select 同一外键出现次数超过 X 次的行，忽略最近的 X 次（时间戳）

Question

我有一个 table，其中一个属性是外键，另一个属性是时间戳

Table

id	fk	timestamp
1	2	16-02-2022
2	2	01-02-2022
3	2	02-02-2021
4	3	24-05-2020
5	3	11-01-2022
6	3	16-09-2021
7	3	01-01-2022

我想select相同外键出现X次以上的行，我想忽略每个外键的X个最近（时间戳）元素

所以基本上 X 为 2 时 select 会 return

id	fk	timestamp
3	2	02-02-2021
6	3	16-09-2021
4	3	24-05-2020

如果 X 为 3，select 将 return

id	fk	timestamp
4	3	24-05-2020

编辑：

非常感谢@forpas 的好方法

解决方案

SELECT id, fk, timestamp 
FROM (
  SELECT *,
         COUNT(*) OVER (PARTITION BY fk) counter,
         ROW_NUMBER() OVER (PARTITION BY fk ORDER BY timestamp DESC) rn
  FROM tablename
) t
WHERE counter > ? AND rn > ?;

Answer 1

使用COUNT()和ROW_NUMBER()window函数：

SELECT id, fk, timestamp 
FROM (
  SELECT *,
         COUNT(*) OVER (PARTITION BY fk) counter,
         ROW_NUMBER() OVER (PARTITION BY fk ORDER BY timestamp DESC) rn
  FROM tablename
) t
WHERE counter > ? AND rn > ?;

将 ? 替换为您想要的值。

但是，如果对每个 fk 的总行数和要取消的行数应用相同的数字 X，则可以简化查询：

SELECT id, fk, timestamp 
FROM (
  SELECT *, ROW_NUMBER() OVER (PARTITION BY fk ORDER BY timestamp DESC) rn
  FROM tablename
) t
WHERE rn > ?;

参见demo。

Answer 2

SELECT 
id, fk, timestamp
FROM
(
SELECT id, fk, timestamp, ROW_NUMBER() OVER(PARTITION BY fk ORDER BY timestamp ASC) AS X,
COUNT(1) OVER(PARTITION BY fk) AS Total
FROM MyTable
) AS S
WHERE A.Total - S.X >= @X

select 同一外键出现次数超过 X 次的行，忽略最近的 X 次（时间戳）

select the rows that have more that X occurrences of the same foreign key and ignore the X most recent (timestamp)

mysql

sql

window-functions