SQL：计算两个时间间隔之间的工作时间

Posted 2023-04-12

【中文标题】SQL：计算两个时间间隔之间的工作时间

【英文标题】：SQL: Calculate working hours between two time intervals

【发布时间】：2020-12-21 19:06:44

【问题描述】：

我正在尝试计算每个送货员的每月工作时间。这是一家快递公司的数据，每个快递员开始轮班和结束轮班的时间都不一样。

这是表格的样例：

Type_ID	Delivery_Guy_ID	Date_Created
5000	210	2020-11-16 16:34:43
7000	210	2020-11-16 16:35:24
3000	210	2020-11-16 16:35:46
3000	210	2020-11-16 16:37:41
4000	210	2020-11-16 16:39:41
3000	210	2020-11-16 16:42:53
4000	210	2020-11-16 16:47:53
3000	210	2020-11-16 16:48:16
4000	210	2020-11-16 16:50:16
3000	210	2020-11-16 16:53:01
2000	210	2020-11-16 18:53:07

类型 ID = 2000 指的是“Shift Ended”，类型 ID = 7000 指的是“Shift Started”。送货员可能会在一天中多次开始轮班和结束轮班。

这是我尝试过的查询：

WITH working_hours_cte AS 
( 
       SELECT * 
       FROM   ( 
                       SELECT   id AS id 
                                , 
                                type_id AS type_id 
                                , 
                                delivery_guy_id AS delivery_guy_id
                                , 
                                Timezone('Africa/Cairo', date_created) as date_created
                       FROM     delivery_guys_event
                       ORDER BY delivery_guy_id,
                                date(date_created), 
                                date_created ) AS t 
       WHERE  t.date_created >= date('11-01-2020') 
       AND    t.date_created <= date('11-30-2020') ), get_shift_start_time_cte AS 
( 
                SELECT DISTINCT 
                ON( 
                                                t.delivery_guy_id) t.* , 
                                t.date_created::timestamptz AS shift_start_time 
                FROM            working_hours_cte  AS t 
                WHERE           t.type_id = 7000 
                GROUP BY        1, 
                                2, 
                                3, 
                                4 
                                
                ORDER BY        t.delivery_guy_id, 
                                date(date_created), 
                                t.date_created), get_shift_end_time_cte AS 
( 
                SELECT DISTINCT 
                ON( 
                                                t.delivery_guy_id) t.* , 
                                t.date_created::timestamptz AS shift_end_time 
                FROM            working_hours_cte  AS t 
                WHERE           t.type_id = 2000 
                GROUP BY        1, 
                                2, 
                                3, 
                                4
                ORDER BY        t.delivery_guy_id, 
                                date(date_created), 
                                t.date_created), get_working_hours_cte1 AS 
( 
          SELECT    base.* , 
                    (date_part('day', ap.shift_end_time::timestamptz - aa.shift_start_time::timestamptz) * 24 + date_part('hour', ap.shift_end_time::timestamptz - aa.shift_start_time::timestamptz)) * 60 + date_part('minute', ap.shift_end_time::timestamptz - aa.shift_start_time::timestamptz) AS working_hours
                     
          FROM      working_hours_cte AS base 
          LEFT JOIN get_shift_start_time_cte   AS aa 
          ON        base.delivery_guy_id = aa.delivery_guy_id
          LEFT JOIN get_shift_end_time_cte AS ap 
          ON        base.delivery_guy_id = ap.delivery_guy_id) 
SELECT    u.NAME                                          AS delivery_guy_name
          , 
          round(sum(timings.working_hours::numeric),0)/60 AS working_hours 
FROM      get_working_hours_cte1                          AS timings 
LEFT JOIN delivery_guys_contacts                          AS u  
ON        timings.delivery_guy_id = u.id 
WHERE     timings.date_created >= date('11/01/2020') 
AND       timings.date_created <= date('11/30/2020') 
GROUP BY  delivery_guy_name 
ORDER BY  delivery_guy_name ASC 

期望的输出：

Delivery_guy_name	Working_hours
John	200
Michael	150
Joe	230

我也尝试自行加入，但似乎没有人给出正确的小时数。谁能告诉我问题出在哪里？

【问题讨论】：

请同时提供所需的输出

我刚刚编辑了问题

【参考方案1】：

将当前行与下一行之间的时间差相加如何，如果当前行是班次结束时除外？这可以通过窗口函数轻松完成：

select delivery_guy_id, sum(diff) as total_diff
from (
    select delivery_guy_id, 
        lead(date_created) over(partition by delivery_guy_id order by date_created) 
            - date_created as diff
    from mytable
) t
wxhere type_id <> 2000
group by delivery_guy_id

如果在船舶开始/结束之外没有记录，这将起作用 - 即，班次结束后总是有班次开始。

【讨论】：

非常感谢您的提醒！我组织了数据，只包括结束班次和第一次班次，结束班次总是跟着开始班次。我无法按 delivery_guy_id 进行分区，因为我想独立查看每个班次。换句话说，从第 1 行中减去第 2 行，从第 3 行中减去第 4 行。当我使用 ntile() 函数并为每两行分配相同的行号时，它起作用了。然后，我按行号进行分区，并按 delivery_guy_id 进行 sum over() 分区。所以谢谢你让我走上正轨！