Oracle SQL - 根据条件删除记录子集

Question

我有以下table

ITEM    LOCATION    EFFECTIVE_DATE  SELLING_RETAIL
00139259    11001   01-JUL-20           2
00139259    11001   08-OCT-19           1
00139259    11001   04-OCT-19           1
00139259    11001   01-JAN-18           2
00139259    11001   04-DEC-17           1
00139259    11001   02-MAR-17           500

我只需要按降序保留项目、位置和生效日期三个记录。

所以我留在：

ITEM    LOCATION    EFFECTIVE_DATE  SELLING_RETAIL
00139259    11001   01-JUL-20           2
00139259    11001   08-OCT-19           1
00139259    11001   04-OCT-19           1

其余全部删除。

table中有很多项目和位置，我需要为item/location

的组合保留3条记录

Answer 1

您可以使用 row_number():

select t.*
from (select t.*,
             row_number() over (partition by item, location order by EFFECTIVE_DATE desc) as seqnum
      from t
     ) t
where seqnum <= 3;

这 returns 3 行所有 item/location 对。如果您只考虑一对，您可以使用：

select t.*
from t
where item = ? and location = ?
order by effective_date desc
fetch first 3 rows only;

Answer 2

您想使用 ROW_NUMBER 分析函数查找最近三行之后的行，然后使用 ROWID 伪列与 DELETE 语句关联：

DELETE FROM table_name
WHERE ROWID IN (
  SELECT ROWID
  FROM   (
    SELECT ROW_NUMBER() OVER (
             PARTITION BY item, location ORDER BY effective_date DESC
           ) AS rn
    FROM   table_name
  )
  WHERE  rn > 3
)

其中，对于您的样本数据：

CREATE TABLE TABLE_NAME ( ITEM, LOCATION, EFFECTIVE_DATE, SELLING_RETAIL ) AS
SELECT '00139259', 11001, DATE '2020-07-01', 2 FROM DUAL UNION ALL
SELECT '00139259', 11001, DATE '2019-10-08', 1 FROM DUAL UNION ALL
SELECT '00139259', 11001, DATE '2019-10-04', 1 FROM DUAL UNION ALL
SELECT '00139259', 11001, DATE '2018-01-01', 2 FROM DUAL UNION ALL
SELECT '00139259', 11001, DATE '2017-12-04', 1 FROM DUAL UNION ALL
SELECT '00139259', 11001, DATE '2017-03-02', 500 FROM DUAL;

然后删除后：

SELECT * FROM table_name;

输出：

ITEM	LOCATION	EFFECTIVE_DATE	SELLING_RETAIL
00139259	11001	2020-07-01 00:00:00	2
00139259	11001	2019-10-08 00:00:00	1
00139259	11001	2019-10-04 00:00:00	1

db<>fiddle here