SELECT 如何将每个唯一列值的两条记录作为 MySQL 中的一行？

Question

我有一个像这样的 MySQL table :

+----+-----+-------+------+------+-------+---------------------+
| ID | GID | Name  |  p1  | p10  | p100  |      createdAt      |
+----+-----+-------+------+------+-------+---------------------+
|  1 | 100 | Item1 |  150 | 1499 | 10245 | 2020-07-04 12:00:00 |
|  2 | 857 | Item2 | 1047 | 9875 | 90000 | 2020-07-04 12:00:10 |
|  3 | 100 | Item1 |  149 | 1495 | 10245 | 2020-07-04 12:15:00 |
|  4 | 857 | Item2 | 1099 | 9875 | 89999 | 2020-07-04 12:15:10 |
|  5 | 100 | Item1 |  149 | 1495 | 10247 | 2020-07-04 12:30:00 |
|  6 | 857 | Item2 |  970 | 9879 | 89998 | 2020-07-04 12:30:10 |
+----+-----+-------+------+------+-------+---------------------+

我尝试的是为每个唯一的 GID 值输出最近两个 createdAt

中的 p1, p10, p100

输出示例：

+-----+-------+------+------+-------+---------+----------+-----------+
| GID | Name  |  p1  | p10  | p100  | p1-last | p10-last | p100-last |
+-----+-------+------+------+-------+---------+----------+-----------+
| 100 | Item1 |  149 | 1495 | 10245 |     149 |     1495 |     10247 |
| 857 | Item2 | 1099 | 9875 | 89999 |     970 |     9879 |     89998 |
+-----+-------+------+------+-------+---------+----------+-----------+

我尝试使用子查询来实现我的目标，但我不同意table。

感谢任何能为我提供信息和帮助的人。

Answer 1

您将为此使用 lag()：

select gid, name, p1, p10, p100, prev_p1, prev_p10, prev_p100
from (select t.*,
             lag(p1) over (partition by gid order by createdAt) as prev_p1,
             lag(p10) over (partition by gid order by createdAt) as prev_p10,
             lag(p100) over (partition by gid order by createdAt) as prev_p100,
             row_number() over (partition by gid order by createdAt desc) as seqnum
      from t
     ) t
where seqnum = 1;

Here 是一个 db<>fiddle.

子查询 returns 每列的先前值。外部查询只是过滤到每个 gid/name 组合的最新行。

Answer 2

没有window个函数就不容易优雅了
这是一种方法，它涉及自连接、2 级聚合和条件聚合：

select t.gid, t.name,
  max(case when c.counter = 1 then t.p1 end) p1,
  max(case when c.counter = 1 then t.p10 end) p10,
  max(case when c.counter = 1 then t.p100 end) p100,
  max(case when c.counter = 0 then t.p1 end) p1_last,
  max(case when c.counter = 0 then t.p10 end) p10_last,
  max(case when c.counter = 0 then t.p100 end) p100_last
from tablename t inner join (
  select t1.gid, t1.createdat, count(t2.createdat) counter
  from tablename t1 left join tablename t2
  on t2.gid = t1.gid and t1.createdat < t2.createdat 
  group by t1.gid, t1.createdat
  having count(t2.createdat) <= 1
) c on c.gid = t.gid and c.createdat = t.createdat
group by t.gid, t.name

参见demo。
结果：

| gid | name  | p1   | p10  | p100  | p1_last | p10_last | p100_last |
| --- | ----- | ---- | ---- | ----- | ------- | -------- | --------- |
| 100 | Item1 | 149  | 1495 | 10245 | 149     | 1495     | 10247     |
| 857 | Item2 | 1099 | 9875 | 89999 | 970     | 9879     | 89998     |

SELECT 如何将每个唯一列值的两条记录作为 MySQL 中的一行？

How to SELECT two records for each unique column value as one row in MySQL?

mysql

sql

greatest-n-per-group

mariadb