------------------------------------------+
| A        | 2       |  2020-01-01 |  2020-01-10      | 10              | 2 Projects were assigned during this time and break in both projects                                              |
+----------+---------+-------------+------------------+                 |                                                                                                                   |
| A        | 3       |  2020-01-01 |  2020-01-10      |                 |                                                                                                                   |
+----------+---------+-------------+------------------+-----------------+-------------------------------------------------------------------------------------------------------------------+
|          |         |             | Total Break Days | 50              |                                                                                                                   |
+----------+---------+-------------+------------------+-----------------+-------------------------------------------------------------------------------------------------------------------+

DB-Fiddle 链接：https ://www.db-fiddle.com/f/c8fMneAUkhb2P3rzjMtVZm/0

Answer 1

使用递归 CTE 获取每位员工的所有工作和所有休息日期。
然后，对于这两种情况下的每个日期，使用聚合将所有项目作为逗号分隔的列表获取GROUP_CONCAT()。
如果这些列表在某个日期匹配，那么这是一个休息日期。

WITH RECURSIVE 
  working_dates AS (
    SELECT `Employee`, `Project Id`, `Project Assignment Date` AS date, `Project Relieving Date`
    FROM `Project Assignment`
    UNION ALL
    SELECT `Employee`, `Project Id`, date + INTERVAL 1 day, `Project Relieving Date`
    FROM working_dates
    WHERE date < `Project Relieving Date`
  ),
  break_dates AS (
    SELECT `Employee`, `Project Id`, `Break Start Date` AS date, `Break End Date`
    FROM `Break`
    UNION ALL
    SELECT `Employee`, `Project Id`, date + INTERVAL 1 day, `Break End Date`
    FROM break_dates
    WHERE date < `Break End Date`
  ),
  working AS (
    SELECT `Employee`, date,
           GROUP_CONCAT(`Project Id` ORDER BY `Project Id`) projects
    FROM working_dates
    GROUP BY `Employee`, date 
  ),
  breaks AS (
    SELECT `Employee`, date,
           GROUP_CONCAT(`Project Id` ORDER BY `Project Id`) projects
    FROM break_dates
    GROUP BY `Employee`, date
  )
SELECT w.`Employee`,
       COUNT(*) assigned_days, 
       COUNT(b.date) AS break_days,
       COUNT(*) - COUNT(b.date) worked_days
FROM working w LEFT JOIN breaks b
ON w.`Employee` = b.`Employee` AND w.date = b.date AND w.projects = b.projects
GROUP BY w.`Employee`

请参阅演示。

Answer 2

将Break Id列添加到Break表后，我可以基于@forpass 建议的聚合技术来推导休息日：

然后，对于这两种情况下的每个日期，使用 GROUP_CONCAT() 将所有项目作为逗号分隔列表获取。

对于每个中断，获取重叠项目的计数和列表（使用 GROUP_CONCAT）。然后再次加入它Break以查找重叠中断的计数和列表以及最小的常见重叠（最新开始和最早结束）。用于ROW_NUMBER消除重复项。

将 Assigned Days 的查询移动到另一个 CTE 中，并与 CTE 一起休息以获得所需的结果。

WITH breaks_summary AS (
    SELECT `Employee`, SUM(break_days) break_days
    FROM (      
        SELECT b.`Employee`, DATEDIFF(b.end_date, b.start_date)+1 break_days, ROW_NUMBER() OVER (PARTITION BY b.break_ids) rn, overlapping_breaks, break_ids, projects_count
        FROM (
            SELECT b_p_cnt.`Employee`, b_p_cnt.`Project Id`, b_p_cnt.projects_count, 
            COUNT(b2.`Break Id`) overlapping_breaks, GROUP_CONCAT(b2.`Break Id`) break_ids, MAX(b2.start_date) start_date, MIN(b2.end_date) end_date
            FROM (
                SELECT b1.`Break Id`, b1.`Employee`, b1.`Project Id`, b1.start_date, b1.end_date, GROUP_CONCAT(pa.`Project Id`) projects, count(pa.`Project Id`) projects_count
                FROM (
                    SELECT `Break Id`, `Employee`, `Project Id`, `Break Start Date` AS start_date, `Break End Date` AS end_date
                    FROM `Break` 
                    ) b1
                LEFT JOIN `Project Assignment` pa ON b1.`Employee` = pa.`Employee`
                    AND ((b1.start_date BETWEEN pa.`Project Assignment Date` AND IFNULL(pa.`Project Relieving Date`,CURDATE()))
                        OR (b1.end_date BETWEEN pa.`Project Assignment Date` AND IFNULL(pa.`Project Relieving Date`,CURDATE())))
                GROUP BY b1.`Break Id`, b1.`Employee`, b1.`Project Id`, b1.start_date, b1.end_date) b_p_cnt
            LEFT JOIN (
                SELECT `Break Id`, `Employee`, `Project Id`, `Break Start Date` AS start_date, `Break End Date` AS end_date
                FROM `Break`
                ORDER BY `Break Id`) b2 ON b_p_cnt.`Employee` = b2.`Employee` 
                    AND ((b_p_cnt.start_date BETWEEN b2.start_date AND b2.end_date)
                        OR (b_p_cnt.end_date BETWEEN b2.start_date AND b2.end_date))
            GROUP BY b_p_cnt.`Break Id`, b_p_cnt.`Employee`, b_p_cnt.`Project Id`, 
                b_p_cnt.start_date, b_p_cnt.end_date, b_p_cnt.projects, b_p_cnt.projects_count
            HAVING count(b2.`Break Id`) = b_p_cnt.projects_count
            ORDER BY b_p_cnt.`Employee`, `Project Id`) b        
            ) breaks
    WHERE rn = 1
    GROUP BY `Employee`),   
assigned AS (
    SELECT merged.`Employee`, SUM(DATEDIFF(merged.EndDate,merged.`Project Assignment Date`)+1) assigned_days
            FROM (SELECT s1.`Employee`, s1.`Project Assignment Date`,
                    MIN(IFNULL(t1.`Project Relieving Date`,CURDATE())) AS EndDate
                FROM `Project Assignment` s1
                INNER JOIN `Project Assignment` t1 ON t1.`Employee` = s1.`Employee`
                    AND s1.`Project Assignment Date` <= IFNULL(t1.`Project Relieving Date`,CURDATE())
                    AND NOT EXISTS( SELECT * FROM `Project Assignment` t2
                        WHERE t2.`Employee` = s1.`Employee`
                            AND IFNULL(t1.`Project Relieving Date`,CURDATE()) >= t2.`Project Assignment Date` 
                            AND IFNULL(t1.`Project Relieving Date`,CURDATE()) < IFNULL(t2.`Project Relieving Date`,CURDATE()))
                WHERE NOT EXISTS( SELECT * FROM `Project Assignment` s2
                    WHERE s2.`Employee` = s1.`Employee`
                        AND s1.`Project Assignment Date` > s2.`Project Assignment Date` 
                        AND s1.`Project Assignment Date` <= IFNULL(s2.`Project Relieving Date`,CURDATE()))
                GROUP BY s1.`Employee`, s1.`Project Assignment Date`
                ORDER BY s1.`Project Assignment Date`) merged
        GROUP BY merged.`Employee`)
SELECT ad.`Employee`,
    ad.assigned_days,
    IFNULL(bs.break_days,0) break_days,
    (ad.assigned_days - IFNULL(bs.break_days,0)) worked_days
FROM assigned ad
LEFT JOIN breaks_summary bs ON ad.`Employee` = bs.`Employee`

使用查询更新了 DB-Fiddle：https ://www.db-fiddle.com/f/c8fMneAUkhb2P3rzjMtVZm/3

感谢所有通过改进问题和提供可能答案做出贡献的人。