如何根据创建日期获取上一条和下一条记录，并在单个查询中按另一列分组？

Question

举个例子，我有一个类似于下面的 table，

+----+--------------+------------+-------+
| id | date_col     | label      | tag   |
+----+--------------+------------+-------+
| 1  | 2010-09-07   | Record 1   | 810   |
| 2  | 2010-09-03   | Record 2   | 810   |
| 3  | 2010-08-23   | Record 3   | 811   |
| 4  | 2010-08-23   | Record 4   | 809   |
| 5  | 2010-08-23   | Record 5   | 810   |
| 6  | 2010-08-12   | Record 6   | 809   |
| 7  | 2010-08-06   | Record 7   | 811   |
| 8  | 2010-08-06   | Record 8   | 809   |
| 9  | 2010-08-02   | Record 9   | 810   |
| 10 | 2010-08-01   | Record 10  | 811   |
+----+--------------+------------+-------+

有没有一种方法可以在单个查询中根据 810 等特定标记获取 2010-08-23 等特定日期前后的 date_col 数据。

这样我就可以得到 2010-08-02 和 2010-09-03，即 Record 9 和 Record 2 作为结果。

Answer 1

您可以使用lag()/lead():

select t.*
from (select t.*, 
             lag(date_col) over (partition by tag order by date) as prev_date,
             lead(date_col) over (partition by tag order by date) as next_date
      from t
      where tag = 810
     ) t
where '2010-08-23' in (prev_date, next_date);

您没有同一日期的重复标签示例，因此我认为这是不可能的。如果可能的话，你需要一个稍微复杂一点的解决方案。

要处理重复标签，您可以使用累积 max()/min() 和 window 帧：

select t.*
from (select t.*, 
             max(date_col) over (partition by tag order by date rows between unbounded preceding and interval 1 day preceding) as prev_date,
             min(date_col) over (partition by tag order by date rows between interval 1 following and unbounded following) as next_date
      from t
      where tag = 810
     ) t
where '2010-08-23' in (prev_date, next_date);

如何根据创建日期获取上一条和下一条记录，并在单个查询中按另一列分组？

How to get the previous and next record based on creation date and group by another column in a single query?

mysql

sql

aggregate

max

min