如何排除周末和节假日日期并在 MySQL 中查找预期日期？

Question

我正在处理我有 2 张桌子的任务。即tickets 和holidays。

现在我也有完成门票的天数。现在我需要通过排除假期（在holidays 表中指定）和weekends 来查找预期日期。

现在我可以使用ticket created date 和days to complete the ticket 查找日期。但无法通过移除节假日和周末来计算预计到期日。

如果机票预计到期日在节假日或周末，我们需要将预计到期日提前。

然后我们需要比较ticket_closed_date和expected_due_date。

如果ticket_closed_date <= expected_due_date 则需要将isSlaMet 作为YES 返回。否则需要将isSlaMet 作为NO 返回。

票务表

假期表

示例： 通常，如果在 2nd October,2020 创建工单并且完成天数是 3，那么预计到期日将是 5th October，我们有一个假期，@ 987654342@。但是在工单创建日期和预计到期日期之间有 1 个假期和 2 个周末。即3rd, 4th and 5th of October 。所以我们需要将预产期延长3 days（因为2个周末+1个假期）。即8th October。票在9th October关闭。

然后我们需要比较ticket closed date(9th october)和expected due date(8th October)，并将isSlaMet作为YES返回。

预期输入

Tickets Table
--------------------------------------------------------------------------------
tid         createdAt         apply_sla    ticket_closed_date     days_to_complete
--------------------------------------------------------------------------------
100    2020-10-02 00:00:00        1       2020-10-09 00:00:00           3       
--------------------------------------------------------------------------------
       
Holidays Table
----------------------------------------------
id        holiday_date          end_date
----------------------------------------------
20         2020-10-05          2020-10-05
----------------------------------------------

Along with the above holiday, we need to exclude Weekends.

预期输出

Tickets Table
--------------------------------------------------------------------------------------------------------------------------
tid       createdAt     apply_sla    ticket_closed_date   days_to_complete   expected_due_date    completedIn   isSlaMet
--------------------------------------------------------------------------------------------------------------------------
100  2020-10-02 00:00:00    1       2020-10-09 00:00:00             3        2020-10-08 00:00:00      4           NO
  
--------------------------------------------------------------------------------------------------------------------------

这是我目前一直在使用的查询。

    SELECT 
        `t`.`tid`, `t`.`createdAt`, `t`.`days_to_complete`,
        `t`.`ticket_closed_date`,`holidays`.`holiday_date`,
        `holidays`.`end_date`, `t.apply_sla`,
        IF(ISNULL(`t`.`ticket_closed_date`),
            NULL,
            IF((`t`.`apply_sla` = 1),
                IF(((CAST(`t`.`createdAt` AS DATE) + INTERVAL (`t`.`days_to_complete` + 1) DAY) BETWEEN `holidays`.`holiday_date` AND `holidays`.`end_date`),
                    IF((CAST(`t`.`ticket_closed_date` AS DATE) <= (`holidays`.`end_date` + INTERVAL `t`.`days_to_complete` DAY)),
                        'YES',
                        'NO'),
                    IF((CAST(`t`.`ticket_closed_date` AS DATE) <= (`t`.`createdAt` + INTERVAL (`t`.`days_to_complete` + 1) DAY)),
                        'YES',
                        'NO')),
                IF(((TO_DAYS(`t`.`ticket_closed_date`) - TO_DAYS(`t`.`createdAt`)) > (`t`.`days_to_complete` + 1)),
                    'NO',
                    'YES')
            )
        ) AS `isSlaMet`
    FROM
        (`tickets` `t`
        LEFT JOIN `holidays` ON (((CAST(`t`.`createdAt` AS DATE) + INTERVAL (`t`.`days_to_complete` + 1) DAY) BETWEEN `holidays`.`holiday_date` AND `holidays`.`end_date`)))
    ORDER BY `t`.`tid` DESC;

Answer 1

这看起来更复杂。最简单的方法可能是蛮力：使用递归 CTE 枚举工单创建日期和关闭日期之间的所有天数（仅在 MySQL 8.0 中可用），然后过滤掉周末和节假日以计算 SLA 天数：

with recursive cte_tickets as (
    select tid, created_at as dt, ticket_closed_date
    from tickets 
    where apply_sla = 1
    union all
    select tid, dt + interval 1 day, ticket_closed_date
    from cte_tickets
    where dt < ticket_closed_date
)
select t.*, 
    t.created_at 
        + interval (t.days_to_complete + sum(weekday(dt) in (5, 6) or h.holiday_date is not null)) day
        as expected_due_date,    
    count(*) - sum(weekday(dt) in (5, 6) or h.holiday_date is not null) - 1 completed_in,
    t.ticket_closed_date <= t.created_at 
        + interval (t.days_to_complete + sum(weekday(dt) in (5, 6) or h.holiday_date is not null)) day
        as is_sla_met
from tickets t
inner join cte_tickets ct on ct.tid = t.tid
left join holidays h on ct.dt between h.holiday_date and h.end_date
group by t.tid

Demo on DB Fiddle：

时间 | created_At | apply_sla |门票关闭日期 | days_to_complete |预计到期日 |完成_in | is_sla_met --: | :----------------- | :-------- | :----------------- | ---------------: | :----------------- | ------------: | ---------: 100 | 2020-10-02 00:00:00 | 1 | 2020-10-09 00:00:00 | 3 | 2020-10-08 00:00:00 | 4 | 0