SQL - 自定义日期组

Posted 2023-05-09

【中文标题】SQL - 自定义日期组

【英文标题】：SQL - Custom Date Groups

【发布时间】：2021-01-29 10:27:30

【问题描述】：

这个真的把我的头撞到墙上了。

我基本上是在尝试在 SQL 中创建周/月组，因此假设团队成员于 2021 年 5 月 1 日开始他们的合同，那么 05/01/2021 - 12/01/2021 将是第 1 周、第 12 周/01/2021 - 19/01/2021 将是第 2 周，以此类推，直到第 26 周。目的是我可以使用这些周组更轻松地与其他团队成员进行比较。

例如，假设我有这样的简单数据。

Staff_Members	Start Date
team Member 1	25/06/2019
team Member 2	30/06/2019

我想用 SQL 把它变成这个。

Staff_Members	Start Date	Week Group	Week Start	Week End
Team member 1	25/06/2019	Week 1	25/06/2019	02/07/2019
Team member 1	25/06/2019	Week 2	02/07/2019	09/07/2019
Team member 1	25/06/2019	Week 3	09/07/2019	16/07/2019
Team member 1	25/06/2019	Week 4	16/07/2019	23/07/2019
Team member 1	25/06/2019	Week 5	23/07/2019	30/07/2019
Team member 1	25/06/2019	Week 6	30/07/2019	06/08/2019
Team member 1	25/06/2019	Week 7	06/08/2019	13/08/2019
Team member 1	25/06/2019	Week 8	13/08/2019	20/08/2019
Team member 1	25/06/2019	Week 9	20/08/2019	27/08/2019
Team member 1	25/06/2019	Week 10	27/08/2019	03/09/2019
Team member 1	25/06/2019	Week 11	03/09/2019	10/09/2019
Team member 1	25/06/2019	Week 12	10/09/2019	17/09/2019
Team member 1	25/06/2019	Week 13	17/09/2019	24/09/2019
Team member 1	25/06/2019	Week 14	24/09/2019	01/10/2019
Team member 1	25/06/2019	Week 15	01/10/2019	08/10/2019
Team member 1	25/06/2019	Week 16	08/10/2019	15/10/2019
Team member 1	25/06/2019	Week 17	15/10/2019	22/10/2019
Team member 1	25/06/2019	Week 18	22/10/2019	29/10/2019
Team member 1	25/06/2019	Week 19	29/10/2019	05/11/2019
Team member 1	25/06/2019	Week 20	05/11/2019	12/11/2019
Team member 1	25/06/2019	Week 21	12/11/2019	19/11/2019
Team member 1	25/06/2019	Week 22	19/11/2019	26/11/2019
Team member 1	25/06/2019	Week 23	26/11/2019	03/12/2019
Team member 1	25/06/2019	Week 24	03/12/2019	10/12/2019
Team member 1	25/06/2019	Week 25	10/12/2019	17/12/2019
Team member 1	25/06/2019	Week 26	17/12/2019	24/12/2019
Team member 2	30/06/2019	Week 1	30/06/2019	07/07/2019
Team member 2	30/06/2019	Week 2	07/07/2019	14/07/2019
Team member 2	30/06/2019	Week 3	14/07/2019	21/07/2019
Team member 2	30/06/2019	Week 4	21/07/2019	28/07/2019
Team member 2	30/06/2019	Week 5	28/07/2019	04/08/2019
Team member 2	30/06/2019	Week 6	04/08/2019	11/08/2019
Team member 2	30/06/2019	Week 7	11/08/2019	18/08/2019
Team member 2	30/06/2019	Week 8	18/08/2019	25/08/2019
Team member 2	30/06/2019	Week 9	25/08/2019	01/09/2019
Team member 2	30/06/2019	Week 10	01/09/2019	08/09/2019
Team member 2	30/06/2019	Week 11	08/09/2019	15/09/2019
Team member 2	30/06/2019	Week 12	15/09/2019	22/09/2019
Team member 2	30/06/2019	Week 13	22/09/2019	29/09/2019
Team member 2	30/06/2019	Week 14	29/09/2019	06/10/2019
Team member 2	30/06/2019	Week 15	06/10/2019	13/10/2019
Team member 2	30/06/2019	Week 16	13/10/2019	20/10/2019
Team member 2	30/06/2019	Week 17	20/10/2019	27/10/2019
Team member 2	30/06/2019	Week 18	27/10/2019	03/11/2019
Team member 2	30/06/2019	Week 19	03/11/2019	10/11/2019
Team member 2	30/06/2019	Week 20	10/11/2019	17/11/2019
Team member 2	30/06/2019	Week 21	17/11/2019	24/11/2019
Team member 2	30/06/2019	Week 22	24/11/2019	01/12/2019
Team member 2	30/06/2019	Week 23	01/12/2019	08/12/2019
Team member 2	30/06/2019	Week 24	08/12/2019	15/12/2019
Team member 2	30/06/2019	Week 25	15/12/2019	22/12/2019
Team member 2	30/06/2019	Week 26	22/12/2019	29/12/2019

提前感谢您提供的任何建议。 汤姆

【问题讨论】：

欢迎来到 SO。请参阅Why should I provide an MCRE for what seems to me to be a very simple SQL query

也就是说，数据显示的问题一般最好在应用代码中解决

【参考方案1】：

您可以使用递归 CTE。在 mysql 中，这看起来像：

with recursive cte as (
      select staff_member, start_date, 1 as week_number, start_date as week_start, start_date + interval 7 day as end_date
      from t
      union all
      select staff_member, start_date, week_number + 1, end_date, end_date + interval 7 day
      from cte
      where year(start_date) = year(week_start)
     )
 select *
 from cte
 order by staff_member, week_start;

Here 是一个 dbfiddle。

【讨论】：

谢谢你！直到现在才听说递归 CTE，所以这很有帮助！

【参考方案2】：

使用 dateadd(day, 7 , "start-date") 它会将您的开始日期增加 7 天并给出结束日期。